Download BrainTrax Algebra: Advanced

version 1.02
Advanced
Malcolm E. Hays
U N I V E R S I T Y
O F
M I S S O U R I – R O L L A
Copyright © 1999 – 2003 Malcolm E. Hays and the University of Missouri – Rolla.
All rights reserved.
No part of this work may be reproduced or transmitted in any form or by any means, electronic
or mechanical, including photocopying, recording, scanning, or by any information storage or
retrieval system without the prior written consent of the University of Missouri – Rolla unless such
copying is permitted by federal copyright law.
Preface
The University of Missouri – Rolla's (UMR's) BrainTrax Algebra system is an online learning tool aimed at
helping struggling students comprehend algebra. You can find it at:
http://braintrax.umr.edu
Click on "Algebra" at the top of the page to enter the online BrainTrax Algebra Brain. The Algebra Brain
is a powerful tool designed to associate different concepts with each other in a visual web of knowledge.
We have enhanced the visual web with substantial content. It is not a textbook. It is a way of learning
mathematics.
What you hold in your hands is the entire body of knowledge contained in the Algebra Brain. Over 1000
pages of pure algebraic goodness, distilled into printer-friendly format. You might be wondering, "If it is
available online, why print it out?"
First and foremost, in an effort to be more accessible to teachers and administrators, we have created this
document so that everyone has a chance to see what the Algebra Brain has to offer. Oftentimes a computer
terminal is not handy, or the right software has not been installed, or for some reason the hardware just does
not work. In that case, we have BrainTrax Algebra in hand to provide instructors and administrators a
glimpse of the many features offered in the Algebra Brain.
As mentioned earlier, the online Algebra Brain contains over 1000 pages of algebra. This is very difficult
to comprehend in a website since you only see a small portion of the site at any given time. This
monstrously thick volume hammers home the idea of 1000 pages of algebra, each page better than the last.
Unfortunately, BrainTrax Algebra lacks one important component: it is not interactive like a website. You
cannot point and click your way through this book.
Instructors can use BrainTrax Algebra as a reference to find information that might be of interest to their
students (or themselves—there are a lot of interesting facts in BrainTrax Algebra not found in any algebra
textbook). Having reviewed a given concept, the instructor can then point students to it in the Algebra
Brain. Each concept in BrainTrax Algebra has a "thought-path" associated with it to guide users to that
concept in the Algebra Brain (concepts in the Algebra Brain are called "thoughts").
We would like to stress that fact that BrainTrax Algebra is not a textbook! It does not contain any problem
sets or exercises, it does not contain information on how to use a graphing calculator (found in many
current textbooks), and it does not operate in a linear fashion like a typical algebra textbook. BrainTrax
Algebra does have lengthy discussions on most topics covered in algebra. It has hundreds, if not
thousands, of examples each with a detailed explanation of the solution. It has three levels of explanation
for most concepts, suitable for the beginning algebra student through the more advanced students. In
addition, it tries to connect algebra with the real world by illustrating how history has played an important
role in the development of mathematics.
Although we have tried to duplicate the many features of the Algebra Brain in this document, some features
are only available online. These include the audio commentary for the Basic-level instruction, the evershifting matrix of the Brain itself, and the instantaneous transfer between levels of instruction.
The BrainTrax website by itself has many features available to students and instructors alike. For students,
we have sample exams, which are actual tests administered at UMR over the years. Each test also has a
detailed solution provided. Students can work the tests, and then see how they did. Instructors can access a
large number of documents, including all of the information here in BrainTrax Algebra, all in printerfriendly format.
We feel that we have created a highly usable resource for instructors and students. Each concept is covered
in-depth in language the student can understand, while training the student to think rigorously about
mathematics. It is never enough to simply fill-in-the-blanks of a formula; we encourage the student to
understand why he or she obtained the given answer, a critical component in problem solving.
The BrainTrax Development Team has worked long and hard under the auspices of the University of
Missouri – Rolla to provide superior mathematics instruction for the students of UMR. However, we feel it
is in the best interests of UMR in particular and the rest of the state of Missouri in general to provide
Preface – iii
superior mathematical instruction to all students preparing to go on to college. If you would like more
information about BrainTrax Algebra, the BrainTrax Development Team, or the Algebra Brain, please do
not hesitate to contact us:
Attn: BrainTrax Development
UMR Computer and Information Services
104 Computer Science Building
1870 Miner Circle
Rolla, MO 65409–1110
Phone: (573) 341–4841
Fax:
(573) 341–4216
Email: [email protected]
iv – Preface
BrainTrax Algebra – Advanced
Contents
Features
ix
Multivariable Linear Systems
87
Unit 1 – Fundamentals of Algebra
1
Systems of Inequalities
93
Introduction to Fundamentals of Algebra
3
Unit 3 – Functions
97
Algebraic Expressions
5
Introduction to Functions
99
Basic Rules of Algebra
7
Function Terminology
103
Properties of Equality
9
Introduction to Polynomial Functions
105
Properties of Exponents
11
The Fundamental Theorem of Algebra
107
Properties of Fractions
15
Polynomial Operations
109
Properties of Inequalities
17
Adding / Subtracting Polynomials
111
Properties of Zero
19
Multiplying Polynomials
113
Cartesian Plane
21
Special Product Patterns
115
Radicals
27
Special Factoring Patterns
117
Properties of Radicals
29
Polynomial Division
121
Real Numbers
31
Synthetic Division
123
Absolute Value
33
Factor Theorem
125
Ordering Real Numbers
35
Remainder Theorem
127
Unit 2 – Linear Equations
39
Real Zeros
129
Introduction to Linear Equations
41
Descartes's Rule of Signs
133
Lines and Slope
45
Rational Zero Test
137
Parallel and Perpendicular Lines
51
Complex Numbers
143
Slope-Intercept Form of a Line
53
Rational Functions
149
Point-Slope Form of a Line
55
Asymptotes
151
Equations of Lines
59
Sketching Rational Functions
155
Linear Inequalities
61
Partial Fractions
161
Absolute Value Inequalities
63
Distinct Linear Factors
165
Polynomial Inequalities
65
Distinct Quadratic Factors
169
Rational Inequalities
69
Repeated Linear Factors
173
Linear Systems
71
Repeated Quadratic Factors
177
Substitution Method
73
Mixed Factors
181
Graphical Approach
77
Transcendental Functions
185
Graphical Interpretation of Solutions
79
Exponential Functions
187
Method of Elimination
83
Logarithmic Functions
191
Contents – v
Properties of Logarithms
195
Unit 5 – Matrices and Determinants
297
Natural Base e
199
Matrices
299
Exponential / Logarithmic Equations
203
Matrix Operations
301
Exponential and Logarithmic Models
207
Identity Matrix
303
Exponential Growth and Decay
209
Matrix Addition
305
Gaussian Model
213
Scalar Multiplication
307
Logarithmic Model
215
Matrix Multiplication
309
Logistics Growth
219
Properties of Matrix Operations
313
Translations and Combinations
223
Linear Systems in a Matrix
317
Arithmetic Combinations
225
Elementary Row Operations
319
Composition of Functions
227
Gaussian Elimination
323
Reflections in Coordinate Axes
229
Gauss-Jordan Elimination
327
Vertical and Horizontal Shifts
231
Inverse of a Square Matrix
329
Inverse Functions
235
Finding Matrix Inverse
333
Finding the Inverse
239
Inverse of 2 x 2 Matrix
337
Horizontal Line Test
243
Determinants
341
Unit 4 – Quadratics
245
Properties of Determinants
347
Quadratics
247
Triangular Matrix
351
Quadratic Equations
249
Applications of Determinants
353
Factoring
253
Area of a Triangle
355
Extracting Square Roots
255
Cramer's Rule
359
Completing the Square
257
Test for Collinear Points
365
Quadratic Formula
259
Unit 6 – Sequences and Probability
369
Conic Sections
261
Sequences and Probability
371
Circles
263
Factorials
373
Parabolas
267
Counting Principle
375
Ellipses
271
Combinations
377
Hyperbolas
275
Permutations
379
Translations of Conics
281
Binomial Theorem
383
Circle Translation
283
Pascal's Triangle
387
Parabola Translation
285
Probability
389
Ellipse Translation
289
Independent Events
393
Hyperbola Translation
293
Union of Two Events
395
Complementary Events
397
vi – Contents
Sequences
399
Arithmetic Sequence
403
Geometric Sequence
407
Summation Notation
413
Appendix
417
Appendix A – Algebra Alphabet
419
Glossary
421
Works Consulted
431
Index
Contents – vii
viii – Contents
General Features:
Three levels of instruction – In an effort to reach as wide an audience as possible, we include three levels
of instruction: Basic, Intermediate and Advanced. Sometimes a student needs a little extra explanation,
therefore we provide the student with the following:.
Basic: Suitable for students of Algebra I and up, Basic level instruction is just that, a
rudimentary explanation of the concept at its most fundamental level. We still assume that the
user understands basic arithmetic, however, along with some fundamental algebraic concepts such
as variables and expressions. Basic level instruction also serves as remediation for higher-level
students who require a little more explanation on a given concept. Some concepts are not even
covered at the Basic level in a normal algebra course, but whenever possible, we still try to include
a brief, non-technical description of the concept as well as showing the student why the given
concept is important in later mathematics courses.
Intermediate: This level is most suitable for students of Algebra II and up. Advanced users
who are struggling to understand the Advanced level of a concept may remediate to this level for a
simpler, more detailed explanation of the given concept. Basic users who wish to increase their
knowledge of a given concept may decide to learn at this level too, thus becoming even more
proficient in algebra. Almost all concepts are covered at the Intermediate level; often with much
more detail than at the Basic or even Advanced levels.
Advanced: The highest level of instruction in the Algebra Brain. The Advanced users are
typically freshman at UMR who require some additional instruction outside the classroom in order
to fully understand a given concept. Other users include students in high school who are taking a
collegiate-level algebra course. Users who are struggling with mathematical comprehension can
remediate themselves down to the Intermediate level, if they so choose. Every concept in the
Algebra Brain is covered at the Advanced level. Some concepts—in particular formula sheets—
are covered exclusively at the Advanced level simply because no additional explanation is
required for Basic and Intermediate users.
Unit Organization – The information is structured into six units, roughly analogous to the algebra
syllabus used on the UMR campus.
Unit 1: Fundamentals of Algebra – This unit covers pre-algebra concepts such as real numbers,
absolute value, the Cartesian plane, algebraic expressions, and properties of arithmetic.
Unit 2: Linear Equations – Linear equations is the study of lines and slope. We introduce those
concepts, along with linear inequalities and linear systems.
Unit 3: Functions – Central to the study of algebra is the concept of a function. We cover
numerous types of functions, including rational functions, polynomial functions, transcendental
functions, and inverse functions.
Unit 4: Quadratics – Although quadratics should technically fall under the category of
functions, we assign the study of quadratic its own unit. In this unit we describe the various ways
of solving a general quadratic equation, concluding with the Quadratic Formula. We also
introduce the conic sections, a subset of quadratic equations. The conic sections include the circle,
ellipse, parabola, and hyperbola.
Unit 5: Matrices and Determinants – Matrices are often used in algebra to represent systems of
linear equations. We demonstrate how to use a matrix to solve a system of linear equations. A
determinant is a real number associated with a square matrix (and only a square matrix!). Using
determinants, we can find the area of a triangle, solve a system of linear equations (via Cramer's
Rule), and determine if three points are all on the same line.
Unit 6: Sequences and Probability – Sequences are special functions in algebra. We cover
arithmetic and geometric sequences, sums of sequences, and summation notation. Probability is
the determination of the likelihood of an event given some constraints. We discuss the Binomial
Theorem, permutations, combination, factorials, complementary events and independent events.
Each unit is further subdivided into numerous concepts arranged in a sequential manner, although they do
not appear sequentially in the Algebra Brain.
Features – ix
Concept Features:
Each concept in BrainTrax Algebra includes the following features.
Color coding – We have color coded BrainTrax Algebra to match the color codes we use in the online
Algebra Brain. Each concept has a title and a level assigned to it at the top of the page. For instance,
"Descartes's Rule of Signs" is green on the Intermediate level pages, and "Intermediate" (also in green)
appears next to the title. The three colors used in the brain are:
Basic – purple
Intermediate – green
Advanced – blue
Intermediate
Goal / Terms / MathTrax – On each page containing detailed information about a given concept, we
have reproduced the information found in the online Algebra Brain in a colored box at the top of the
concept.
Goal – the learning objective for the concept
Terms – a list of words and phrases, along with the definitions, the student is expected to learn in
the given concept
MathTrax – to add some historical perspective to the study of algebra, we include an interesting
factoid about the development of mathematics over the course of several millennia
To find the number of real (and imaginary) roots of a polynomial function.
Descartes's Rule of Signs – a rule determining upper bounds to the number of positive
zeros and to the number of negative zeros of a polynomial function.
positive real zeros – positive real x-values for which the function equals zero.
negative real zeros – negative real x-values for which the function equals zero.
variation in sign – two consecutive coefficients in a polynomial function having opposite
signs.
In 1649, René Descartes (1590-1650) traveled to Sweden to tutor 20-year old Queen
Christina. She preferred to start her days at 5 am. The combination of extremely cold
winters and early rising resulted in Descartes contracting a fatal case of pneumonia. He
died early spring of 1650. In 1666, his remains were exhumed and returned to France.
The French ambassador received permission to remove Descartes's right forefinger
from the corpse. Descartes's skull is said to have been removed by a guard and sold
several times. Supposedly, Descartes's skull is currently on display at the Museé de
l'Homme in the Palais de Chaillot in France.
Thought Path – After the Goal / Terms/ MathTrax box, we include a thought path. This is the route the
student would travel to find the page in the Algebra Brain. Following this path in the Algebra Brain will
reinforce the relationships between algebraic concepts. In addition, the thoughts in the Algebra Brain
contain more features than are found here, simply because the online experience offers much more in terms
x – Features
of interactivity. Basic users, for instance, can have the text read to them, along with descriptions of the
graphics to help students understand what they see.
Functions → Polynomial Functions → Real Zeros → Descartes's Rule of Signs
To get to "Descartes's Rule of Signs" in the Algebra Brain, the user clicks on "Functions" followed by
"Polynomial Functions" followed by "Real Zeros" and finally on "Descartes's Rule of Signs".
Conceptual information – Following the thought path, the user enters the realm of conceptual
information. Here we explain the concepts in detail, often with fully-worked proofs. We also provide the
relevant formulas the student needs to understand the material. Much of the conceptual information is
given in plain (often very plain) English with little mathematical terminology. Often the most obscure
mathematical formulas are better understood if redefined in terms the user can understand. We also include
historical information to help the student realize that mathematics is continually being refined over the
centuries.
However, we can use Descartes's Rule of Signs to find not only which roots are real and
which roots are imaginary, but also the signs of those roots. Many problems in the real
world only concern themselves with positive solutions. A good example is any projectile
example involving time. Time is always a positive quantity (time moves forward, not
backward). Thus, if we have a polynomial equation involving time that gives us negative
roots, we can automatically disregard them because they do not make sense in the context
of a time-based problem.
We must note the qualifier that the roots given by Descartes's Rule of Signs are all real.
According to the Fundamental Theorem of Algebra, the degree of the polynomial tells us
the total number of roots. Therefore, if Descartes's Rule of Signs yields us three real roots
(positive, negative, or both) for a 5th degree polynomial, then we must conclude that the
remaining 2 roots are imaginary.
Formulas / Definitions – Many ideas in algebra can be summed up in a simple formula or phrase. For
instance, the Pythagorean Theorem is written as, "Given a right triangle with sides of lengths a, b, and c,
where the side of length c is opposite the right angle, the square of c is the sum of the squares of a and b, or
in other words, a2 + b2 = c2." Students are expected to not only memorize the formula, but also understand
why it works, although sometimes that is well beyond the scope of algebra.
Descartes's Rule of Signs
Let f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 be a polynomial with real
coefficients and a0 ≠ 0.
1.
The number of positive real zeros of f is either equal to the number of
variations in signs of f(x) or less than that number by an even integer.
This means that for every change of sign (from +an to –an – 1 to + an – 2, etc),
you will have a zero for each change in sign or two. If you do not have a
zero for each change in sign, then you will have two less than the maximum
number of zeros. If you do not have that many, then you will have four less
than the maximum, etc… These are all for positive real zeros.
2.
The number of negative real zeros of f is either equal to number of
variations of sign of f(–x) or less than that number by an even integer.
The same rule that applies to the positive real zeros applies to the negative
zeros as well.
Features – xi
Notes – Many formulas and definitions are followed by notes. These are additional pieces of information
the student needs to keep in mind when applying the formula or definition.
Notes: Variation in sign means that two consecutive coefficients have opposite signs.
Where there is only one variation in sign, Descartes’s Rule of Signs guarantees
the existence of exactly one positive (or negative) real root.
Descartes's Rule of Signs does not tell us what the zeros are. It simply states
how many zeros we will find, as well as whether they will be positive or
negative. We need to use other methods to actually find the zeros.
Examples – Most pages contain at least one example problem after a formula or definition is given. This
is so the student can see for him or herself how the formula or definition is applied. Each and every
example problem is accompanied by a step-by-step explanation of how the problem is solved. Many
example problems are real-world situations, sometimes with disastrous consequences if the problem is
solved incorrectly! Besides simply showing the student how to work a problem, we also strive to teach the
student how to answer the question that is asked. Oftentimes, the mathematics is only a small part of the
problem. If we obtain two answers, one positive and one negative, which one do we use?
Use Descartes's Rule of Signs to determine how many real solutions the following
function has as well as if they are positive or negative.
x3 – 2x2 – 5
=
0
given equation
solution
Our given equation has coefficients that are positive, negative, and negative again. This
translates to only 1 variation in sign.
Descartes's Rule of Signs tells us that the number of positive real solutions is equal to the
number of changes in sign or less than that by an even number. Thus, we can have either
1 positive real root or 0 positive real roots (we can't have less than 0 roots. That just
wouldn't make sense).
In order to determine our negative real roots, we have to replace "x" with "–x" in the
function and simplify, noting any variations in sign again:
(–x)3 – 2(–x)2 – 5
=
–x3 – 2x – 5
replace "x" with "–x"
All the coefficients are negative. Thus, we conclude from Descartes's Rule of Signs that
there are no negative real roots, since there are no variations in sign.
Thus we have at most, 1 positive real root.
answer
Graphs / Pictures – Each graph and diagram in the Algebra Brain has been hand-crafted for maximum
effect. Pictures really can be worth a thousand words if designed efficiently. We have strived to provide
clear, concise images that relay as much information about a given situation as possible. Most diagrams
include color to enhance the effect. For instance,
indicates how we determine the number of changes of sign in a polynomial expression so that we can then
apply Descartes's Rule of Signs.
xii – Features
Connections – Finally, we try to enhance the learning experience by providing connections between
concepts. Many concepts have little or no meaning to the student now, but may become more and more
important if the student wishes to learn higher mathematics. For instance, exponential functions may not
seem to have any importance to a student learning algebra in 10th grade, but that same student will be very
glad he or she learned about exponential functions in 10th grade when he or she studies calculus. Since the
Algebra Brain was originally designed for algebra students on the UMR campus, we feel it is only fitting
that we inform students that the concepts they learn today will be of great benefit to them when they study
calculus later in their academic careers.
Property 4 is somewhat misleading. When you study calculus, you will find that it is
appropriate to say that (a / 0) "approaches" infinity. To see why, consider dividing a by a
really small decimal number that is close to zero, such as 0.00001. Then you have:
a
0.00001
a
=
−5
rewrite denominator in exponential form
10
=
a(105)
property of exponents
=
a(100,000)
rewrite in standard notation
Dividing a by a really small decimal number is the same as multiplying a by a really
large number. Thus, as the denominator gets closer and closer to zero, the evaluation of
the expression becomes larger and larger, eventually "reaching" infinity when the
denominator equals zero.
Features found only in the Algebra Brain:
All of the features described above can be found in the online Algebra Brain—located at
braintrax.umr.edu—but the following features are found only in the Algebra Brain, since they are
unique to the web-based learning environment and cannot be represented in a paper-based model.
The Brain – The online Algebra Brain contains a navigation feature known simply as "The
Brain". This navigation tool allows students to quickly find the algebra concept they seek.
Students need no more than a dozen clicks to go from any point in the Algebra Brain to any other
point in the Algebra Brain.
Interactive Example and Testing System (IETS) – Authorized students can access our
database of interactive example problems for each concept. Instructors who contact us can request
userids and passwords for their students. Students are able to test their knowledge of the various
concepts and applications, obtaining instantaneous feedback, as well as detailed explanations
when they fail to solve a problem correctly.
Transfer between levels – Each concept in the Algebra Brain contains a menu with three
buttons indicating "Basic", "Intermediate", or "Advanced" levels of knowledge. Students can
click on one of the three buttons to be instantly transferred to the level of instruction they selected.
Furthermore, from the BrainTrax homepage, they can choose a default level of instruction.
Audio – Basic level pages in the Algebra Brain contain an audio icon that—when clicked—reads
the information on the page to the student. Graphs and diagrams are explained to the student so
they understand what the graph or diagram represents.
"i" and "e" buttons – The "i" button on a page indicates that if a student clicks on the button,
he or she will be transferred to another page on the same concept with more information about the
concept. The "e" button is placed next to an example on a page and will transfer the student to
another example of the same concept, though at a different instruction level.
Mouseover definitions – Bolded terms on a web page in the Algebra Brain can be moused over
to reveal the definition of the word or phrase. This provides the student with instantaneous
Features – xiii
information about the term he or she is unfamiliar with instead of having to refer to a glossary. In
addition, all terms are located at the top of a given web page for easy access.
Access 24/7 – The Algebra Brain is available to all Missouri K–12 students 24-hours a day, 7
days a week. To access the online Algebra Brain, simply enter "braintrax.umr.edu" (NO "www")
into the address window of Internet Explorer 5.0+. Due to complex technical issues, the Algebra
Brain can only be viewed on a PC-compatible machine running Internet Explorer 5.0 or higher.
Once on our homepage, click on "Algebra" at the top of the page to enter the Algebra Brain.
xiv – Features
Unit 1
Fundamentals of Algebra
Fundamentals of Algebra – 1
2 – Fundamentals of Algebra
Before we can really get into algebra, we first need to guarantee that we actually understand the very
foundations of algebra.
Algebra is really only the manipulation of numbers, variable, or expressions using the basic operations of
addition, subtraction, multiplication, and division. Of course, this is a very oversimplified explanation. We will
see that this can become quite complicated at times.
Algebraic Expressions
An algebraic expression is a number, variable or combination of numbers and variables joined together by
addition, subtraction, multiplication or division symbols. Do not mistake an expression for an equality. They
are quite different. An equality is composed of two or more expressions separated by an equals sign "=". An
expression does not contain an equals sign.
Basic Rules of Algebra
We also need to understand some of the most common properties that are shared by real numbers and variables.
These include the Commutative Properties of Addition and Multiplication as well as the infamous Distributive
Property. It is necessary to memorize these properties as they will make life much easier during later portions
of the course.
Radicals
Radicals, or roots of numbers, are very common in algebra. They crop up in a wide variety of applications and
sometimes the solution to a particular problem can only be expressed by a radical. The followers of the Greek
mathematician Pythagoras originally discovered radicals, but they did not fully understand their implication, so
they ignored them.
Real Numbers
Finally, we have real numbers. A real number is any number that can be expressed in decimal form. This
includes all rational and irrational numbers. For the most part, the only numbers we will be dealing with in
algebra are real numbers. It is still necessary to state when we are dealing with real numbers to avoid any
ambiguity. There are other numbers called complex, or imaginary, numbers that we will deal with at a certain
point in the Brain.
Fundamentals of Algebra – 3
4 – Fundamentals of Algebra
Advanced
To learn the basic terms associated with algebraic expressions.
variable – a symbol which can represent any one of a set of numbers or other objects (including other algebraic
expressions).
algebraic expression – a combining of numbers, variables and operations in a way that stands for a number.
constant – a numerical symbol that never changes its values.
term – a constant number of a variable or the product of numbers and variables.
constant term – term in an expression that does not contain any variables.
variable term – term in an expression that contains variables.
coefficient – the numerical factor of a monomial expression.
evaluate – to find the value of an expression for a given value.
The Arabic word al-jabr, from which we get the word "algebra", is translated into English as "restoration" or
"completion". It refers to the operation of moving known or unknown quantities of the same power to one side or
the other of an equation so that neither side has a negative quantity. For example, x = 7 – 2x would be
"restored" or "completed" to 3x = 7.
Fundamentals of Algebra → Algebraic Expressions
One of the most basic (and famous) characteristics of algebra is the use of letters to represent numbers. A letter
is a variable because it can represent a variety of numbers. A combination of letters and numbers is called an
algebraic expression. Here are a few examples:
3x + 2
4x2
2−x
7
5x
Definition of an Algebraic Expression
A collection of letters (variables) and real numbers (constants) combined using the operations of
addition, subtraction, multiplication, division and exponentiation is called an algebraic expression.
Note:
An algebraic expression does not contain an equals sign (=). In algebra we use the equals sign (=) to
equate two expressions together.
An algebraic term is a part of an algebraic expression that is separated from the other parts by addition or
subtraction. For example, in the expression:
3x2 + 4x – 1
we have three terms: 3x2 and 4x are variable terms and 1 is a constant term. The numerical factor in front of
the variable is the coefficient of the variable term. The coefficient of 3x2 is 3. If there is no number in front of
the variable, then it has an implied coefficient of 1. Thus, the coefficient of x3 is simply 1.
Fundamentals of Algebra
Algebraic Expressions – 5
Oftentimes, we are asked to evaluate a particular algebraic expression. This means substituting in numerical
values into each of the variables and simplifying the resulting arithmetic expression. For example,
2x – 4
Value of
Variable
x=3
3x2 – x + 5
x = –1
Expression
6 – Algebraic Expressions
Substitute
Value of Expression
2(3) – 4
6–4=2
3(–1)2 – (–1) + 5
3+1+5=9
Fundamentals of Algebra
Advanced
To review the most elementary properties obeyed by real numbers, variables, and expressions.
additive inverse – the opposite of a number. If the number a is negative, then the additive inverse of a is a
positive number.
multiplicative inverse – the reciprocal of a number.
numerator – signifies the number of parts of the denominator that is taken. In a fraction, the numerator lies
above the line representing division.
denominator – lies below the line of division in a fraction. This is the divisor. If the numerator equals 1, then the
denominator determines in how many parts the unit is to be divided.
The French mathematician Nicolas Chuquet (ca. 1500) wrote Triparty en la science des nombres, in which he
uses a form of exponent notation. Our expressions 3x2 and 4x3 were written as .3.2 and .4.3, respectively. Zero
and negative exponents were also represented using a similar notation, so x0 would be written as .1.0 and 5x-2 as
.5.2.m. Chuquet even used this notation to perform calculations such as .72.1 divided by .8.3 is .9.2.m. In modern
terms, this is equivalent to 72x ÷ 8x3 = 9x-2.
Fundamentals of Algebra → Basic Rules of Algebra
The four fundamental operations of algebra are:
•
Addition
+
•
Subtraction
–
•
Multiplication
x
()
·
•
Division
÷
/
Of these four operations, the primary ones are addition and multiplication. Subtraction and division are inverses
of addition and multiplication, respectively. We use subtraction to "undo" addition and division to "undo"
multiplication.
Consider that we can write subtraction as:
Subtraction
a – b = a + (–b)
and division as:
Division
If b ≠ 0, then
a÷b
=
a ⎛⎜
1⎞
⎟
⎝b⎠
=
a
b
Using these definitions, –b is the additive inverse (or opposite) of b. Similarly, 1 / b is the multiplicative
inverse (or reciprocal) of b. When we write division in its fractional form, a is the numerator and b is the
denominator. Note that we cannot let b equal 0 because then we would have division by zero, which is
undefined.
Fundamentals of Algebra
Basic Rules of Algebra – 7
The following list of rules of algebra is true not only for real numbers, but for algebraic expressions and
variables.
Basic Rules of Algebra
Let a, b, and c be real numbers, variables, or algebraic expressions.
Property
Commutative Property of Addition
a+b=b+a
Example
3x + 2x2 = 2x2 + 3x
Commutative Property of Multiplication
ab = ba
(1 – x)(2x) = (2x)(1 – x)
Associative Property of Addition
(a + b) + c = a + (b + c)
(x + 5) + 2 = x + (5 + 2)
Associative Property of Multiplication
(ab)c = a(bc)
(2x • 4x2)(7) = (2x)(4x2 · 7)
Distributive Properties:
a(b + c) = ab + ac
3x(5 + x) = 3x · 5 + 3x · x
(a + b)c = ac + bc
(y + 2)y = y · y + 2 · y
Additive Identity Property
a+0=a
2y2 + 0 = 2y2
Multiplicative Identity Property
a·1=a
3x · 1 = 3x
Additive Inverse Property
a + (–a) = 0
7x3 + (–7x3) = 0
Multiplicative Inverse Property
a≠0
In addition to the above properties of operators, we also have the following properties of negation:
Properties of Negation
Let a and b be real numbers, variables, or algebraic expressions:
Note:
Property
1.) (–1)a = –a
Example
(–1)6 = –6
2.)
–(–a) = a
–(–5) = 5
3.)
(–a)b = –(ab) = a(–b)
(–4)2 = –(4 · 2) = 4(–2)
4.)
(–a)(–b) = ab
(–2)(–x) = 2x
5.)
–(a + b) = (–a) + (–b)
–(3 + x) = (–3) + (–x) = –3 – x
Be sure that you understand the difference between the opposite of a number and a negative number.
If a is already negative, then its opposite, –a, is positive. For example, if a = –3, then –a = –(–3) = 3.
8 – Basic Rules of Algebra
Fundamentals of Algebra
Advanced
To review the basic properties of equality used in algebra.
No definitions on this page.
Robert Recorde (1510-1558) was the first to introduce the equals sign, "=", because nothing could be more equal
than two parallel straight lines. The symbol is found in his book The Whetstone of Witte.
Fundamentals of Algebra → Basic Rules of Algebra → Properties of Equality
The following properties are used extensively throughout algebra to help us solve problems. In general, for
both sides of an equation to be equal, the sum total of both sides must be equivalent. Furthermore, whenever
we do something to one side of an equation (add, subtract, multiply, or divide by a quantity), then we have to
perform the exact same procedure on the other side of the equation in order to maintain balance.
Properties of Equality
Let a, b, and c be real numbers, variables, or algebraic expressions.
1.)
If a = b, then a + c = b + c
add c to both sides
2.)
If a = b, then ac = bc
multiply both sides by c
3.)
If a + c = b + c, then a = b
subtract c from both sides
4.)
If ac = bc and c ≠ 0, then a = b
divide both sides by c
Fundamentals of Algebra
Properties of Equality – 9
10 – Properties of Equality
Fundamentals of Algebra
Advanced
To review properties of exponents and their use in scientific notation.
exponent – in a power, the number of times the base occurs as a factor. In the expression 23, 3 is the exponent
and tells us that 2 occurs as a factor three times.
base – the repeated factor in a power. In a2, a is the base.
power – a product of equal factors. The repeated factor is the base. A positive exponent tells the number of
times the base occurs as a factor.
Nicolas Chuquet (c.1455-c.1500) anticipated Napier's logarithms by almost 100 years. In his analysis of the
powers of 2, he makes it very clear that if x = 32 = 25, then, in hindsight, 5 = log2 x. Chuquet unfortunately failed
to realize the connection so we had to wait a century until Napier could complete Chuquet's thought.
Fundamentals of Algebra → Basic Rules of Algebra → Properties of Exponents
An exponent is nothing more than a shorthand form of multiplication.
repeated multiplication
x·x·x·x
exponential form
x4
(2)(2)(2)(2)(2)
(2)5
(3x)(3x)(3x)
(3x)3
In more general terms, if a is a real number and n is a positive integer than
an
=
a·a·a·a·...·a
n factors
where n is the exponent and a is the base. The expression an is read as "a raised to the nth power."
Exponents obey a number of useful properties that we can exploit when working with algebraic equations.
They also have a number of applications when working with exponential and logarithmic equations.
Fundamentals of Algebra
Properties of Exponents – 11
Properties of Exponents
Let a and b be real numbers, variables, or algebraic expressions. Let m and n be integers
(positive or negative, it makes no difference). All denominators and bases are nonzero.
1.)
Property
a ma n = a m + n
2.)
3.)
=
a–n
am – n
=
x5 – 2
=
t–3
=
=
=
a0 = 1, a ≠ 0
(x + 5z)0 = 1
5.)
(ab)m = ambm
(5y)2 = 52y2 = 25y2
6.)
(am)n = amn
8.)
(y3)–4
y(3)(–4)
=
=
=
| a2 | = | a |2 = a2
x3
=
4.)
7.)
Note:
Example
23 · 24 = 23 + 4 = 27 = 128
=
y–12
=
=
|(–4)|2 = | –4 |2 = (4)2 = 16
It is vitally important to understand the difference between (–2)4 and –24. In (–2)4, the parentheses
indicate the exponent applies to everything inside the parentheses, not just the 2. However, in –24, the
exponent only applies to the 2. Hence, (–2)4 = 16, whereas –24 = –16.
Scientific Notation
Oftentimes it is extremely inefficient and time consuming to write down very large numbers. For instance, one
famous number is Avogadro's number, which is:
602,200,000,000,000,000,000,000
Avogadro's number
This would become extremely tedious to write down every time we needed to use it, so scientists and
mathematicians rely on a short hand method of writing really large and really small numbers. This shortcut is
called scientific notation. Using scientific notation, Avogadro's number is reduced to:
602,200,000,000,000,000,000,000
=
6.022 x 1023
Avogadro's number in scientific notation
Now, which number is more compact? I think the choice is obvious. We can also use scientific notation for
really small numbers. For instance, the mass (in grams) of one proton is approximately:
1. 67 x 10–24
=
0.000000000000000000000000167
mass of proton in grams
Again, it would be really quite tedious to write out the mass of a proton in standard decimal form.
In general, scientific notation takes the form:
±c x 10n
1 ≤ c < 10, n is an integer
scientific notation
A positive exponent indicates that the absolute value of the number is large (sometimes very, very large). A
negative exponent indicates that the absolute value of the number is less than 1.
12 – Properties of Exponents
Fundamentals of Algebra
Some more useful large and small numbers, both in normal and scientific notation:
speed of light
gravitational constant
Planck constant
mass of the sun
distance to the nearest star
distance to the Andromeda galaxy
scientific notation
3 x 108 m/s
–11
6.67 x 10 Nm2/kg2
6.63 x 10–34 Js
1.99 x 1030 kg
4.04 x 1016 m
2.1 x 1022 m
normal decimal notation
300,000,000 m/s
0.0000000000667 Nm2/kg2
0.000000000000000000000000000000000663 Js
1,990,000,000,000,000,000,000,000,000,000 kg
40,400,000,000,000,000 m
21,000,000,000,000,000,000,000 m
After comparing the two notations, which would you rather use?
Fundamentals of Algebra
Properties of Exponents – 13
14 – Properties of Fractions
Fundamentals of Algebra
Advanced
To review the basic properties of fractions.
No definitions on this page.
Simon Stevin (1548-1620), a Dutch mathematician, physicist, and engineer, is credited with explaining the
concept of decimal fractions. At the time, it was unfamiliar to most mathematicians. He didn't actually view
decimal fractions as fractions and wrote the numerators without the denominators. The number π was written as
301
4
1
6
. . . The circled digit of each pair after the zero (which stands for the decimal point)
represents the power of 10 in the denominator: 1
= 1/10, 4
= 4 / 100 = 4 / 102.
Fundamentals of Algebra → Basic Rules of Algebra → Properties of Fractions
Fractions serve a number of uses in algebra. For one thing, they indicate a small portion of the whole, such as
1
/2. They can also indicate a ratio such as 3:2 = 3/2. Finally, they can be used to represent division of two
numbers as in 7 ÷ 5 = 7/5.
The following properties apply to all fractions, everywhere, regardless of the purpose they are trying to convey:
Fundamentals of Algebra
Properties of Fractions – 15
Properties of Fractions
Let a, b, c, and d be real numbers, variables, or algebraic expressions
such that b ≠ 0 and d ≠ 0.
1.)
Equivalent Fractions
=
2.)
if and only if ad = bc
Rules of Signs
=
3.)
=
=
Generate Equivalent Fractions
=
4.)
and
c≠0
Add or Subtract with Like Denominators
=
5.)
Add or Subtract with Unlike Denominators
=
6.)
Multiply Fractions
=
7.)
Divide Fractions
=
16 – Properties of Fractions
=
c≠0
Fundamentals of Algebra
Advanced
To review the properties of inequalities.
No definitions on this page.
Charles Babbage (1792-1871) is considered to be the father of modern computers. He is most famous for his
design of the Difference Engine, which would have greatly speeded arithmetic calculations. Unfortunately,
Babbage was never able to obtain the funding to build a full-scale production model. He attempted to get
funding for a smaller-scale, more efficient version and was again turned down. Finally, nearly a century and a
half later, the Science Museum of London constructed Babbage's second Difference Engine. It weighed three
tons and worked flawlessly.
Fundamentals of Algebra → Basic Rules of Algebra → Properties of Inequality
When we solve linear inequalities in one variable, we use procedures remarkably similar to those used to solve
liner equations. That is, we try to isolate the variable on one side of the inequality symbol (<, >, ≤, ≥).
However, there are two very important considerations we have to keep in mind when we work with inequalities.
When we multiply or divide both sides of an inequality by a negative number, we reverse the direction of the
inequality. Here is an example:
–2
<
4
original inequality
(–3)(–2)
>
4(–3)
multiply both sides by –3
reverse inequality symbol
6
>
–12
simplify both sides
As we can see, when we multiplied both sides of the inequality by –3, the left hand side suddenly became a
positive number and the right hand side suddenly became a negative number. Since positive numbers are
always greater than negative numbers, we had to reverse the inequality symbol.
If two inequalities have the same solution set, then they are equivalent. For instance,
x+3
<
6
and
x
<
3
are equivalent because values of x less than 3 will satisfy both inequalities.
Here is a list of the operations that we can use to create equivalent inequalities:
Fundamentals of Algebra
Properties of Inequalities – 17
Properties of Inequalities
Let a, b, c, and d be real numbers.
1.)
Transitive Property
a<b
2.)
Note:
implies
a<c
and
c<d
implies
a+c<b+d
Addition of a Constant
a<b
4.)
b<c
Addition of Inequalities
a<b
3.)
and
implies
a+c<b+c
Multiplication by a Constant
for c > 0, a < b
implies
ac < bc
for c < 0, a < b
implies
ac > bc
Each of the properties given above is true if we replace the symbol "<" with "≤" and if we replace the
symbol ">" with "≥". For instance, the multiplication property above could just as easily be written as:
for c > 0, a < b
implies
ac ≤ bc
for c < 0, a < b
implies
ac ≥ bc
18 – Properties of Inequalities
Fundamentals of Algebra
Advanced
To examine why the number zero plays such an important part in mathematics and to learn about
some of its basic properties.
inclusive or – in the statement a or b, the answer may be a, it may be b, and the answer could also be both a
and b.
While we in the United States are used to using a period (.) as a decimal indicator, other countries, such as those
in Continental Europe, use a comma (,) instead.
Fundamentals of Algebra → Basic Rules of Algebra → Properties of Zero
Zero is by far one of the strangest numbers in all of existence. It is both nothing and everything. Since it is so
unique, it should come as no surprise that it exhibits a number of unique characteristics when combined with
other real numbers in algebraic expressions.
Properties of Zero
Let a and b be real numbers, variables, or algebraic expressions.
1.)
a + 0 = a and a – 0 = a
2.)
a·0=0
3.)
Note:
a≠0
4.)
(a / 0) is undefined
5.)
Zero Factor Property: If ab = 0, then a = 0 or b = 0
The "or" in the Zero Factor Property includes the possibility that either or both factors may be zero.
This is an inclusive or and it is the way "or" is generally used in mathematics.
Property 4 is somewhat misleading. When you study calculus, you will find that it is appropriate to
say that (a / 0) "approaches" infinity. To see why, consider dividing a by a really small decimal
number that is close to zero, such as 0.00001. Then you have:
a
0.00001
=
a
−5
10
rewrite denominator in exponential form
=
a(105)
property of exponents
=
a(100,000)
rewrite in standard notation
Dividing a by a really small decimal number close to zero is the same as multiplying a by a really large
number. Thus, as the denominator gets closer and closer to zero, the evaluation of the expression
becomes larger and larger, eventually "reaching" infinity when the denominator equals zero.
Fundamentals of Algebra
Properties of Zero – 19
20 – Properties of Zero
Fundamentals of Algebra
Advanced
To plot points in the Cartesian plane.
To find the distance between two points in the Cartesian plane.
To find the midpoint of a line segment connecting two points in the Cartesian plane.
rectangular coordinate system - a means of plotting points using two perpendicular axes, one designated as x
and the other designated as y. Each point is assigned an (x, y) coordinate that determines its location relative to
a central point in the coordinate system.
Cartesian plane - the more common term for the rectangular coordinate system. It is named after Rene
Descartes (1596-1650).
x-axis - the horizontal real number line in the Cartesian plane.
y-axis - the vertical real number line in the Cartesian plane.
origin - the point of intersection of the two axes, designated with the ordered pair (0, 0) and labeled as O.
quadrant - one of four equal pieces of the Cartesian plane, formed by the perpendicular intersection of the axes.
coordinate - the general term for either of the two numbers in the ordered pair (x, y).
abscissa - the specific term for the x-coordinate in the ordered pair (x, y).
ordinate - the specific term for the y-coordinate in the ordered pair (x, y).
Descartes (1596-1650), like many mathematicians of his day, also delved extensively into the realm of natural
philosophy (what we call science today). He postulated that there are ten natural laws to the universe, the sun is
in the middle of the solar system and the planets create small whirlpools or vortices in the medium of the
universe. These vortices affect all other nearby objects. His theories have long since been disproved by Isaac
Newton (1642-1727) and others, but he did lay some of the groundwork for others who followed him.
Interestingly, the first two of his natural laws are nearly identical to Newton's first two laws of motion.
Fundamentals of Algebra → Cartesian Plane
We often use a real number line to indicate how two real numbers relate to each other. For instance, the
numbers x = 4 and x = –1 can be indicated on the following number line:
This is fine if we are only dealing with one variable. But oftentimes we are forced by circumstances to deal
with two variables, one of which is dependent on the other. The most familiar two variables are x and y. The
physical representation of an ordered pair of x and y variables is a two-dimensional model often called the
rectangular coordinate system or the Cartesian plane.
The Cartesian Plane is named after René Descartes (1596–1650), who is responsible for a large portion of
analytic geometry and is also credited with the existential phrase, "I think, therefore I am" (cogito ergo sum in
Latin).
Fundamentals of Algebra
Cartesian Plane – 21
The Cartesian plane
The horizontal real line is called the x-axis. The vertical real line is called the y-axis. The point of intersection
of the two axes is called the origin. The two axes divide the plane into four equal parts called quadrants. The
first quadrant is always the top right portion of the graph above the x-axis and to the right of the y-axis. The
order of the remaining quadrants goes around in a counterclockwise direction. That is, the second quadrant is
above the x-axis and to the left of the y-axis; the third quadrant is below the x-axis and to the left of the y-axis;
the fourth quadrant is below the x-axis and to the right of the y-axis.
Each point in the Cartesian plane can be identified by a unique ordered pair (x, y) of real numbers x and y.
These are called the coordinates of the point. The number x represents the directed distance from the y-axis to
the point. The number y represents the directed distance from the x-axis to the point. We call the first
coordinate of the ordered pair the x-coordinate, or abscissa. The second coordinate of the ordered pair is called
the y-coordinate, or ordinate. In practice, we usually just refer to them as x- and y-coordinates, seldom calling
them the abscissa and ordinate.
Note:
Although we use an ordered pair (a, b) to represent an open interval on the real number line, this
should not be confused with the ordered pair (a, b) representing a point in the Cartesian plane. The
context of the given problem should give us enough clues to determine if we are discussing an open
interval or a point in the plane.
Two formulas are extremely useful when discussing points in the Cartesian plane. They are the Distance
Formula and the Midpoint Formula:
The Distance Formula
The distance d between the point (x1, y1) and (x2, y2) in the plane is given by
d
=
The Distance Formula is derived from the Pythagorean Theorem. Recall that the Pythagorean Theorem states
that for any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the
22 – Cartesian Plane
Fundamentals of Algebra
lengths of the two sides. That is, if the right triangle has hypotenuse of length c and sides of length a and b,
then c2 = a2 + b2.
Suppose we want to find the distance d between two points (x1, y1) and (x2, y2) in the plane. With these two
points, we can construct a right triangle, as shown below:
The length of the vertical side is the absolute value of the difference between the y-coordinates: | y2 – y1 |.
Similarly, the length of the horizontal side is the absolute value of the difference of the x-coordinates: | x2 – x1 |.
When we apply the Pythagorean Theorem, we obtain:
d2
=
d
=
| x2 – x1 |2 + | y2 – y1 |2
(
x2 − x1
)2 + (
y2 − y1
Pythagorean Theorem
)2
take square roots of both sides
We can remove the absolute value operators because we are squaring numbers and that always leads to a
positive result. Thus, we can simplify our equation to:
d
=
( x 2 − x 1) 2 + ( y 2 − y 1) 2
remove absolute value operators
Distance Formula
We choose the positive square root for d because the distance between two points in the plane is not a directed
distance. We have just established the Distance Formula.
Given the following ordered pairs (2, 1) and (4, 5)
a.)
Plot the points represented by these ordered pairs.
b.)
Find the distance between the two points.
solution
a.)
The first ordered pair can be represented by plotting a point on the Cartesian plane where x = 2 and
y = 1. The second ordered pair can be represented by a point at x = 4 and y = 5, as shown below:
Fundamentals of Algebra
Cartesian Plane – 23
answer
b.)
The distance between the two points is found by applying the Distance Formula, with (x1, y1) = (2, 1)
and (x2, y2) = (4, 5):
d
d
=
( x 2 − x 1) 2 + ( y 2 − y 1) 2
=
( 4 − 2) + ( 5 − 1)
=
( 2) + ( 4)
=
4 + 16
simplify
=
20
simplify under radical operator
=
2 5
2
2
2
2
Distance Formula
substitute in x2 = 4, x1 = 2, y2 = 5 and y1 = 1
simplify inside parentheses
simplify radical
answer
The other useful formula is the Midpoint Formula and is useful for finding the midpoint of a line segment in the
plane.
The Midpoint Formula
The midpoint of the line segment joining the
points (x1, y1) and (x2, y2) is given by
Find the midpoint of the line segment connecting the points (2, 1) and (4, 5). Plot this point on the same graph
we used in the previous example.
solution
The Midpoint Formula is pretty straightforwa rd. We simply substitute in (x1, y1) = (2, 1) and (x2, y2) = (4, 5).
24 – Cartesian Plane
Fundamentals of Algebra
Midpoint
=
⎛ x1 + x2 , y1 + y2 ⎞
⎜
⎟
2
⎝ 2
⎠
Midpoint Formula
=
⎛ 2 + 4, 1 + 5⎞
⎜ 2
⎟
2 ⎠
⎝
substitute in x1 = 2, x2 = 4, y1 = 1 and y2 = 5
=
⎛ 6, 6⎞
⎜ 2 2⎟
⎝
⎠
simplify numerators
=
(3, 3)
simplify fractions
answer
This point can be represented as follows:
answer
Fundamentals of Algebra
Cartesian Plane – 25
26 – Cartesian Plane
Fundamentals of Algebra
Advanced
To recognize that we can write radicals in exponential notation and to review normal radical notation.
square root – a square root of a number b is the solution of the equation x2 = b.
cube root – a cube root of a number c is the solution of the equation x3 = c.
principle nth root of a – the nth root that has the same sign as a.
radical symbol – an operator symbol that indicates we are taking the nth root of a real number, variable, or
algebraic expression.
index (radical) – a real number placed above and to the left of a radical symbol to indicate what root is sought.
radicand – the quantity under a radical symbol.
A fragment of a table of square roots was unearthed in Nippur, approximately 100 miles southeast of Baghdad,
Iraq. It was written by the ancient Mesopotamians and dates back to 1800 B.C.E. It is currently located in the
Babylonian section of the University of Pennsylvania.
Fundamentals of Algebra → Radicals
Every once in a while, we are asked to find the "root" of a number, whether it’s the square root or cube root or
some other root. These kinds of numbers are called radicals and are denoted with the symbol "√". In general,
the root of a number a is another number, which, when multiplied by itself n times, will produce our original
number a. For example, the third root of 8 is 2 because if we multiply 2 by itself 3 times, we end up with 8, i.e.
2 x 2 x 2 = 8 so 3√8 = 2.
Definition of the nth Root of a Number
Let a and b be real numbers and let n ≥ 2 be a positive integer. If
a
=
bn
then b is an nth root of a. If n = 2, the root is a square root. If n = 3, the root is a cube root.
Many numbers have more than one nth root. For example, both –4 and 4 are square roots of 16. The principle
nth root of a number is defined as follows:
Definition of Principle nth Root of a Number
Let a be a real number that has at least one nth root. The principle nth root of a is the nth root that
has the same sign as a. It is denoted by a radical symbol:
principle nth root of a
The positive integer n is the index of the radical, and the number a is the radicand. If n = 2, we omit
the index n and simply write √a rather than 2√a.
Fundamentals of Algebra
Radicals – 27
In looking at the nth root of various real numbers, we can arrive at the following conclusions:
1.
If a is a positive real number and n is a positive even integer, then a has exactly two real nth roots
denoted by n√a and –n√a.
2.
If a is any real number and n is an odd integer, then a has only one real nth root denoted by n√a.
3.
If a is a negative real number and n is an even integer, then a has no real nth root. (It has a
complex root instead.)
4.
n
28 – Radicals
√0 = 0
Fundamentals of Algebra
Advanced
To review basic properties of radical numbers.
No definitions on this page.
In the 6th century B.C.E., the Greek mathematicians, following Pythagoras, considered irrational numbers an
inexplicable error on the part of the Supreme Architect. Thus, they had to be kept secret, otherwise one might
incur divine wrath.
Fundamentals of Algebra → Radicals → Properties of Radicals
Radicals are essentially exponents in disguise. Thus, they obey many of the same properties obeyed by
conventional integer exponents.
Properties of Radicals
Let a and b be real numbers, variables, or algebraic expressions such that the indicated roots
are real numbers. Let m and n be positive integers.
Property
1.)
Example
=
2.)
=
4.)
=
=
=
5
For n even
=
|a|
=
| –6 |
For n odd
=
a
=
–6
Fundamentals of Algebra
4
=
=
a
=
=
=
=
5.)
(2)2
=
=
=
3.)
6.)
=
=
6
Properties of Radicals – 29
30 – Properties of Radicals
Fundamentals of Algebra
Advanced
To review the concept of real numbers and how they include other sets of numbers such as the
rational and irrational numbers.
real number – a rational or an irrational number, which can be represented by a finite or infinite decimal.
rational number – a number that is the ratio, or quotient, of two integers.
irrational number – any number that cannot be expressed as a terminating or repeating decimal.
integers – the set of whole numbers and their negative counterparts.
whole numbers – all positive integers and zero.
natural numbers – all positive integers.
real number line – a line with numbers evenly spaced throughout the line.
origin – the point 0 on the real number line.
one-to-one correspondence – every point on the number line corresponds to exactly one real number.
In the first half of the fifteenth century (ca. 1450 B.C.E), the Persian mathematician Ghiyat ad din Ghamshid ibn
Mas'ud al-Kashi did important work on algebra, especially concerning the binomial theorem, decimal fractions,
exponential powers of whole numbers, and the set of irrational numbers.
Fundamentals of Algebra → Real Numbers
For most applications in the real world, we use a concept known as real numbers. When you count your
pocket change, wonder how many more minutes until lunch, or decide to invest in a lucrative stock, you are
using real numbers.
Real numbers come in a wide variety of symbols, such as:
9, –3, 5/2, 0.66666666…, 34.32, √5, π, and e.
We can separate out different groups of the set of real numbers into smaller subsets. The two largest subsets of
real numbers are the rational and irrational numbers. A rational number is any number that can be expressed
as the ratio of two integers p/q where q ≠ 0. For example, the numbers:
0.25
=
0.34343434….
=
1
4
rational numbers
34
99
are both rational numbers.
The decimal representation of a rational number either repeats (as in 2.734734734….) or terminates (as in 0.4)
If a real number cannot be written as the ratio of two integers, then it is an irrational number. Irrational
numbers have infinite non-repeating decimal representations. Some of the most famous irrational numbers
include:
Fundamentals of Algebra
Real Numbers – 31
2
≈
1.4142135623731
.π
≈
3.14159265358979
e
≈
2.71828182845905
irrational numbers
The symbol "≈" means "approximately to". We can never actually write down the exact value of π because as
far as we have been able to determine, π has no end.
Besides rational and irrational numbers, we can further subdivide the rational numbers into integers and nonintegers. An integer is basically a rational number whose denominator is "1". This include both positive and
negative numbers, such as {…–2, –1, 0, 1, 2…}. The set of integers can also be subdivided into whole
numbers (which includes both positive integers and 0) and natural numbers (which includes positive integers,
but does not include zero!).
If we want to actually "look" at a real number, we plot the number on a real number line. The point 0 on the
number line is called the origin, as all other numbers originate from this central point. By convention, numbers
to the right of 0 are positive numbers and numbers to the left of zero are negative, as shown below:
Every real number corresponds to exactly one point on the real number line. Every point on the number line
corresponds to exactly one real number. This relationship is called one-to-one correspondence.
32 – Real Numbers
Fundamentals of Algebra
Advanced
To review the concept of absolute value and a few of the properties of the absolute value function
absolute value – the absolute value of a number is its distance from 0 on a number line.
distance – a measure of how far apart two things are from each other.
The ancient Mayans used more than one number system. The "ordinary" system was a base 20 system and
contained a true "0". The "learned" system was used exclusively by the priests in astronomical calculations as a
means of reckoning time.
Fundamentals of Algebra → Real Numbers → Absolute Value
One way to define the absolute value of a real number is its distance from the origin. For example, the number
–4 is located 4 units to the left of the origin, so its absolute value is 4. We disregard the negative sign when we
are talking about absolute value, since a magnitude (distance) is always positive. The negative (or positive)
sign only indicates the number's direction with respect to the origin.
A formal definition of absolute value is as follows:
Definition of Absolute Value
If a is a real number, then the absolute value of a is:
|a|
=
a
–a
if a ≥ 0
if a < 0
From this definition, we can see that the absolute value of a real number is never negative. For example, if
a = –3, then | –3 | = –(–3) = 3 > 0.
Like many operations, the absolute value operator has a number of properties associated with it that follow from
the definition.
Properties of Absolute Value
If a and b are any two real numbers,
then the following properties are true:
1.
|a|
≥
0
2.
| –a |
=
|a|
3.
| ab |
=
|a||b|
4.
Fundamentals of Algebra
=
b≠0
Absolute Value – 33
Finally, we can use the absolute value operator to define the distance between two points on the real number
line. For example, the distance between –3 and 5 is:
| –3 – 5 |
=
| –8 |
=
8
as shown below:
Distance Between Two Points on the Real Number Line
Let a and b be any two real numbers. The distance between a and b is given by:
d(a, b)
34 – Absolute Value
=
|b–a|
=
|a–b|
Fundamentals of Algebra
Advanced
To review how real numbers relate to one another.
less than - one value is smaller than a compared value.
greater than - one value is larger than a compared value.
greater than or equal to - one value is larger than or the same as a compared value.
less than or equal to - one value is smaller than or the same as a compared value.
interval - any subset of real numbers.
bounded - any interval which has a finite upper and lower boundary.
unbounded - an interval that has infinity as one or both of its endpoints.
Although we call our numerals "Arabic numerals", the Arabs actually borrowed them from the Hindu number
system. The original Arabic numerals were much more similar in nature to the Greek and Roman systems,
which used letters to represent numerals and had no concept of place-value (where the position of a number
indicates its value).
Fundamentals of Algebra → Real Numbers → Ordering Real Numbers
When we talk about the relationships between real numbers, we usually like to talk about one number being
"larger" or "smaller" than another number. In a sense, we determine the order of the numbers. 3, for
example, is a larger quantity than 2, so we say that 3 is "greater than" 2 and denote this using a mathematical
symbol: 3 > 2. This can be extended to other types of relationships, such as "less than", "greater than or
equal to", and "less than or equal to".
Definition of Order on the Real Number Line
If a and b are real numbers, a is less than b if b – a is positive. This order is denoted by the inequality:
a
<
b
a is less than b
This can also be described by saying that b is greater than a and is denoted by the inequality
b
>
a
b is greater than a
a
b is greater than or equal to a
The inequality
b
≥
means that b is greater than or equal to a, while the inequality
a
≤
b
a is less than or equal to b
means that a is less than or equal to b. The symbols <, >, ≤, ≥ are inequality symbols.
If we want to look at this definition geometrically, this implies that a < b only if a is to the left of b on the real
number line. Then the next statement is obviously that b is greater than a if b is to the right of a on the real
number line, as shown below:
Fundamentals of Algebra
Ordering Real Numbers – 35
Interval Notation
We can use the power of inequalities to describe subsets of real numbers called intervals. For example, if we
wanted to look at all the numbers between 4 and 5 (but not including 4 and 5), then we would write this as the
interval (4, 5). The parentheses tell us that the points 4 and 5 are not included in our interval. If, on the other
hand, we really did want to have 4 and 5 in our interval, then we would write this with straight brackets instead
of parentheses [4, 5]. Of course, we could also include 4 and exclude 5 or vice versa. If we include one point
but not the other, we write one side with a parentheses and the other side with a straight bracket: (4, 5] or [4, 5).
This can easily be generalized to any two real numbers (including negative real numbers).
All of these intervals are called bounded intervals because they involve an upper and lower boundary. This is
summarized below:
Bounded Intervals on the Real Number Line
Notation
Interval Type
Inequality
[a, b]
Closed
a≤x≤b
(a, b)
Open
a<x<b
(a, b]
Half-open
a<x≤b
[a, b)
Half-open
a≤x<b
Graph
Besides bounded intervals, we can talk about the notion of unbounded intervals. This is an interval that
essentially has infinity as one of its endpoints. For example, the interval described by "all real numbers greater
than 7" means that 7 is the lower bound of the interval and there is no real upper bound of the interval. This can
be written as (7, ), where the symbol for infinity, ∞, indicates that this interval is unbounded. Note that we
always close of the unbounded side of an interval with a parenthesis instead of with a square bracket. You
cannot write (7, ], because infinity is not bounded.
36 – Ordering Real Numbers
Fundamentals of Algebra
This is summarized below:
Unbounded Intervals on the Real Number Line
Notation
Interval Type
Inequality
[a,
)
Half-open
x≥a
(a,
)
Open
x>a
(–
, b]
Half-open
x≥b
(–
, b)
Open
x<b
(–
,
)
Graph
Entire real line
The Law of Trichotomy states that for any two real numbers a and b, precisely one of the following relationships
is possible:
a
a
a
=
or
<
or
>
Fundamentals of Algebra
b
b
Law of Trichotomy
b
Ordering Real Numbers – 37
38 – Ordering Real Numbers
Fundamentals of Algebra
`
Unit 2
Linear Equations
Linear Equations – 39
40 – Linear Equations
Advanced
To learn the basics of linear equations and how to solve them
solve – to find the result by the use of certain given data, previously known facts or methods, and newly
observed relations.
solution – a replacement of the variable(s) in an algebraic sentence that makes the sentence true.
identity – an equation whose sides are equivalent expressions.
conditional equation – an equation that is true for only a few members in the domain.
linear equation – an equation in which each term is either a constant or a monomial of degree 1.
equivalent equations – equations having the same solution set over a given domain.
Aryabhata the Elder (c.476-550) was a Hindu astronomer and mathematician. He was often correct about the
solution to one particular problem, but incorrect about the solution to a related problem. For instance, in his
poem Garite (Hindu mathematics was expressed in poetic form), he correctly states the formulae for the areas of
a triangle and a circle. However, he fails to correctly extrapolate the formulae to find the volume of threedimensional shapes. An Arab commentator, al-Biruni, described all of Hindu mathematics as a mixture of
"common pebbles and costly crystals."
Linear Equations
An equation is a statement that two algebraic expressions are equal to each other. Examples include: 3x – 4 =
5, x2 – 4 = 0, and 3 – 2x = x2. If we wish to solve an equation, we want to find all values of x that will satisfy
the equation or in other words make it true. If we let x = 3, then 3x – 4 = 5 can also be written as 3(3) – 4 = 5.
If we add 4 to both sides of the equation, we end up with 3(3) = 9, which is a true statement.
Therefore we say that x = 3 is the solution to the equation 3x – 4 = 5.
An equation that is true for every value of x in the domain of the variable is called an identity.
For example,
x2 – 4
=
(x – 2)(x + 2)
identity
is an identity because it is true for any real value of x. We are restricting the domain of x to real numbers only,
excluding complex numbers.
An equation that is true for only a few of the numbers in the domain of x is called a conditional equation .
For example,
0
=
(x + 3)(x – 1)
conditional equation
is a conditional equation because it is only true for x = –3 and x = 1. Learning how to solve conditional
equations is one of the primary purposes of algebra.
Linear Equations in One Variable
The focus of this section of the Algebra Brain is in solving Linear Equations. A linear equation has exactly one
solution. Period. In general, a linear equation in one variable is defined as:
Linear Equations
Introduction to Linear Equations – 41
Definition of Linear Equation
A linear equation in one variable x is an equation that can be written in the standard form
ax + b
=
0
linear equation
where a and b are real numbers and a ≠ 0.
To solve a conditional equation in x (such as a linear equation), the idea is to isolate x on one side of the
equation and group all the real numbers on the other side. Then we simplify the equation by adding,
subtracting, multiplying or dividing the real numbers. When all we have is an x on one side and a single real
number on the other side, the equation has been solved for x. This process takes the form of a number of steps.
We transform the original linear equation that we are given into a simpler, equivalent equation, which has the
same solution as our original equation.
Generating Equivalent Equations
An equation can be transformed into an equivalent equation by one or more of the following steps:
Guideline
Given Equation
Equivalent
Equation
1.
Remove symbols of grouping, combine like terms,
or reduce fractions on one or both sides of the equation
3x – 2x = 5
x=5
2.
Add (or subtract) the same quantity to (or from)
both sides of the equation
2x + 7 = 4
2x = –3
3.
Multiply (or divide) both sides of the equation
by the same nonzero quantity.
4x = 8
x=2
4.
Interchange the two sides of the equation
3=x
x=3
Sometimes we find out that a linear equation has no solution because we end up with a contradictory statement,
such as 0 = 4 or 3 = 12. An example of a contradictory equation is x = x + 1. If we solve it for x, we end up
with 0 = 1, which cannot possibly be true.
Solve the equation and check the solution.
7 – 2x
=
15
solution
In order to isolate the x-term, we need to get rid of the 7. We subtract 7 from both sides. Then, in order to get
rid of the –2 in front of the x, we divide both sides by –2. It is important that we do this in the stated order;
otherwise we will get the wrong answer.
7 – 2x
=
15
given equation
7 – 2x – 7
=
15 – 7
subtract 7 from both sides
–2x
=
8
simplify
−2x
=
−2
x
=
8
−2
–4
42 – Introduction to Linear Equations
divide both sides by –2
simplify
answer
Linear Equations
Now we need to check our answer. We substitute in our answer into the original equation:
7 – 2x
=
15
original equation
7 – 2(–4)
=
15
substitute in x = –4
7+8
=
15
simplify
15
=
15
check
Since 15 = 15 is a true statement, we conclude that x = –4 is indeed the solution to this equation.
Let's look at another, more complicated example.
Solve the equation (if possible) and check the solution.
8(x + 2) – 3(2x + 1)
=
2(x + 5)
solution
The first thing we need to do is apply the Distributive Property to get rid of all the parentheses. They only get
in the way right now.
8(x + 2) – 3(2x + 1)
=
2(x + 5)
given equation
8x + 16 – 6x – 3
=
2x + 10
Distributive Property
(note that the negative sign gets carried inside the parentheses)
Next, we simplify the equation, attempting to get all the real numbers on one side, and the x-term isolated by
itself on the other side:
8x + 16 – 6x – 3
=
2x + 10
result from above
2x + 13
=
2x + 10
simplify by combining like terms
13
=
10
subtract 2x from both sides
Whoops! What happened here? We ended up with a statement that is clearly not true. How do we interpret
this? Graphically, we can interpret the left and right hand sides of the equation as parallel lines. On the left
hand side, we have y1 = 2x + 13. On the right hand side, we have y2 = 2x + 10. This means that both equations
have the same slope, but different y-intercepts. They will never intersect, thus they have no common points. In
other words, our given equation has no real solution.
answer
Equations Involving Fractional Expressions
Another common type of equation is when a linear expression is in the denominator of a fraction. An example
of this would be:
1
x
+
2
x−5
=
0
equation with a linear expression as a denominator
We solve these by finding the Least Common Denominator (LCD). We then multiply every term by the LCD.
Then we solve the resulting equation just as we did before.
Linear Equations
Introduction to Linear Equations – 43
Continuing with our example above, the LCD is x(x – 5). We multiply both terms by the LCD:
⎛ 1 + 2 ⎞ ( x) ( x − 5)
⎜ x x − 5⎟
⎝
⎠
1
x
( x) ( x − 5) +
=
0(x)(x – 5)
multiply both sides by LCD
( x) ( x − 5)
=
0
Distributive Property
(x – 5) + 2(x)
=
0
cancel like terms in numerators and denominators
x – 5 + 2x
=
0
remove parentheses
3x – 5
=
0
combine like terms
3x – 5 + 5
=
0+5
add 5 to both sides
3x
=
5
simplify
2
x−5
3x
=
3
=
x
5
3
5
3
divide both sides by 3
simplify
In order to check this, we let x = 5/3 in our original equation and get:
1
5
3
+
2
5
3
−5
3
=
5
+
2
5
−
3
3
=
5
+
15
3
dividing by a fraction is same as multiplying by reciprocal
convert whole number in denominator in second term to a
fraction
2
simplify secondary fraction in second term
− 10
3
=
=
=
=
3
5
3
5
3
5
+ 2⎛⎜ −
⎝
−
−
0
44 – Introduction to Linear Equations
6
10
3
5
3
⎞
⎟
10 ⎠
dividing by fraction is same as multiplying by its reciprocal
multiply 2 inside parentheses
reduce second fraction to lowest terms
simplify
check
Linear Equations
Advanced
To learn about the relationship between lines and slope.
No definitions on this page.
Thales of Miletus (c.640-546 B.C.) is not only famous for introducing geometry to the Greeks, but also for
establishing the need for deductive reasoning and proofs to demonstrate the validity of a theorem or statement.
The Greek philosopher Proclus credits Thales with five basic theorems of geometry:
(1) The circle is bisected by its diameter.
(2) The base angles of an isosceles triangle are equal.
(3) Pairs of vertical angles formed by two intersecting straight lines are equal.
(4) An angle inscribed in a semi-circle is a right angle.
(5) Two triangles are congruent if they have two angles and one side that are equal.
Thales reportedly used this last theorem to find the distance from a ship to the shore using a tower of known
height, a plum line, and a carpenter's square (called a gnomon).
Linear Equations → Lines and Slope
In real life problems the word slope can mean several different things. The slope of a line can refer to a ratio or
a rate. The slope is a ratio if the x-axis and y-axis have the same unit of measure. For example, the steepness of
Mt. St. Helens is a ratio of height to length and both are measured in feet. The slope is said to be a rate, or rate
of change if the x-axis and y-axis have different units. Economists calculate marginal cost, the rate of cost per
unit, which is measured in dollars and the number of units.
The slope of a line is also an important tool in sketching graphs. We can easily draw a graph given any two
points on the line or by using one point and the slope of the line. Making a graph is important because you can
use it to determine other points on the line. When looking at the graph, you can determine the behavior of the
line at a glance.
Definition of the Slope of a Line
In a coordinate plane, the slope of a line is the ratio of its vertical rise to the horizontal run.
slope
=
rise
run
definition of slope
The rise is simply the change in the y direction of the line and the run is the change in the x direction.
The slope is usually denoted by m and is determined by any two points on a particular line.
Using the points (x1, y1) and (x2 , y2), we can calculate the slope of the line.
slope
=
=
rise
run
y2 − y1
x2 − x1
Let's consider a line passing through the points (0, 3) and (4, 5).
Linear Equations
Lines and Slope – 45
If we let (x1, y1) = (0, 3) and (x2, y2) = (4, 5), we can find the slope of the line using the definition of the slope:
m
=
=
=
=
y2 − y1
x2 − x1
5−3
4−0
2
4
1
2
two point form of slope
this is read as "y sub 2 minus y sub 1 divided by x sub 2 minus x sub 1"
substitute in x1 = 0, x2 = 4, y1 = 3 and y2 = 5
simplify numerator and denominator
reduce fraction to lowest terms
The change in the y direction (or vertical change) is the rise and the change in the x direction (horizontal
change) is the run. The slope, or rate of change, is the quotient of these changes.
A slope of –3/4 would mean for every change of 4 units to the right there is a change of 3 units down. Also, this
means that for every change of 1 horizontal unit there is a vertical change of –3/4 of a unit.
Note:
When calculating the slope of a line, it is important to remember to subtract the coordinates in the
same order. Therefore, when practicing you should always label your first point as (x1, y1) and your
second point as (x2, y2). Then make sure that you form the numerator and denominator using the same
order of subtraction:
Correct
y2 − y1
Incorrect
y2 − y1
x2 − x1
x1 − x2
When finding the slope of the line, 4 different possibilities can occur. The line can rise, fall, have no slope, or
be undefined. Let's take a look at the 4 cases and their graphs.
Case 1: A Line with a Positive Slope
A line with a positive slope, as in the previous example, rises from left to right. That is, the right end of the line
is higher on the graph than the left end of the line, as indicated below:
46 – Lines and Slope
Linear Equations
Case 2: A Line with a Negative Slope
Find the slope of the line passing through (0, 3) and (4, 1).
m
=
=
=
=
y2 − y1
x2 − x1
1−3
4−0
−2
4
−
1
2
two point slope formula
substitute in y2 = 1, y1 = 3, x2 = 4, and y1 = 0
simplify
reduce fraction to simplest form
slope is negative
A negative slope indicates that the end of the line is lower on the right than it is on the left, as shown below:
Case 3: A Line with a Zero Slope
Find the slope of the line passing through (5, 4) and (10, 4).
m
=
=
=
=
Linear Equations
y2 − y1
x2 − x1
4−4
10 − 5
0
5
0
two point slope formula
substitute in y2 = 4, y1 = 4, x2 = 10 and x1 = 5
simplify
slope is 0
Lines and Slope – 47
A zero slope means that the line is completely horizontal. It neither rises nor falls as we move along the line
from left to right. A graph of a flat line is given below:
The equation for this horizontal line is y = 5. In general, horizontal lines have equations in the form
y
=
constant
general form of a horizontal linear equation
Case 4: A Line with an Undefined Slope
Find the slope of the line passing through (3, 2) and (3, 7).
m
=
=
=
=
y2 − y1
x2 − x1
7−2
3−3
5
0
undefined
two point slope formula
substitute in y2 = 7, y1 = 2, x2 = 3, and x1 = 3
simplify
division by zero cannot be defined
Because division by zero is not possible, this expression has no meaning. We say that the slope of a vertical line
is undefined. A graph of a vertical line is shown below:
48 – Lines and Slope
Linear Equations
The equation for this vertical line is x = 3. In general, vertical lines have equations of the form
x
=
constant
general form of a vertical linear equation
Given the point (10, 2) and slope of 1/5, sketch the graph.
solution
Plot the point (10, 2) on a coordinate plane. Then, move up 1 unit in the y direction and to the right 5 units in
the x direction. You should be at the point (15, 3).
Keep in mind a slope of 1/5 is the same as a slope of –1/–5. So, if you move down 1 unit in the y direction, then
you also move left 5 units in the x direction. Also, the calculated slope for a given line will always be the same
no matter which pair of points is chosen.
answer
The church of St. Paul is a mortuary chapel, which stands 32 feet wide. The design is similar to that of a miniCathedral. The main roof and the bell tower both rise 5 feet for every 4 feet of horizontal distance. The bell
tower is 4 feet wide and 15 feet tall. If the walls of the chapel are 50 feet high and the walls of the bell tower
are 12 feet high, find the distance from the ground to the peak of the bell tower.
solution
First, we find the slope of the roof. It rises 5 feet for every 4 feet of horizontal distance it travels, so we can
determine the slope as follows:
m
=
=
rise
run
5
4
definition of slope
substitute in rise = 5 and run = 4
Next, we need to find the height of the main roof. The distance from one side of the chapel to the peak of its
roof is one-half the total width of the chapel. The chapel is 32 feet wide, so half that distance is 16 feet. Now to
find the height of the roof, we set up ratio relating the slope to the actual height of the chapel roof, which we
will call c.
Linear Equations
Lines and Slope – 49
5
4
=
c
16
ratio of slope to actual height of chapel
80
=
4c
cross multiply
20
=
c
divide both sides by 4
Thus, the height of the chapel roof is 20 feet.
Next, determine the height of the roof on the bell tower and call it b. The distance from one side of the chapel to
the peak of its roof is 4 divided by 2, which is 2 ft. Again, set up a ratio.
5
b
=
4
2
ratio of slope to actual height of bell tower
10
=
4b
cross multiply
2.5
=
b
divide both sides by 4
Thus, the roof of the bell tower is only 2.5 feet high.
To find the total height of the chapel from the stone floor to the very top of the bell tower, we add c = 20 and
b = 2.5 to height of the walls w = 50 and the height of the bell tower a = 12:
h
=
w+c+a+b
total height is sum of four heights
=
50 + 20 + 12 + 2.5
substitute in w = 50, c = 20, a = 12, and b = 2.5
=
84.5 ft
simplify
answer
50 – Lines and Slope
Linear Equations
Advanced
To define parallel and perpendicular lines.
parallel lines – two lines are parallel if there is a plane in which they both lie and they do not cross at any point
within the plane.
perpendicular lines – two straight lines in a plane, the intersection of which forms right angles.
Euclid's (c.300 B.C.) Fifth Postulate is one of the most important contributions to geometry. This postulate lays
the foundation for the theory of parallel lines, including the idea that only one parallel to a given line can be
drawn through a given point external to that line. Oddly enough, the Fifth Postulate cannot be proved using the
previous four postulates. Attempts to prove the Fifth Postulate have led to the creation of non-Euclidean
geometries. The alternate geometries have contributed greatly to the study of physics and relativity.
Linear Equations → Lines and Slope → Parallel and Perpendicular Lines
Recall from geometry that one definition of parallel lines is two lines in the same plane that never cross. The
algebraic definition is a bit more technical. In algebra, we define lines in terms of slope, so it should come as
no surprise that we define both parallel and perpendicular lines in terms of their slopes.
Parallel and Perpendicular Lines
1.)
Two distinct nonvertical lines are parallel if and only if their slopes are equal. That is, m11 = m2.
2.)
Two nonvertical lines are perpendicular if and only if their slopes are negative reciprocals of
each other. That is, m1 = –1 / m2.
Write an equation of the line through the given point (a) parallel to the given line and (b) perpendicular to the
given line.
4x – 2y
=
3
(2, 1)
solution
First, we need to rewrite the equation of the line in some form that is easier for us to work with. Slope-intercept
form is always a safe bet so let's rearrange the given equation a bit.
4x – 2y
=
3
given equation
–2y
=
–4x + 3
subtract 4x from both sides
y
=
y
=
Linear Equations
−4x + 3
−2
2x −
3
2
divide both sides by –2
simplify
slope-intercept form
Parallel and Perpendicular Lines – 51
a.)
b.)
Any line that is parallel to the given line above must have the same slope of 2. Thus, the line through
(2, 1) that is parallel to the given line must also have a slope of 2 and its equation can be found using
the point-slope form for the equation of a line:
y – y1
=
m(x – x1)
point-slope form
y–1
=
2(x – 2)
substitute in m = 2, x1 = 2, and y1 = 1
y–1
=
2x – 4
Distributive Property
y
=
2x – 3
add 1 to both sides
slope-intercept form
answer
Any line that is perpendicular to the given line must have a slope of –1/ 2, since this is the negative
reciprocal of 2. Thus, the line through (2, 1) that is perpendicular to the given line can be found by
again applying the point-slope form for the equation of a line:
y – y1
=
m(x – x1)
point-slope form
y–1
=
1
− ( x − 2)
2
substitute in m = –1/2, x1 = 2, and y1 = 1
y–1
=
1
− x+1
2
Distributive Property
y
=
1
− x+2
2
add 1 to both sides
slope-intercept form
answer
The graph of all three lines is given below:
52 – Parallel and Perpendicular Lines
Linear Equations
Advanced
To find the equation of a line given its slope and its y-intercept.
No definitions on this page.
Alan Turing (1912-1954) was an English algebraist, logician, and code-breaker. He is also a leading pioneer in
computer theory. In one of his papers he describes a theoretical automatic machine that could solve any
mathematical problem as long as the machine was given the correct problem-solving instructions or algorithm. It
was called the Turing machine after its creator. Turing worked with other mathematicians to develop a system of
equations called logic gates. This system relied on the use of binary numbers (1 and 0) to produce the problem
solving algorithms that would be used by an automatic calculating machine. Modern-day computers still use the
same logic to function; they just do it better and faster than Turing could ever dream of. He died in 1954 after
eating a cyanide-laced apple.
Linear Equations → Lines and Slope → Slope-Intercept Form of a Line
One of the simplest equations for modeling two variables is the linear equation
y
=
mx + b
linear equation
This equation is called linear because when we graph it on the Cartesian plane, it takes the shape of a line. In
mathematics, the term line refers exclusively to straight lines, not curves, whirls, loops, or any other sort of
non-straight lines.
If we let x = 0 in the above equation, we end up with
y
=
b
let x = 0
We call the point (0, b) the y-intercept because this is the point at which the line crosses the y-axis, as shown
below.
positive slope, line rises
negative slope, line falls
If b = 0, then the line passes through the origin. If m = 0, then we have a horizontal line given by the equation
y
=
Linear Equations
b
horizontal line
Slope-Intercept Form of a Line – 53
The Slope-Intercept Form of the Equation of a Line
The graph of the equation
y
=
mx + b
slope-intercept form
is a line whose slope is m and whose y-intercept is (0, b).
Once we have determined the slope of a line and its y-intercept, it is a relatively simple matter to sketch the
graph of the line. If we have a vertical line, then that line has the equation
x
=
equation of a vertical line
a
The slope of a vertical line cannot be written in the form y = mx + b because the slope of a vertical line is
undefined.
Find the slope and y-intercept of the equation of the line. Sketch a graph of this line.
2x + 3y – 9
=
0
solution
First, let's transform the given equation so that it is in slope-intercept form. In other words, we solve the
equation for y:
2x + 3y – 9
=
0
given equation
3y
=
–2x + 9
isolate the term containing y
y
=
y
=
−2x + 9
3
2
− x+3
3
divide both sides by 3
simplify
slope-intercept form
Based on the slope-intercept form of the linear equation, we can see that the slope is m = –2/3 and the y-intercept
is (0, 3), as indicated in the graph below:
answer
54 – Slope-Intercept Form of a Line
Linear Equations
Advanced
To write a linear equation using the slope of a line and a point on that line.
point-slope form – one form of a linear equation derived from knowing one point on the line and its slope.
Among the many secrets Thales of Miletus (c.640-546) learned from the Babylonians was the secret of
predicting solar eclipses. Using these secrets, Thales was able to predict the solar eclipse of 585 B.C. He still
had a tremendous amount of luck to help him, however, as the eclipse just happened to be visible from his
school in Miletus. This prediction bolstered Thales's fame and reputation throughout the lands of Greece.
Linear Equations → Lines and Slope → Point-Slope Form of a Line
One way in which we defined the slope of a line was in terms of two coordinate points (x1, y1) and (x, y). The
slope m was defined as:
y − y1
=
x − x1
m
definition of slope
Now, suppose we multiply both sides by the quantity (x – x1):
y – y1
=
m(x – x1)
multiply both sides by (x2 – x1)
This equation is known as the point-slope form of the equation of a line because it involves two points on the
line and their slope.
The Point-Slope Form of the Equation of a Line
The equation of the line with slope m passing through the point (x1, y1) is:
y – y1
=
m(x – x1)
point-slope form of the equation of a line
One of the most useful applications of this form of the equation of a line is actually finding the equation of a
line. That is, given a point and the slope of the line, we can easily find an equation for that line. Then we can
convert from this form to the slope-intercept form to help us graph the line.
Find an equation of the line that passes through the given point and has the indicated slope. Then graph the
line.
point
(–3, 5)
slope
m = –2
Linear Equations
Point-Slope Form of a Line – 55
solution
We use the point-slope formula with m = –2 and (x1, y1) = (–3, 5).
y – y1
=
m(x – x1)
point-slope form
y–5
=
–2(x – (–3))
substitute in m = –2, x1 = –3, and y1 = 5
y–5
=
–2(x + 3)
simplify inside parentheses
y–5
=
–2x – 6
Distributive Property
y
=
–2x – 1
add 5 to both sides
slope-intercept form
answer
Based on this equation, we can see that the slope is –2 and the y-intercept is –1, so we have the following graph:
answer
Two-Point Form of the Equation of a Line
If we have two points instead of one point, we can still find the equation of a line passing through those two
points. If we have the two points (x1, y1) and (x2, y2), then the slope of the line connecting these two points is
given by the difference in the y-coordinates divided by the difference in the x-coordinates:
m
y2 − y1
=
x2 − x1
x1 ≠ x2
When we plug this into the point-slope form of the equation of a line, we obtain:
y – y1
=
y2 − y1
x2 − x1
(x − x1)
two-point form
This is called the two-point form of the equation of a line.
56 – Point-Slope Form of a Line
Linear Equations
Find an equation of the line passing through the two points. Then sketch the graph of the line.
(–1, 3)
(3, 5)
solution
We use the two-point form of the equation of a line as given above with (x1, y1) = (–1, 3) and (x2, y2) = (3, 5):
y – y1
=
y–3
=
y–3
=
y–3
=
y–3
=
y
=
y2 − y1
x2 − x1
5−3
3+ 1
2
4
1
2
1
2
1
2
(x − x1)
( x + 1)
two-point form
substitute in for x1, y1, x2 and y2
( x + 1)
simplify fraction
( x + 1)
simplify fraction
x+
x+
1
2
7
2
Distributive Property
add 3 to both sides
slope-intercept form
answer
Thus, we have a line with slope m = 1/2 and y-intercept (0, 7/2), as shown on the graph below:
answer
Linear Equations
Point-Slope Form of a Line – 57
58 – Point-Slope Form of a Line
Linear Equations
Every line can be written in the form
Ax + By + C
=
0
standard form of the equation of a line
where A, B, and C are all real numbers. Every other form of the equation of a line can be derived from the
standard form. If we let A = 0, then we have:
By + C
=
0
y
=
−
let A = 0
solve for y
equation of horizontal line
C
B
If we let B = 0, then we have:
Ax + C
=
0
x
=
−
let B = 0
solve for x
equation of vertical line
C
A
Recall that a vertical line has an undefined slope.
Here we just want to summarize the different equations for lines. Using them is largely a matter of preference,
depending on the situation. If we know one point and the slope, then we usually use the point-slope form. If
we are given the slope and the y-intercept, we use the slope-intercept form. If all we know is two points, then
we use the two-point form, and so on.
Equations of Lines
1.)
General (standard) form:
Ax + By + C
2.)
=
=
=
b
horizontal line
mx + b
slope-intercept form
=
m(x – x1)
point-slope form
Two-point form:
y – y1
Linear Equations
vertical line
Point-slope form:
y – y1
6.)
a
Slope-intercept form:
y
5.)
general form
Horizontal line:
y
4.)
0
Vertical line:
x
3.)
=
=
two-point form
Equations of Lines – 59
60 – Equations of Lines
Linear Equations
Advanced
To solve linear inequalities for the independent variable.
No definitions on this page.
In 1854, George Boole (1815-1864) wrote An Investigation of the Laws of Thought, wherein he applied
mathematics to the study of logic. He reduced logical relationships to statements about equality, inequality,
exclusion and inclusion. These statements were expressed symbolically using a two-digit or binary code. The
algebraic rules which governed logical relationships became known as Boolean algebra. They have since had a
tremendous impact on the fields of electronics and computer science.
Linear Equations → Linear Inequalities
The simplest type of inequality is a linear inequality in a single variable. That is, we have an algebraic
expression with only one variable raised to the first power being compared to some real number. An example
of a linear inequality is:
x–5
<
8
sample linear inequality
Solving a linear inequality is very similar to solving a linear equation. The only thing we have to remember is
that when we multiply or divide both sides of an inequality by a negative number, we have to reverse the
inequality symbol.
Solve the inequality and sketch the solution on the real number line.
5x – 3
<
2x + 9
solution
In order to solve this inequality, we try to rearrange the terms so that the variable is on one side and everything
else is on the other side, just as we do when solving linear equations.
5x – 3
<
2x + 9
original inequality
5x – 2x
<
3+9
add 3 to both sides
subtract 2x from both sides
3x
<
12
simplify both sides
x
<
4
divide both sides by 3
we divide by a positive number, so the inequality symbol remains the same
answer
Linear Equations
Linear Inequalities – 61
The graph of the solution looks like the following.
answer
Note:
We use an open circle at x = 4 to indicate that x = 4 is not a part of the solution!
Double Inequalities
Sometimes it is convenient to express two inequalities as a single expression, combining them into a double
inequality. For instance, we can write 2 ≤ 3x + 5 and 3x + 5 < 13 as the following:
2
≤
3x + 5
<
13
To solve such an inequality, we work on all three parts simultaneously. Whatever we do to one part of the
inequality, we have to do to all three parts.
Solve the inequality and sketch its graph on the real number line.
1
≤
3x – 5
<
13
original double inequality
6
≤
3x
<
18
add 5 to all three parts
2
≤
x
<
6
divide all sides by 3
answer
So x is greater than or equal to 2 and x is less than 6. We indicate this on the number line below:
answer
We could also have solved this problem as two inequalities:
1
≤
3x – 5
6
≤
2
≤
and
3x – 5
<
13
solve as two inequalities
3x
3x
<
18
add 5 to both sides of each inequality
x
x
<
6
divide both sides of each inequality by 3
We still obtain the same solution set. The solution consists of all values of x that satisfy both inequalities. That
is, the solution set is 2 ≤ x < 6, which is exactly what we obtained in the example above.
Be very careful when combining inequalities, as the inequalities have to satisfy the Transitive Property of
Inequalities in order to be combined. That is, if a < b and b < c, then a < b < c. For instance, it is not correct to
combine the inequalities 4 < 3x and 3x < –2 into 4 < 3x < –2 because 4 is not less than –2.
62 – Linear Inequalities
Linear Equations
Advanced
To examine inequalities involving the absolute value operator.
No definitions on this page.
George Boole (1815-1864) married Mary Everest (Boole) in 1855. She was the niece of Sir George Everest, the
man for whom the tallest mountain in the world is named. Boole died of pneumonia at the age of 49.
Apparently, he had walked two miles through a drenching downpour and caught pneumonia. His wife believed
that the cure for the illness should resemble the cause, so she put him to bed and repeatedly doused him with
buckets of cold water. Needless to say, this did not help Boole in any way and he died shortly thereafter.
Linear Equations → Linear Inequalities → Absolute Value Inequalities
Inequalities that involve the absolute value operator "| |" have to be treated carefully. Recall that absolute
value is defined as the distance from the origin. The absolute value of a real number is always a positive
quantity.
When we deal with inequalities, we are actually concerned with finding a range of values for the independent
variable x. An absolute value inequality, therefore, involves looking at values of x that are either within a
certain distance of the origin (indicated by the < or ≤ symbols) or are located outside a certain distance from the
origin (indicated by the > or ≥ symbols). All absolute value inequalities can be broken down into either a
conjunction (use of the word "and") or a disjunction (use of the word "or").
For instance,
|x|
≥
5
sample absolute value inequality
actually means all values of x that are more than 5 units away from the origin. Since this can include positive or
negative values of x, we can break this inequality down into the following two inequalities:
x
≥
5
or
x
≤
–5
We can summarize these principles with the following guidelines:
Solving an Absolute Value Inequality
Let x be a variable of an algebraic expression and let a be a real number such that a ≥ 0.
1.)
The solutions of | x | < a are all values of x that lie between –a and a.
|x|<a
2.)
–a < x < a
conjunction
The solutions of | x | > a are all values of x that are less than –a or greater than a.
|x|>a
Linear Equations
if and only if
if and only if
x < –a or x > a
disjunction
Absolute Value Inequalities – 63
Solve the inequality and sketch the solution on the real number line.
| 4x |
>
12
solution
Since we have the greater than symbol (>), we have to apply the second guideline given above. We rewrite this
as two inequalities linked by the term "or":
4x
<
–12
or
4x
>
12
rewrite inequality as disjunction
x
<
–3
or
x
>
3
divide both sides by 4
answer
To graph the solution set, we plot all those points on the number line that are greater than 3 or less than –3, as
shown below. Notice that we use open circles at x = –3 and x = 3 to indicate that they are not part of the
solution set:
answer
64 – Absolute Value Inequalities
Linear Equations
Advanced
To find the solution set of a polynomial inequality.
No definitions on this page.
Georg Cantor (1845-1918), a Russian-born German algebraist and analyst, was instrumental in the development
of set theory. His first paper dealt with algebraic numbers. An algebraic number is any real number that is a
solution to an equation with integer coefficients. He proved there is a one-to-one correspondence between the
set of all algebraic numbers and the set of all positive integers. Furthermore, the set of all reals cannot be put
into the same one-to-one correspondence. As he explained it, the set of positive integers and the set of
algebraic numbers have the same power, but the power of the set of all reals is different from either.
Linear Equations → Linear Inequalities → Polynomial Inequalities
In the previous thought, we only concerned ourselves with solving linear inequalities. Now we are going to
take it up a notch and solve inequalities involving higher degree polynomials.
Suppose we have the following polynomial inequality:
x2 + 2x – 3
<
0
sample polynomial inequality
A polynomial can only change its sign at its zeros (points where the graph of the polynomial crosses the x-axis).
Between two consecutive zeros, a polynomial has to be entirely positive (above the x-axis) or entirely negative
(below the x-axis). When we arrange the zeros of a polynomial in order from least to greatest, we are in effect
dividing the number line into manageable intervals. Inside each interval, the polynomial has no sign changes.
The zeros for the polynomial are called the critical numbers of the inequality. The resulting intervals are called
the test intervals for the inequality. For instance, the polynomial
x2 + 2x – 3
=
(x + 3)(x – 1)
factor the polynomial
has two zeros, located at x = –3 and x = 1. These zeros divide the number line into three test intervals:
(–∞, –3)
first test interval
(–3, 1)
second test interval
(1, ∞)
third test interval
Once we have determined our test intervals, we choose a value for x from each interval and plug that value into
our given inequality. If the inequality holds true for the value of x we choose, we can conclude that the interval
that contains that value of x is a valid solution set for the inequality. If the inequality does not hold true for the
value of x we choose, then we conclude that the interval that contains that values of x is not a valid solution set
for the inequality. This same approach can be used for any polynomial inequality, including rational or linear
inequalities.
Linear Equations
Polynomial Inequalities – 65
Finding Test Intervals for a Polynomial
To determine the intervals on which the values of a polynomial are entirely negative or entirely
positive, use the following steps.
1.)
Find all real zeros of the polynomial, and arrange the zeros in increasing order (from smallest
to largest). The zeros are the critical numbers of the polynomial.
2.)
Use the critical numbers of the polynomial to determine its test intervals.
3.)
Choose one representative value of x in each test interval and evaluate the polynomial at that
value. If the value of the polynomial is negative, then the polynomial will have negative
values for every x-value in the test interval. If the value of the polynomial is positive, then the
polynomial will have positive values for every x-value in the test interval.
Solve the inequality and graph the solution on the real number line.
x2 – 3
≥
1
solution
First, we have to rearrange the terms of this inequality so that we have "0" on one side and everything else on
the other side. Then we can proceed with finding the zeros of the polynomial using standard techniques.
x2 – 3
≥
1
given inequality
x2 – 4
≥
0
subtract 1 from both sides
Now we factor the left hand side to find the zeros:
x2 – 4
=
(x – 2)(x + 2)
factor the polynomial expression
This polynomial expression only equals zero when x = –2 or x = 2. Thus, the polynomial's test intervals are:
(– , –2]
Note:
first test interval
[–2, 2]
second test interval
[2,
third test interval
)
If the degree of a polynomial is given by n, then the number of test intervals we obtain will be given by
n + 1.
Inside each test interval, we choose representative values of x and plug them into our polynomial:
Interval
(–∞, –2]
x-value
x = –4
Polynomial Value
(–4)2 – 4 = 16 – 4 = 12 ≥ 0
Conclusion
positive
[–2, 2]
x=0
0 – 4 = –4 ≤ 0
negative
[2, ∞)
x=4
42 – 4 = 16 – 4 = 12 ≥ 0
positive
66 – Polynomial Inequalities
Linear Equations
We want values of x such that the polynomial inequality
x2 – 4
≥
0
holds true. This only happens when x is inside the intervals (– , –2] and [2, ). Furthermore, because we use
the symbol ≥, we include the endpoints of the intervals. Therefore, we have the following solution sets for the
inequality:
x
≤
–2
or
x
≥
2
answer
The solution sets of this inequality can be graphed as follows:
answer
Note:
In this example, we used the symbol ≥, which means we had to include the endpoints in our intervals,
thus we used square brackets "[" and "]" to indicate that –2 and 2 were included in the solution set.
Had we been given the symbol > instead, we would have had to use "(" and ")" around the points –2
and 2. Infinity is always written with parentheses because it is never included as part of the solution
set.
Linear Equations
Polynomial Inequalities – 67
68 – Polynomial Inequalities
Linear Equations
Advanced
To solve inequalities involving rational expressions.
No definitions on this page.
Georg Cantor (1845-1918) held little influence in his home country of Germany, but was widely recognized in the
international mathematical community. His theories of sets led to whole new branches of mathematics, such as
topology, measure theory, and set theory itself. In 1884, Cantor suffered a nervous breakdown and continued to
suffer from many such mental crises over the course of the rest of his life. One of Cantor's more remarkable
creations is the concept of a transfinite number. Transfinite numbers describe the number of objects in a set.
Linear Equations → Linear Inequalities → Polynomial Inequalities → Rational Inequalities
Solving rational inequalities involves many of the principles covered in the previous thought concerning
polynomial inequalities. However, since rational inequalities by definition involve dividing a polynomial by
another polynomial, we have to pay very close attention to the denominator of our rational expression.
As we did for polynomial inequalities, we need to use the concept of critical numbers and test intervals in order
to solve a rational inequality. The values of a rational expression can change sign only at its zeros (the x-values
for which the numerator is zero) and its undefined values (the x-values for which the denominator is zero).
Solve
3x − 5
x−5
>
4
solution
First, we need to get all the terms on one side of the inequality and the other side of the inequality is set equal to
zero. Then we look for the critical numbers for this inequality that are located at points where the numerator is
equal to zero and where the denominator is equal to zero. This establishes the endpoints of our test intervals.
We test one number from each interval and see if the inequality is true. If it is true, then that interval forms a
solution set for the inequality.
Linear Equations
Rational Inequalities – 69
3x − 5
x−5
3x − 5
x−5
−4
3x − 5 − 4( x − 5)
x−5
3x − 5 − 4x + 20
x−5
−x + 15
x−5
>
4
original inequality
>
0
subtract 4 from both sides
>
0
combine left hand side into a single rational expression
>
0
Distributive Property
>
0
simplify numerator
At this point, we need to find our critical numbers. When x = 15, the numerator equals zero, so this is one
critical number. When x = 5, the denominator is 0, so this is another critical number. Thus, our critical
numbers are:
x
=
15
one critical number
x
=
5
another critical number
Our test intervals then become:
first test interval
(–∞, 5)
(5, 15)
second test interval
(15, ∞)
third test interval
We test three points, one from each interval, as shown on the line below. We graph the solution set on the line
below:
Based on our tests, we discover that the following values of x are the only ones that make the given inequality at
the top of the page true.
5
<
x
<
15
answer
All rational inequalities can be solved using this same process, although the details will depend exactly on the
inequality in question.
70 – Rational Inequalities
Linear Equations
In this section of the Algebra Brain, we are going to discuss systems of linear equations and inequalities. A
linear system consists of two or more linear equations containing any number of variables all raised to the first
power.
For instance:
3x – 4y
=
5
2x + 5y
=
–2
and
is a linear system.
There are numerous ways we can solve systems of linear equations.
1.)
Substitution Method – Solve the system of equations for one of the variables, then start backsubstituting into the equations until all the variables have been solved for.
2.)
Graphical Approach – Graph the system of linear equations (assuming that you only have the
variables x and y). Then the intersection of the lines is the solution of the linear system.
3.)
Method of Elimination – Eliminate one or more variables from the system of linear equations, then
solve for the remaining variable. Back-substitute the solution for that variable into the linear equations
and solve for the remaining variables.
4.)
Multivariable Linear Systems – A method for solving systems of linear equations involving three or
more variables. This method is closely related to the method of elimination. Another name for this
technique is Gaussian elimination.
5.)
Systems of Inequalities – Graph inequalities. The solution to the system of inequalities consists of
those regions in the plane that satisfy the inequalities.
Linear Equations
Linear Systems – 71
72 – Linear Systems
Linear Equations
Advanced
To solve a system of linear equations using the Method of Elimination.
No definitions on this page.
Among Regiomontanus's (1436-1476) accomplishments is his presentation of the first trigonometrical formula for
the area of a triangle. It appeared in his 1464 work De triangulis omnimedis libri guingre. This same work also
presented innovative notations for the sine and cosine. Another of his works, Tabulae Directionum, or "table of
directions" introduced the tangent function.
Linear Equations → Linear Systems → Substitution Method
In the other sections of the Algebra Brain, we develop methods for solving a single equation involving a single
unknown quantity. Now it is time to learn to solve multiple equations involving multiple unknown quantities.
Many types of problems in science, engineering, and business involve two or more equations containing two or
more variables.
To solve these kinds of problems, we need to find solutions of systems of equations. For the time being, we are
going to only focus on systems of linear equations.
Suppose we have the following system of linear equations:
3x + y
=
9
Equation 1
2x – 3y
=
–5
Equation 2
The solution of this system of linear equations is an ordered pair (x, y) that satisfies both linear equations. In
this case, the solution for this system of linear equations is the ordered pair (2, 3). To check, you can substitute
in x = 2 and y = 3 into each equation and see if each equation holds:
3x + y
=
9
Equation 1
3(2) + 3
=
9
substitute in x = 2, y = 3
6+3
=
9
simplify
9
=
9
both sides are equal
check
2x – 3y
=
–5
Equation 2
2(2) – 3(3)
=
–5
substitute in x = 2, y = 3
4–9
=
–5
simplify
–5
=
–5
both sides are equal
check
Now, how did we know that the solution to this system is (2, 3)? We applied the following set of guidelines for
solving a system of linear equations. We call this technique the method of substitution (or simply Substitution
Method).
Linear Equations
Substitution Method – 73
Substitution Method
Note:
1.)
Solve one of the equations for one variable in terms of the other.
2.)
Substitute the expression found in Step 1 into the other expression to obtain an equation of
one variable.
3.)
Solve the equation obtained in Step 2.
4.)
Back-substitute the solution in Step 3 into the expression obtained in Step 1 to find the value
of the other variable.
5.)
Check that the solution satisfies each of the original equations.
The term back-substitution implies that we work backwards. First, we solve for one of the variables,
and then we substitute that value back into one of our original equations in the system to find the value
of the other variable.
Solve the system of linear equations using the Substitution Method.
2x – y + 2 = 0
4x + y – 5 = 0
solution
Our first step is to solve one of the equations for one of the variables. We will solve the first equation for y.
2x – y + 2
=
0
Equation 1
2x + 2
=
y
add y to both sides
Now that we have an expression for y, we can substitute this expression into the second equation:
4x + y – 5
=
0
Equation 2
4x + (2x + 2) – 5
=
0
substitute in y = 2x + 2
6x – 3
=
0
combine like terms
6x
=
3
add 3 to both sides
x
=
x
=
3
6
1
2
divide both sides by 6
reduce fraction to lowest terms
Now that we have a value for x, we can back-substitute this value into either one of our original equations. We
will substitute in x = 1/2 into our first equation and solve the equation for y:
74 – Substitution Method
Linear Equations
2x – y + 2
2⎛⎜
=
0
Equation 1
=
0
substitute in x = 1/2
1–y+2
=
0
simplify
3
=
y
add y to both sides
1⎞
⎟− y+2
⎝ 2⎠
So now we have found that y = 3. Therefore, the solution to this system of linear equations is the ordered pair:
(x, y)
=
⎛ 1 , 3⎞
⎜2 ⎟
⎝
⎠
answer
Now we need to double-check to make sure that this is indeed the correct solution. We substitute in
(x, y) = (1/2, 3) into both equations and see if they hold true:
2x – y + 2
2⎛⎜
=
0
Equation 1
=
0
substitute in x = 1/2 and y = 3
1–3+2
=
0
simplify
0
=
0
simplify
check
4x + y – 5
=
0
Equation 2
=
0
substitute in x = 1/2 and y = 3
2+3–5
=
0
simplify
0
=
0
simplify
check
1⎞
⎟ −3+ 2
⎝ 2⎠
4⎛⎜
1⎞
⎟ + 3−5
⎝ 2⎠
Note:
When solving systems of linear equations it is extremely important to double-check your solution.
Arithmetic errors are extremely common.
Linear Equations
Substitution Method – 75
76 – Substitution Method
Linear Equations
Advanced
To solve systems of linear equations graphically.
No definitions on this page.
Subrahmanyan Chandrasekhar (1910-1995), also known as Chandra, was a brilliant astrophysicist and
mathematician. His first love was mathematics but he became a physicist at his father's insistence. He is most
famous for his work on the origins, dynamics, and structures of stars. In 1983, he was awarded the Nobel Prize
for his research into the deaths of aged stars.
Linear Equations → Linear Systems → Graphical Approach
On occasion, solving a system of equations can be very difficult using straight algebraic techniques. It may be
that the equations just do not have anything in common for us to work with. For instance, suppose we have the
system
y
=
1 + ex
y
=
x2
and
At first glance, this looks to be an unfriendly system of equations. In order to solve this system, we have to set
the equations equal to each other and then solve for x. This is not at all easy to do with this particular system of
equations. However, if we graph the equations, then we can determine the solutions to this system by looking
for places where the graphs of the equations intersect, as shown below:
As we can see from the graph, the given system of equations only has 1 solution located between x = –1 and
x = –2.
We can easily apply this to systems of linear equations as well. If we have a system of linear equations, then
the solution to the equation is the point on the graph where the lines cross.
If the graphs of a system of equation have no intersection, then the system of equation has no solution.
Linear Equations
Graphical Approach – 77
Solve the system of equations graphically.
4x – 5y
=
17
2x + 2y
=
4
and
solution
To solve this problem graphically, we have to use a handy graphing tool. You can use a calculator or a software
program to draw the graph. As always, we use MathCAD 2000®.
Most graphing tools will only graph linear equations in slope-intercept form, so we first need to transform the
equations so that they are in the form y = mx + b.
y
=
y
=
4
5
x−
17
5
–x + 2
Equation 1 in slope-intercept form
Equation 2 in slope-intercept form
Then the graph of these two lines is given by the following:
Based on the graphs of these two lines, we can see that they intersect at the point (3, –1). Therefore, the
solution to this linear system is:
(x, y)
=
(3, –1)
78 – Graphical Approach
answer
Linear Equations
Advanced
To identify the number of solutions of a linear system by looking at the graph of the system.
consistent – a system of linear equations that has at least one solution.
inconsistent – a system of linear equations that has no solution.
Subrahmanyan Chandrasekhar (1910-1995) did important research into white dwarf stars. He concluded that
the radius of a white dwarf is directly proportional to its mass. Furthermore, he theorized that the white dwarf
stage is only one possible final stage of a star's evolution. The primary factor in determining a star's final stage is
its mass. Stars larger than 1.2 times the mass of our sun will most likely become neutron stars. These are only
9 miles in diameter but are more massive than our own sun. Supermassive stars, many times the mass of our
sun, sometimes end up as black holes. Stars with a mass less than or equal to 1.2 solar masses become white
dwarfs, roughly the size of Earth but with all the mass of the sun (or a little bit more).
Linear Equations → Linear Systems → Graphical Approach → Graphical Interpretation of Solutions
In general, for any given system of equations (linear or otherwise) there are three possibilities for the solutions
to the system. There can be only one solution, or there can be two or more solutions, or there can be no
solution.
If we have a system of equations in which all the equations are linear, then if the system has two different
solutions, then it has infinitely many solutions. This stems from the fact that two lines can only intersect at
more than one point if they are the exact same line! That is, the two lines are exactly identical. For instance,
the system of equations below produces only a single line on the same graph:
2x + 2y
=
4
Equation 1
x+y
=
2
Equation 2
Equation 1 is simply a multiple of Equation 2. When we solve Equation 1 to put it in slope-intercept form, we
will find out that it has the same slope as Equation 2 as well as the same y-intercept. Therefore, Equation 1 is
identical to Equation 2 and the solution for this system consists of all real numbers:
2x + 2y
=
4
Equation 1
2y
=
–2x + 4
subtract 2x from both sides
y
=
–x + 2
divide both sides by 2
Equation 1 in slope-intercept form
x+y
=
2
Equation 2
y
=
–x + 2
subtract x from both sides
Equation 2 in slope-intercept form
Linear Equations
Graphical Interpretation of Solutions – 79
Graphical Interpretation of Solutions
For a system of two linear equations in two variables,
the number of solutions is given by one of the following:
1.)
Number of Solutions
Exactly one solution
Graphical Interpretation
The two lines intersect at one point
2.)
Infinitely many solutions
The two lines are identical
3.)
No solution
The two lines are parallel
Each of the three cases is illustrated below:
Exactly ONE solution
Infinitely MANY solutions
NO solution
Consistent
Two lines that intersect
One point of intersection
Consistent
Two lines that coincide
Infinitely many points of intersection
Inconsistent
Two parallel lines
No point of intersection
A system of linear equations is called consistent if it has at least one solution (remember that if it has two, then
it also has infinitely many others). A linear system is called inconsistent if it has no solution.
Use a graphing utility to graph the lines in the system. Determine whether the solution is consistent or
inconsistent. If it is consistent, determine the number of solutions.
2x + y
=
5
Equation 1
x – 2y
=
–1
Equation 2
solution
To graph the equations, we will first need to put them into slope-intercept form. This may also give us a clue as
to whether the solution is consistent or not.
2x + y
=
5
Equation 1
y
=
–2x + 5
subtract 2x from both sides
slope-intercept form
x – 2y
=
–1
Equation 2
–2y
=
–x – 1
subtract x from both sides
y
=
1
2
x+
1
2
divide both sides by –2
slope-intercept form
80 – Graphical Interpretation of Solutions
Linear Equations
Ah ha! We have a clue! Notice that the slopes of the lines are negative reciprocals of each other. That right
there tells us that these two lines are perpendicular. Two lines that are perpendicular can only intersect at one
point. Therefore, we know even before we graph the lines that the solution will be consistent. Let's go ahead
and verify this with the graph:
As you can see from the graph, our prediction based on the slopes was correct. The solution is consistent.
Furthermore, because the lines only intersect at one point, there can be only one solution to this system of linear
equations. answer
To actually find the solution to the system of equations, we would have to use one of the methods described in
this section of the Algebra Brain, such as the Substitution Method or the Method of Elimination.
Linear Equations
Graphical Interpretation of Solutions – 81
82 – Graphical Interpretation of Solutions
Linear Equations
Advanced
To solve systems of linear equations by eliminating one or more variables.
No definitions on this page.
Chandrasekhar's (1910-1995) theories of stellar evolution were ridiculed at first by another prominent
astronomer, Sir Arthur Stanley Eddington. Eddington had greater status in the scientific community, so he
managed to suppress Chandra\s theories about neutron stars or the more massive black holes. It took 20 years
before Chandra's theories resurfaced and gained wide acceptance. It was 50 years before Chandra was
awarded the Nobel Prize for his white dwarf research. Oddly, despite the scientific disputes between Chandra
and Eddington, they retained a close personal relationship.
Linear Equations → Linear Systems → Method of Elimination
When we solve systems of linear equations, we can use a variety of methods to find the solution. Which
method we use is largely a matter of preference, although some methods are more efficient than others.
If we have a system of two linear equations with only two unknowns, then we can use the Method of
Elimination to solve the system. This method involves "eliminating" one of the variables in the system by
adding one equation to a multiple of the other equation so that one of the variables cancels out. Consider the
following simple linear system of equations.
x+y
=
2
x–y
=
0
and
If we add the two equations together, we get:
x+y
=
2
Equation 1
x–y
=
0
Equation 2
2x
=
2
add Equation 1 to Equation 2
x
=
1
divide both sides by 2 to solve for x
Notice that we eliminated the variable y and obtained a single equation in x. This principle even works if we
have "messy" linear equations involving coefficients of x and y. We add the equations in such a way as to get
rid of one of the variables. Once we have found one of the variables, we can back-substitute into either one of
the original equations to find the other variable:
x+y
=
2
Equation 1
1+y
=
2
substitute in x = 1
y
=
1
subtract 1 from both sides
Thus, the solution to the linear system given above is:
(x, y)
=
(1, 1)
solution to the linear system
In general, we can apply the following guidelines to help us solve systems of linear equations using the Method
of Elimination:
Linear Equations
Method of Elimination – 83
Method of Elimination
To use the Method of Elimination to solve a system of linear equations in x and y, use the following
steps:
1.)
Obtain coefficients for x (or y) that differ only in sign by multiplying all terms of one or both
equations by suitably chosen constants.
2.)
Add the equations to eliminate one variable and solve the resulting equation.
3.)
Back-substitute the value obtained in Step 2 into either of the original equations and solve for
the other variable.
4.)
Check your solution in both of the original equations.
Solve the following system of linear equations by the Method of Elimination.
2x – y
=
3
Equation 1
4x + 3y
=
21
Equation 2
solution
According to Step 1, above, we first need to multiply one of the equations by some constant so that the
coefficients of one variable are the same in both equations. Let's multiply Equation 1 by 3 and see what
happens:
3(2x – y)
=
3(3)
multiply both sides of Equation 1 by 3
6x – 3y
=
9
simplify
Notice that we have obtained an equivalent equation to Equation 1. Furthermore, the coefficient of y is the
same magnitude as the coefficient of y in Equation 2, but has a different sign. Therefore, when we add the
equivalent equation to Equation 2, we obtain:
6x – 3y = 9
equivalent to Equation 1
4x + 3y = 21 Equation 2
10x
= 30 add the two equations
= 3
x
divide both sides by 10 to solve for x
Now that we have solved the system for x, we can use this value in either of the original linear equations to
solve them for y. We will use Equation 1 because it is slightly easier to solve for y:
2x – y
=
3
Equation 1
2(3) – y
=
3
substitute in x = 3
6–y
=
3
simplify
3
=
y
add y to both sides
subtract 3 from both sides
84 – Method of Elimination
Linear Equations
Once we have solved one of the equations for y, we can go ahead and conclude that the solution is:
(x, y)
=
(3, 3)
answer
We are not quite done, though. Many students feel compelled to stop when they have found the solution.
However, we also need to verify our solution. Arithmetic errors can lead to all kinds of wrong answers, so we
have to check our solution in both of our original equations.
2x – y
=
3
Equation 1
2(3) – 3
=
3
substitute in x = 3, y = 3
3
=
3
check
4x + 3y
=
21
Equation 2
4(3) + 3(3)
=
21
substitute in x = 3, y = 3
21
=
21
check
The key to properly solving systems of linear equations using the Method of Elimination is to choose the right
constant to multiply one of the equations by. As long as you end up with the coefficients of one variable having
the same magnitude and different signs, you should have no problem. Also remember that when we multiply
the equation by a constant, we have to multiply both sides of the equation by the constant, including each and
every term on both sides. Finally, when we add equations, we add like terms. That is, we add the terms
involving x together, then we add the terms involving y-together, and finally we add the constant terms together.
Linear Equations
Method of Elimination – 85
86 – Method of Elimination
Linear Equations
Advanced
To solve systems of linear equations that involve several variables.
row-echelon form (linear system) – a linear system in a stair-step form with the leading coefficients of each
line being 1.
Regiomontanus (1436-1476), a German trigonometrist and astronomer, also practiced astrology, observing the
close ties between celestial objects and events previously linked to superstition. He predicted that sailors would
someday use the moon for navigation, which did occur during the Age of Exploration after his death. He brought
both plane and spherical trigonometry into the realm of pure mathematics and out of its apprenticeship to
astronomy. Regiomontanus is actually his pen name. It is Latin for his hometown of Königsberg, which
translates into English as "King's Mountain".
Linear Equations → Linear Systems → Multivariable Linear Systems
In the other thoughts in this section of the Algebra Brain, we deal strictly with only two equations involving at
most two unknowns (x and y). The processes that we use to solve those equations can be extended to linear
equations involving three or more variables.
The Method of Elimination is extremely useful for solving multivariable linear systems. Ideally, we use
transformations to turn one of the equations of the system into an equation involving a single variable, which
we can easily solve for. Using that nugget of information, we proceed to back-substitute that value into one of
the other equations so that we can solve for another variable. We continue this process of back-substitution
until we have found solutions for each and every variable. The total solution to the system consists of all real
numbers that satisfy all of the linear equations in the system.
Use back-substitution to solve the following system of linear equations:
3x + 4y – 2z = 13 Equation 1
–x
+ 2z = –3 Equation 2
z = –1 Equation 3
solution
Since we are given that z = –1, we can back-substitute this value into Equation 2:
–x + 2z
=
–3
Equation 2
–x + 2(–1)
=
–3
substitute in z = –1
–x – 2
=
–3
simplify
–x
=
–1
add 2 to both sides
x
=
1
divide both sides by –1
Linear Equations
Multivariable Linear Systems – 87
Now that we know x, we can substitute x = 1 and z = –1 to in Equation 1 and then solve the equation for y:
3x + 4y – 2z
=
13
Equation 1
3(1) + 4y – 2(–1)
=
13
substitute in x = 1 and z = –1
3 + 4y + 2
=
13
simplify
4y
=
8
subtract 5 from both sides
y
=
2
divide both sides by 4
We now have values for x, y and z, so the solution is the ordered triple:
(x, y, z)
=
(1, 2, –1)
answer
We leave it up to you to verify that this is the correct solution for this linear system.
In the example above, we illustrated the power of back-substitution. Unfortunately, very few linear systems are
so neat and tidy. The good news is that we can transform linear systems into a form that will allow us to use
back-substitution to solve the system.
Gaussian Elimination
One method we can use to solve multivariable linear systems is called Gaussian elimination, named after Carl
Friedrich Gauss (1777–1855). Two systems are said to be equivalent if the same solution satisfies both
systems. In order to solve a system of linear equations, we convert the system into an equivalent system that is
in row-echelon form. This refers to the fact that we end up with a "stair-step" pattern in our new linear system.
The leading coefficient of each equation in the new system is 1. In order to produce equations that are
equivalent to our original system of equations, we apply the following operations on our system:
Operations that Produce Equivalent Systems
Each of the following row operations on a system of linear equations
produces an equivalent system of linear equations.
1.)
Interchange two rows
2.)
Multiply one of the equations by a non-zero constant.
3.)
Add a multiple of one equation to another equation
to replace the latter equations
Let's see how these operations can help us solve a system of linear equations. Strap yourself in and feel the g's!
Solve the system of linear equations and check the solution algebraically.
2x –
y + 4z = –8
Equation 1
x + 2y – 3z = 11
Equation 2
4x + 2y +
z = 6
Equation 3
solution
Allll righty, then. Let's get started. The first thing we notice is that the leading coefficient of Equation 2 is 1.
Remember that row-echelon form has the leading coefficient of each equation being 1. Therefore, we will
move Equation 2 up to the top of the stack. That is, we will interchange Equation 1 with Equation 2:
88 – Multivariable Linear Systems
Linear Equations
x + 2y – 3z = 11
2x –
y + 4z = –8
4x + 2y +
interchange Equation 1 with Equation 2
z = 6
Our next steps will involve eliminating the x-variables in the second and third rows. From now on, we will
refer to our operations using the notation Rn, where R represents that we are operating on a row of the linear
system and n indicates which row is being manipulated. Our first step above, then, would be represented by
"interchange R1 with R2". Now when we refer to Row 1, we will be referring to the current top row of the linear
system. We write the operation we are performing next to the row that is being changed.
Back to our problem. To eliminate the x-variable in Row 2, we subtract 2 times Row 1 from Row 2 (we write
this as R2 – 2R1). Note that we multiply all the terms in Row 1 by the constant 2:
x +
2y –
3z = 11
–5y + 10z = –30
4x +
2y +
R2 – 2R1 (subtract 2 times Row 1 from Row 2)
z = 6
To eliminate the x-variable in Row 3, we subtract 4 times Row 1 from Row 3 and place the result in Row 3:
x + 2y –
3z = 11
– 5y + 10z = –30
– 6y + 13z = –38
R3 – 4R1 (subtract 4 times Row 1 from Row 3)
At this point we notice that we have completely eliminated the variable x from both the second and third rows
of the linear system. Now we can work on one of the other variables. Let's start getting rid of the y's in the
third row. First, though, we need to make sure that the leading coefficient of the second row is 1. To do that,
we multiply Row 2 by –1/5 (or divide Row 2 by –5, however you prefer to look at it).
x + 2y –
3z = 11
y –
(–1/5)R2 (multiply Row 2 by –1/5)
2z = 6
– 6y + 13z = –38
Next, we are going to add 6 times Row 2 to Row 3 to get rid of the variable y in Row 3:
x + 2y – 3z = 11
y – 2z = 6
z = –2
R3 + 6R2 (add 6 times Row 2 to Row 3)
Let's take a moment to see what we have done so far. Right now, the leading coefficient of each equation is 1.
Also, the equations are arranged in a "stair step" pattern. Therefore, we conclude that we indeed have
transformed our original linear system into row-echelon form. Furthermore, we can see that we have solved the
system for one of the variables, namely z = –2. Thus, we can start back-substituting and work our way up
through the equivalent linear system. We let z = –2 in Row 2 and solve that equation:
y – 2z
=
6
Row 2
y – 2(–2)
=
6
substitute in z = –2
y+4
=
6
simplify
y
=
2
subtract 4 from both sides
Linear Equations
Multivariable Linear Systems – 89
We now know two of the three variables in our linear system: z = –2 and y = 2. We can back-substitute again
by putting these values in for y and z into Row 1 and solving for x:
x + 2y – 3z
=
11
Row 1 (originally this was Equation 2)
x + 2(2) – 3(–2)
=
11
substitute in y = 2 and z = –2
x+4+6
=
11
simplify
x + 10
=
11
simplify
x
=
1
subtract 10 from both sides
We have just found the last piece of the puzzle. Our solution can be written as the ordered triple:
(x, y, z)
=
(1, 2, –2)
answer
Of course, we will never be absolutely sure that this is the correct solution unless we can verify that it works for
all three original equations:
2x – y + 4z
=
–8
Equation 1
2(1) – 2 + 4(–2)
=
–8
substitute in x = 1, y = 2, and z = –2
–8
=
–8
simplify
check
x + 2y – 3z
=
11
Equation 2
1 + 2(2) – 3(–2)
=
11
substitute in x = 1, y = 2, and z = –2
11
=
11
simplify
check
4x + 2y + z
=
6
Equation 3
4(1) + 2(2) + (–2)
=
6
substitute in x = 1, y = 2, and z = –2
6
=
6
simplify
check
Since the solution fits all three linear equations, we can conclude that this is indeed the correct solution.
answer
Note:
It is actually possible to check our answer graphically if we move from a two-dimensional plane to
three-dimensional space. The solution (1, 2, –2) is represented by the point of intersection of three
lines in space.
Although we have shown a system where we have a unique solution, not all linear systems are so fortunate as to
have one. If, for some reason, we are clicking merrily along and come to a point where we end up with an
equation that doesn't make sense. For instance:
0
=
–2
inconsistency
doesn't make sense because this statement is clearly FALSE. Therefore, any system that results in the statement
just given must have no solution.
On the other hand, if a linear system does not have one solution or no solutions, then it must, by definition, have
infinitely many solutions. If we have a system of linear equations where one equation is just a constant multiple
of another equation (exactly a constant multiple), then we have a system where we have fewer equations than
we have unknowns. In that case, we have infinitely many solutions.
90 – Multivariable Linear Systems
Linear Equations
In general, a linear system has exactly ONE solution if and only if we have the same number of equations as we
have variables. This kind of linear system is called square.
One final thought: As we are sure you realize, systems of linear equations are very tricky beasts to handle. One
of the most effective methods for solving systems of linear equations involves Matrices (described elsewhere in
the Algebra Brain). When we plug the coefficients of a linear system into a matrix, it becomes much, much
easier to find the solutions performing exactly the same sort of operations as we perform above.
Linear Equations
Multivariable Linear Systems – 91
92 – Multivariable Linear Systems
Linear Equations
Advanced
To solve systems of linear inequalities.
No definitions on this page.
In 1474, Regiomontanus (1436-1476) printed his Ephermides, an almanac of daily planetary rotations and
eclipses for the years 1475-1506. Christopher Columbus used this work as a reference to impress the Jamaican
natives with awe-inspiring magical powers. One of the celestial objects Regiomontanus researched was the
comet of 1472, which would be renamed Halley's Comet 210 years later upon its third return.
Linear Equations → Linear Systems → Systems of Inequalities
In the rest of the Algebra Brain, when we refer to inequalities, we usually refer to inequalities involving a single
variable. Now we are going to expand our discussion to inequalities involving two variables. Furthermore, we
are going to talk about systems of inequalities. This is very similar to systems of equations.
Consider the system of inequalities:
–x + y
≤
1
Inequality 1
2x + 3y
>
13
Inequality 2
If we were talking about equations, then the solution to this system would be all values (a, b) such that when a
and b are substituted into each part of the system, the entire system is satisfied. However, these are inequalities.
This means that the solution set of the system will be a range of values that have to satisfy both of the
inequalities.
To sketch the graph of a system of inequalities, we first sketch the graph of the corresponding equations. That
is, we replace the inequality symbol with "=" and draw those equations. If we replace "<" or ">" with "=", then
we draw a dotted line for the equation. The solution set then consists of all points that fall inside a region in the
plane. For instance, in the example above, we would want all points that are below the line x + y = 5 and above
the line 2x + 3y = 10, as shown in the shaded region on the graph below:
Linear Equations
Systems of Inequalities – 93
Let's test the two points given in the graph above and see what happens when we plug them into the system of
inequalities. First, we will use the point (–2, 3):
–x + y
≤
1
Inequality 1
–(–2) + 3
≤
1
substitute in x = –2 and y = 3
2+3
≤
1
simplify
5
≤
1
FALSE
2x + 3y
>
13
Inequality 2
2(–2) + 3(3)
>
13
substitute in x = –2 and y = 3
–4 + 9
>
13
simplify
5
>
13
FALSE
Both statements are false, so we conclude that the point (–2, 3) cannot be a solution to the system of equation.
In order for a given point to be a solution, both statements have to be TRUE. Let's test the other point, (4, 3):
–x + y
≤
1
Inequality 1
–4 + 3
≤
1
substitute in x = 4 and y = 3
–1
≤
1
TRUE
2x + 3y
>
13
Inequality 2
2(4) + 3(3)
>
13
substitute in x = 4 and y = 3
8+9
>
13
simplify
17
>
13
TRUE
Both statements are true, so we conclude that all points inside this region are indeed solutions to the system of
inequalities.
Note:
In this system, the point of intersection of the two lines (2, 3) is not a solution to the system of
inequalities because we cannot use points that are on the line 2x + 3y = 13.
The lines given by the system of inequalities divide the plane into several regions. Inside each region of the
plane, exactly one of the following must be true:
1.)
All points in the region are solutions of the inequality.
2.)
No point in the region is a solution of the inequality.
Therefore, we can test each region of the plane by simply trying a point inside that region. If a single point in a
region satisfies the inequality, then all points in that region must also satisfy the inequality.
Sketching the Graph of an Inequality in Two Variables
1.)
Replace the inequality sign by an equal sign, and sketch the graph of the resulting equation.
Use a dashed line for < or > and a solid line for ≤ or ≥.
2.)
Test one point in each of the regions formed by the graph in Step 1. If the point satisfies the
inequality, shade the entire region to denote that every point in the region satisfies the
inequality.
94 – Systems of Inequalities
Linear Equations
Sketch the graph of the solution to the system of inequalities.
–x + 3y
≤
12
Inequality 1
2x + y
>
7
Inequality 2
solution
First, we will replace the inequality symbol in each inequality by "=" so that we can graph the corresponding
lines:
–x + 3y
≤
12
Inequality 1
3y
≤
x + 12
add x to both sides
y
≤
y
=
2x + y
>
7
Inequality 2
y
>
–2x + 7
subtract 2x from both sides
y
=
–2x + 7
replace > with = to find the corresponding equation
x
3
x
3
+4
divide both sides by 3
+4
replace
with = to find the corresponding equation
We graph the corresponding equations as shown below. However, we use a dashed line to graph the second
corresponding equation because we replaced > with =. This means that points that fall on the line y = –2x + 7
are not part of the solution. Once we have graphed both lines, we shade the points in the plane that satisfy both
inequalities. That is, we want points that satisfy y ≤ x/3 + 4 and points that satisfy y > –2x + 7. The shaded
region shown below gives us these points.
answer
Linear Equations
Systems of Inequalities – 95
96 – Systems of Inequalities
Linear Equations
Unit 3
Functions
Functions – 97
98 – Functions
Advanced
To learn the basic definition of a function.
function – an association of exactly one object from one set (the range) with each object from another set (the
domain). A relationship in which different ordered pairs have different first coordinates.
domain (of a function) – the set of values which are allowable substitutions for the independent variables.
range – the set of values of the function evaluated at points contained within the domain.
The function notation f(x) was first introduced by Leonhard Euler in 1734 in Commentarii Academiae Scientiarum
Petropolitanae. Other mathematicians later adapted the same sort of notation to represent other commonly
occurring types of functions in mathematics, such as gamma functions, beta functions, and Riemann Zeta
functions.
Functions
What is a function?
According to the mathematical definition of a function:
Definition of Function
A function f from a set A to a set B is a rule of correspondence that assigns to each element x in
the set A exactly one element y in the set B. The set A is the domain (or set of inputs) of the
function f, and the set B contains the range (or set of outputs) of the function f.
What does this mean, exactly?
A function establishes a relationship. That means that we want a connection between two numbers. One
number is going to be dependent on the other number.
An element of the domain is paired with an element of the range. This is the connection that we want to
establish. But what makes up the domain and what makes up the range? The domain determines what numbers
will be in the range. Furthermore, each element inside the domain will produce one and only one element
inside the range.
The typical example is with two groups of numbers, one designated as the domain and the other designated as
the range:
function
Functions
Introduction to Functions – 99
Note that we have "mapped" two elements of the domain onto the same element on the range. This is perfectly
legal. However, it is not permitted to map the same element of the domain onto two elements of the range and
still call it a function. In graphic terms:
not a function
Let's look at how this works from a more practical standpoint. Imagine that One-Eyed Squirrel Industries has a
stock that sells for $34 a share in March. In April, the stock has raised to $52 a share. In May, the stock market
takes a major hit and One-Eyed Squirrel Industries stock plummets to a mere $6 a share. However, in June, it
picks itself back up again and climbs up to its original value of $34 a share.
Now let's put this scenario in terms of domains and ranges. In this case, the amount the stock is worth is
dependent on the month of the year. So that means that the month is the domain and the worth of the stock is
the range:
We can plot this information on a graph:
One-Eyed Squirrel Industry Stock Prices
60
52
50
price
40
30
34
34
Series1
20
10
0
March
6
April
May
June
month
If we were just given the graph and not the domain and range that we plotted above, how could we tell if this
graph was a function or not? There is a really simple way to tell. If we can draw a vertical line through any
100 – Introduction to Functions
Functions
point of the graph and it only intersects our graph at exactly one point, then this is a function. This is known as
the Vertical Line Test:
Vertical Line Test
If a vertical line intersects a graph at more than one point, then the graph is not the graph of a function.
For most practical applications, a function takes in a number, manipulates it, and spits another number back out
the other end. For example, if we plug the numbers 2, 3, 4, and 5 into the function f(x) = 3x + 2, we will get the
numbers 8, 11, 14, 17 back out:
f(x)
=
3x + 2
example of a function
f(2)
=
=
3(2) + 2
8
substitute in 2 for x
simplify
f(3)
=
=
3(3) + 2
11
substitute in 3 for x
simplify
f(4)
=
=
3(4) + 2
14
substitute in 4 for x
simplify
f(5)
=
=
3(5) + 2
17
substitute in 5 for x
simplify
We could easily plug in any number for x, and get a different number as a result.
Functions
Introduction to Functions – 101
102 – Introduction to Functions
Functions
Advanced
To review the basic terminology associated with functions.
dependent variable – a variable whose value(s) always depend on the values of other variable(s).
independent variable – in a formula, a variable upon whose value other variables depend.
Leonhard Euler (1707-1783) was a Swiss geometer and number theorist. Among his countless contributions to
mathematics is the notation f(x) for a function of x. At the age of 19, he entered a contest involving the best
place on a ship to set the masts. He only won honorable mention for his efforts, not bad considering he had
never seen a ship before. The greatest accomplishments of his career occurred during the last quarter of his life,
when he was totally blind. Euler's final achievement was the calculation of the orbit of the newly discovered
planet Uranus.
Functions → Function Terminology
There is a lot of terminology associated with functions, so we have come up with a handy reference list of the
common terms associated with functions.
Function Terminology
Function:
A function is a relationship between two variables such that to each value of the
independent variable there corresponds exactly one value of the dependent variable.
Another way to put it is:
Given two sets, M and N, let f denote a rule that assigns each element in M one and
only one element in N.
Function Notation:
y = f(x), read as "y equals f of x".
f is the name of the function
y is the dependent variable
x is the independent variable
f(x) is the value of the function at x
Domain:
The domain of a function is the set of all values (inputs) of the independent variable
for which the function is defined. If x is in the domain of f, we say f is defined at x.
If x is not in the domain of f, we say that f is undefined at x.
Range:
The range of a function is the set of all values (outputs) assumed by the dependent
variable (that is, the set of all function values).
Implied Domain: If f is defined by an algebraic expression and domain is not specified, the implied
domain consists of all real numbers for which the expression is defined.
Functions
Function Terminology – 103
104 – Function Terminology
Functions
Advanced
To review the definition of polynomial functions and why they are useful.
constant function – a function in which f(x) is simply equal to a real (or complex) number.
linear function – a polynomial function of the first degree.
quadratic function – a polynomial function of the second degree.
Niels Henrik Abel (1802-1829), a Norwegian algebraist, proved that fifth and higher order equations have no
algebraic solution. He also begun a general proof of the binomial theorem, which at the time had only been
proven for specific cases. He died on April 6, 1829, from tuberculosis.
Functions → Polynomial Functions
Simply put, a polynomial function is any function that is composed of the sum or difference of monomials all
set equal to some value determined by the value of x in the polynomial.
A formal definition is as follows:
Definition of a Polynomial Function
Let n be a nonnegative integer and let an, an-1, . . . , a2, a1, a0 be real numbers with an ≠ 0.
The function given by:
f(x)
anxn + an-1xn-1 + . . . + a2x2 + a1x + a0
=
polynomial function of degree n
is called a polynomial function of x with degree n.
Polynomial functions are classed by degree. For instance, the polynomial given by:
f(x)
=
a
a≠0
is called a constant function and has degree 0, because n = 0. The graph of this type of function is a horizontal
line.
The polynomial
f(x)
=
ax + b
a≠0
is called a linear function and has degree 1. The graph of any function of this sort is a line whose slope is
given by a and whose y-intercept is given by (0, b).
Polynomial functions of degree 2 are also called quadratic functions. They take the form:
f(x)
=
ax2 + bx + c
a≠0
The word "quadratic" comes from the Latin word for "square", as in "x-squared". Quadratic functions are so
important there is an entire section of the Algebra Brain devoted to nothing but quadratics.
Polynomial functions of higher degree do not really have special names, although we sometimes refer to thirddegree polynomials (f(x) = ax3 + bx2 + cx + d) as "cubic" functions.
Functions
Introduction to Polynomial Functions – 105
In general, we use polynomial functions to model different relationships. For example, we might use a
quadratic equation to model how much a company spends on advertising versus how much it makes in profit.
We might want to model cigarette consumption in the United States to see if putting a warning label on
cigarettes is actually doing any good. If we are into marketing, we might want to model how many people
bought our product over a course of time.
The advantage of modeling data is that we can use the model to make a prediction of the future. Modeling is
used extensively in manufacturing to test products before they are even built. It is far cheaper in the long run to
test a rocket's trajectory on a computer than it is to actually build one and see if it goes where we want it to go.
This is where polynomial functions come in.
106 – Introduction to Polynomial Functions
Functions
Advanced
To learn how to determine just how many solutions a given polynomial has.
multiplicity of a root r – for a root r of a polynomial equation P(x) = 0, the highest power of x – r that appears as
a factor of P(x).
repeated roots (or zeros) – a solution to a polynomial equation that occurs more than once.
The Indian astronomer Ǻryabhata had the idea in 510 C.E. to use the thirty-three letters of the Indian alphabet to
represent all the numbers between 1 and 1018. This sort of notation is what is known as a 'numeral alphabet'.
Functions → Polynomial Functions → Fundamental Theorem - Algebra
One of the most elementary operations performed in algebra is finding solutions to polynomial equations. Most
of the time, it is convenient if we can tell at a glance how many solutions we will find when we actually try and
solve it. The brilliant mathematician Karl Gauss (1777 – 1855) demonstrated a theorem that greatly simplifies
the process of finding zeros or roots. We call his result the Fundamental Theorem of Algebra. We present it
without proof because in order to prove it, we would need to resort to complex variables, which is far beyond
the scope of this current level of instruction.
The Fundamental Theorem of Algebra
If f(x) is a polynomial of degree n, where n > 0,
then f(x) has at least one zero in the complex number system.
As a consequence of the Fundamental Theorem of Algebra, any polynomial of degree n > 0 can be written as
the product of n factors. This property is called the Linear Factorization Theorem and is written formally as:
Linear Factorization Theorem
If f(x) is a polynomial of degree n, where n > 0,
f(x) = anxn + an – 1xn – 1 + . . . + a1x + a0
then f(x) has precisely n linear factors
f(x) = an(x – c1)(x – c2) . . . (x – cn)
where c1, c2, . . . , cn are complex numbers and an is the leading coefficient of f(x).
Each of the linear factors of f(x) corresponds to each zero of f(x).
This theorem can be proved using the Fundamental Theorem of Algebra to conclude that f must have at least
one zero, c1. Therefore, (x – c1) is a factor of f(x) and we can rewrite f(x) as
f(x)
=
(x – c1)f1(x)
where f1(x) is formed from the remaining factors of f(x)
If the degree of f1(x) is greater than 0, then we can apply the Fundamental Theorem of Algebra again to
conclude that f1(x) must have a zero, which we will call c2. Thus, we can rewrite f(x) as:
f(x)
Functions
=
(x – c1)(x – c2)f2(x)
Fundamental Theorem of Algebra – 107
By this time, we should start to see a pattern emerging. The degree of f1(x) is n – 1, where n is the degree of our
original polynomial f(x). Then for each successive function, we subtract 1 from the degree of the preceding
function. Thus the degree of f2(x) is n – 2, the degree of f3(x) is n – 3 and so on. By repeatedly applying the
Fundamental Theorem of Algebra n times, we obtain:
f(x)
=
an(x – c1)(x – c2) . . . (x – cn)
where an is the coefficient of the polynomial f(x)
Note:
Neither of these two theorems tells us how to find the solutions of a polynomial equation nor do they
tell us how many solutions are real and how many are complex. Instead they simply tell us that such
solutions exist. Thus, if we are solving a cubic equation, then we better end up with three solutions
(real, complex, or some combination of real and complex), or else!
1.
f(x) = x – 4 has exactly 1 zero located at x = 4.
2.
f(x) = x2 – 8x + 16 = (x – 4)(x – 4) has exactly 2 zeros, both located at x = 4. These are known as
repeated zeros (or double zeros if only two are the same).
Example 2 illustrates the multiplicity of a root r. In this case, we say that 4 is a root with multiplicity
2 (because it occurs twice). In general, the multiplicity of a root r is the highest power of (x – r) that
appears as a factor of the polynomial.
3.
f(x) = x3 + 9x = x(x2 + 9) = x(x + 3 i)(x – 3 i) has exactly 3 zeros located at x = 0, x = –3 i and x = 3 i.
Note: 2 of these zeros are complex.
4.
f(x) = x4 – 1 = (x2 – 1)(x2 + 1) = (x – 1)(x + 1)(x – i)(x + i) has exactly 4 zeros located at x = 1, x = –1,
x = i and x = –i.
Note: 2 of these zeros are complex.
The French mathematician Jean Le Rond d'Alembert (1717 – 1783) worked independently of Karl Gauss to
prove the Fundamental Theorem of Algebra. As a result of his efforts, in France, the Fundamental Theorem of
Algebra is often called the Theory of d'Alembert (not to be confused with any other theories of d'Alembert that
you may have heard of).
108 – Fundamental Theorem of Algebra
Functions
We deal with four fundamental operations in algebra:
1.
2.
3.
4.
Addition
Subtraction
Multiplication
Division
We can apply these same operations to any combination of polynomial expressions. In the thoughts below, we
will discuss how to add and subtract two polynomials, how to multiply two polynomials together, and two
distinct methods of dividing polynomials.
In general, adding or subtracting polynomials is very similar to adding and subtracting real numbers.
Multiplying two polynomials requires one to fully understand how the Distributive Property works. Dividing
two polynomials is either a matter of long division, as we do with real numbers, or synthetic division for special
cases of polynomial division. Synthetic division is a handy short cut that eliminates all the messiness associated
with long division.
Functions
Polynomial Operations – 109
110 – Polynomial Operations
Functions
Advanced
To simplify, add and subtract polynomials.
like terms – terms that have exactly the same variables to the same powers.
Isaac Newton (1643-1727) is one of the most famous mathematicians of all time. He invented the calculus,
which relies on comparing the rates of change between two or more quantities. With calculus, we can analyze
the physics of moving objects using pure mathematics. Newton also did important work in astronomy and optics.
His telescope so impressed the Royal Society in London that they invited him to join. After tiring of the academic
world, Isaac Newton became Master of the Royal Mint in 1700. He became wealthy from this position because
he received a commission on the coins minted.
Functions → Polynomial Functions → Polynomial Operations → Adding / Subtracting Polynomials
Just as we can add, subtract, multiply and divide real numbers, we can perform the same operations on
polynomial expressions.
In order to add two polynomial expressions, we have to see if there are any like terms , or terms having the
same variables to the same powers. For example, 3x2 and 2x2 are like terms, but 3x2 and 2x3 are not, because
one term has x risen to the second power and the other term has x risen to the third power. Once we see that we
have like terms in our expression then we can simply add the coefficients of like terms together. Thus, if we
have 3x2 and 2x2, which we have already stated as being like terms, then their sum is:
3x2 + 2x2
=
(3 + 2)x2
add coefficients of like terms
=
5x2
simplify
=
(3 – 2)x2
subtract coefficients of like terms
=
1x2
simplify
And their difference is:
3x2 – 2x2
Simplify the following expression
(5x2 + 2x) – (x2 + 3x – 6)
solution
First, remove the parentheses, distributing the negative sign in front of the second set of parentheses. This
makes it easier to rearrange the terms so all the like terms are together.
Then add the coefficients of like terms.
Functions
Adding / Subtracting Polynomials – 111
(5x2 + 2x) – (x2 + 3x – 6)
5x2 + 2x – x2 – 3x + 6
remove parentheses (easier to simplify)
5x2 – x2 + 2x – 3x + 6
group like terms
(5 – 1)x2 + (2 – 3)x + 6
4x2 – x + 6
Note:
given expression
add coefficients of like terms to simplify
answer
When simplifying polynomial expressions, it is generally convenient to list the x-terms in order of
decreasing exponents. Thus when grouping the terms together to simplify, place all the terms of the
highest-power of x first, then group the terms of the next highest power of x and so on down to the
lowest power of x, x0 = 1.
A common mistake when distributing the negative sign inside the parentheses is to fail to change the
sign of each term inside the parentheses. For instance,
–(x2 + 3x – 6 )
=
–x2 – 3x + 6
–(x2 + 3x – 6)
≠
–x2 + 3x – 6
112 – Adding / Subtracting Polynomials
and
Functions
Advanced
To multiply two or more polynomials together.
No definitions on this page.
Claudius Ptolemy (c.70-c.130 A.D.) was a famous Greek astronomer, geometer, and geographer. Ptolemy's
great work, the Almagest, provides a comprehensive overview of Greek geometry and trigonometry. Ptolemy's
geocentric (Earth-centered) view of the universe was so compelling that it remained in force for over 1400 years,
until Copernicus set forth his heliocentric (sun-centered) view of the solar system. Ptolemy the geographer
attempted to map the Known World at that time, but since his knowledge of geography was effectively limited to
the Roman Empire, his maps were very inaccurate. He did, however, introduce the first systematic use of
longitudes and latitudes. His directions for creating maps were still being used by cartographers as late as the
Renaissance.
Functions → Polynomial Functions → Polynomial Operations → Multiplying Polynomials
The Distributive Property is used to multiply two polynomials together. We treat one of the polynomials as a
single quantity and multiply that single quantity by both terms in the second polynomial. Essentially, we apply
the Distributive Property twice. For example, if we want to multiply (3x + 7) by (4x – 1), then we have the
following:
(3x + 7)(4x – 1)
=
3x(4x – 1) + 7(4x – 1)
let (4x – 1) be a single quantity
and apply the Distributive Property
=
(3x)(4x) – (3x)(1) + (7)(4x) – (7)(1)
apply the Distributive Property again
=
12x2 – 3x + 28x – 7
multiply terms
=
12x2 + 25x – 7
add like terms to simplify
If we take a closer look at what we did, we will see a pattern emerge. We multiplied the First terms of the
polynomials together, followed by the Outside terms of the polynomials, followed by the Inside terms, and
followed finally by the Last terms. We call this the FOIL method of multiplying two binomials together. Note
that the outer (O) and inner (I) terms are like terms and can thus be combined into one term.
FOIL Method
First terms:
(3x)(4x)
Outside terms:
(3x)(–1)
Inside terms:
(7)(4x)
Last terms:
(7)(–1)
This method works great for simple binomials, but what happens when we try to multiply two trinomials
together? It turns out that we perform a similar series of operations. In effect, we need to multiply each term of
the first trinomial by each term of the second trinomial.
Functions
Multiplying Polynomials – 113
Extended Distributive Property
To multiply two polynomials multiply each term in the first polynomial
by each term in the second polynomial.
Multiply (x2 – 4x + 6) by (5x2 + 2x – 3).
solution
(x2 – 4x + 6)(5x2 + 2x – 3)
Note:
=
x2(5x2 + 2x – 3) – 4x(5x2 + 2x – 3) +
6(5x2 + 2x – 3)
multiply each term
of the first trinomial by
the second trinomial
=
(5x4 + 2x3 – 3x2) – (20x3 + 8x2 – 12x) +
(30x2 + 12x – 18)
apply Distributive Property
=
5x4 + 2x3 – 3x2 – 20x3 – 8x2 + 12x +
30x2 + 12x – 18
remove parentheses
=
5x4 – 18x3 + 19x2 + 24x – 18
combine like terms together
answer
When multiplying out polynomials, it is a good idea to not skip any steps. By taking care to multiply
each and every term together, it is far less likely that you will make any sign errors or other mistakes.
We can easily (so to speak) extend the multiplication of trinomials to polynomials with even more terms. The
trick is to be careful and take your time to avoid any inadvertent errors. The most common errors in
multiplying any two polynomials are sign errors.
In general, if one polynomial has m terms and the other polynomial has n terms, then the total number of terms
obtained by multiplying the two polynomials together will be m · n terms. If we multiplied this result by a
polynomial with q terms, then we would obtain a polynomial that contains m · n · q terms.
114 – Multiplying Polynomials
Functions
Advanced
To recognize patterns when multiplying polynomials.
No definitions on this page.
Christiaan Huygens (1629–1695) was a Dutch astronomer, mathematician, and physicist. He was a Cartesian,
which meant he followed the ideas of René Descartes rather than the ideas of Isaac Newton. He is generally
known as for two important contributions to science. The first is a pendulum clock inspired by the work of
Galileo. The principles behind his clock are still used today in grandfather and cuckoo clocks. Huygens is also
famous for his wave theory of light, which went against Newton's theories of optics. As it turns out, light is both a
wave and a particle, but neither Huygens nor Newton was aware of this fact at the time. Finally, Huygens
discovered the rings of Saturn, disproving Galileo's idea that Saturn was actually three planets. He was the first
to notice markings on the surface of Mars and discovered Saturn's largest moon, Titan.
Functions → Polynomial Functions → Special Product Patterns
Oftentimes in algebra, we have to multiply two (or more) binomials together. Obviously, we can always resort
to the FOIL method for multiplication, but that can become tedious very quickly. In addition, it is easy to make
mistakes, particularly sign errors. However, if we can instead recognize patterns amongst binomials, then we
can simply apply one of the following formulas. This requires you to actually memorize the formulas.
Sum and Difference
(u + v)(u – v)
=
u2 – v2
=
x2 + 4x – 4x – 16
FOIL method
=
x2 – 16
simplify
answer
Multiply.
(x – 4)(x + 4)
solution
(x – 4)(x + 4)
Square of a Binomial
Functions
(u+ v)2
=
u2 + 2uv + v2
(u – v)2
=
u2 – 2uv + v2
Special Product Patterns – 115
Multiply.
(5x + 7)2
solution
(5x + 7)2
=
25x2 + 2(5x)(7) + 49
apply Square of Binomial formula
=
25x2 + 70x + 49
simplify
answer
Cube of a Binomial
(u + v)3
=
u3 + 3u2v + 3uv2 + v3
(u – v)3
=
u3 – 3u2v + 3uv2 – v3
Multiply.
(2x – 4)3
solution
(2x – 4)3
=
(2x)3 – 3(2x)2(4) + 3(2x)(4)2 – (4)3
apply Cube of Binomial formula
=
8x3 – 12(4x2) + (6x)(16) – 64
simplify
=
8x3 – 48x2 + 96x – 64
simplify
answer
116 – Special Product Patterns
Functions
Advanced
To recognize patterns when factoring polynomials.
No definitions on this page.
Friedrich Wilhelm Bessel (1784-1846) was a self-taught mathematical and astronomical genius. He was the first
to accurately calculate the distances between stars. He is also the first to use the term 'light year', which refers
to the distance light travels in a year, roughly 6 trillion miles (give or take a few million miles). In 1840, he
discovered odd movements in Uranus's orbit. He theorized that there was another celestial body exerting
gravitational influence on Uranus. Unfortunately for him, Neptune, the cause of the disturbances, was not
discovered until after Bessel had already died.
Functions → Polynomial Functions → Special Factoring Patterns
There are a number of useful formulas to help in remembering how to factor. These formulas make it easy to
factor a polynomial if we can recognize its pattern. Each is also fairly straightforward to prove and follows
from the Distributive Property of Multiplication.
Perfect Square Trinomial
u2 + 2uv + v2
=
(u + v)2
u2 – 2uv + v2
=
(u – v)2
Proof:
(u + v)2
(u – v)2
=
(u + v)(u + v)
rewrite
=
u2 + uv + uv + v2
use FOIL method
=
u2 + 2uv + v2
simplify
=
(u – v)(u – v)
rewrite
=
u2 – uv – uv + v2
use FOIL method
=
u2 – 2uv + v2
simplify
proof
Factor.
25 – 10x + x2
solution
25 – 10x + x2
Functions
=
52 – 2(5)(x) + x2
rewrite
=
(5 – x)2
answer
Special Factoring Patterns – 117
Difference of Two Squares
u2 – v2
=
(u + v)(u – v)
Proof:
(u + v)(u – v)
=
u2 – vu + vu – v2
use FOIL method
=
u2 – v2
simplify
proof
Factor.
16 – 9x2
solution
16 – 9x2
=
(4)2 – (3x)2
rewrite
=
(4 – 3x)(4 + 3x)
answer
Sum of Two Cubes
u3 + v3
(u + v)(u2 – uv + v2)
=
Proof:
(u + v)(u2 – uv + v2)
=
u3 – u2v + uv2 + u2v – uv2 + v3
multiply out all terms
=
u3 + v3
simplify
proof
Factor.
8x3 + 27
solution
8x3 + 27
=
(2x)3 + (3)3
rewrite
=
(2x + 3)((2x)2 – (2x)(3) + 32)
apply Sum of Two Cubes formula
=
(2x + 3)(4x2 – 6x + 9)
simplify
answer
118 – Special Factoring Patterns
Functions
Difference of Two Cubes
u3 – v3
=
(u – v)(u2 + uv + v2)
Proof:
(u – v)(u2 + uv + v2)
=
u3 + u2v + uv2 – u2v – uv2 – v3
multiply out all terms
=
u3 – v3
simplify
proof
Factor.
64 – 27x3
solution
(64 – 27x3)
=
(4)3 – (3x)3
rewrite
=
(4 – 3x)(42 + (4)(3x) + (3x)2)
apply Difference of Two Cubes formula
=
(4 – 3x)(16 + 12x + 9x2)
simplify
answer
Factoring by Grouping
au3 + acu2 + bu + bc
=
au2(u + c) + b(u + c)
=
(au2 + b)(u + c)
where a and c are constants
Proof:
(au2 + b)(u + c)
=
au3 + acu2 + bu + bc
use FOIL method
proof
Factor.
5x3 – 10x2 + 3x – 6
solution
5x3 – 10x2 + 3x – 6
Functions
=
5x3 – (2)(5)x2 + 3x – (2)(3)
factor coefficients
=
5x2(x + (–2)) + 3(x + (–2))
a = 5, b = 3, c = –2
=
(5x2 + 3)(x – 2)
Distributive Property in reverse
answer
Special Factoring Patterns – 119
120 – Special Factoring Patterns
Functions
Advanced
1.) Divide a polynomial by a monomial or binomial factor.
2.) Use polynomial long division.
No definitions on this page.
Multiplication and division in Ancient Egypt only required one to memorize one\'s two-times tables. They used an
ingenious method of doubling and addition to multiply and divide numbers.
Functions → Polynomial Functions → Polynomial Operations → Polynomial Division
Dividing numbers is a tricky business, and dividing polynomials can be even more so. When we divide two
numbers, either one of two things can happen: the numbers divide cleanly, or we obtain a remainder. Consider
the example of 9 divided by 3. Everybody knows that the answer is 3, so that tells us that (3)(3) = 9. Now what
if we divided 9 by 4? Then we know that 4 goes into 9 evenly 2 times, 2 x 4 only equals 8, so we end up with a
remainder of 1. That is, 9 divided by 4 equals 2 R 1. This tells us that 4 and 2 are not factors of 9.
In more familiar terms, we can write this as:
2
2
9
–8
1
R
1
We can attribute the same behavior to division of polynomials. If we divide the polynomial f(x) by another
polynomial d(x) and obtain a remainder of zero, then we can conclude that the divisor d(x) and the quotient q(x)
are both factors of the original polynomial f(x).
In the same vein, if f(x) does not divide by d(x) evenly, then we will be left with a remainder function r(x),
which, when added to the product of the divisor d(x) and the quotient q(x), will total our original function f(x).
Using the same familiar model above, we can write this as:
d(x)
q(x)
f(x)
R
r(x)
This leads us to the following mathematical process of division formally known as the Division Algorithm and
is valid for ANY function f(x) (including when f(x) is simply a real number such as the aforementioned 9).
Functions
Polynomial Division – 121
The Division Algorithm
If f(x) and d(x) are polynomials such that
d(x)
≠
0
and the degree of d(x) is less than or equal to the degree of f(x),
there exist unique polynomials q(x) and r(x) such that
f(x) = d(x)q(x) + r(x)
f(x):
d(x):
q(x):
r(x):
dividend
divisor
quotient
remainder
where r(x) = 0 or the degree of r(x) is less than the degree of d(x).
If the remainder r(x) is zero, d(x) divides evenly into f(x) so d(x)
and q(x) are both factors of f(x) in that instance.
As usual when presenting an algorithm or formula, it is often helpful to see the theorem in action, so let's do a
quick example of how this might work in practice:
Divide x3 – 1 by x – 1.
solution
Because there is no x2-term or x-term in the dividend, you need to line up the subtraction by using zero
coefficients (or leaving spaces) for the missing terms.
divisor
x2
x – 1 x + 0x2
– (x3 – x2)
x2
– (x2
3
+ x + 1 quotient
+ 0x – 1 dividend
subtract x3 – x2 from dividend
+ 0x
bring down the 0x term from the divisor
– x)
subtract x2 – x
x – 1 bring down 1 from dividend
– (x – 1) subtract x – 1
0 remainder
Thus, x – 1 divides evenly into x3 – 1 and you can write
3
x −1
x−1
=
x2 + x + 1
x≠1
answer
If we want to check our answer, then we multiply x2 + x + 1 by x – 1 and see if we obtain x3 – 1:
(x2 + x + 1)(x – 1)
Note:
=
x3 + x2 + x – x2 – x – 1
multiply together using Distributive Properties
=
x3 + (x2 – x2) + (x – x ) – 1
regroup like terms
=
x3 – 1
cancel like terms
check
In this example, we basically do the same thing that we do anytime we divide two numbers. We first
note that the term (x – 1) goes into the term (x3) x2 times, just as 2 goes into 9 four times. Multiplying
x2 by (x – 1) gives us x3 – x2, which we subtract from f(x) = x3 + 0x2 + 0x – 1. We continue this process
until there we either end up with 0 or a factor which x – 1 is unable to go into evenly.
122 – Polynomial Division
Functions
Advanced
To simplify rational expressions using a special form of polynomial division.
No definitions on this page.
Thales of Miletus (c.640-546 B.C.) was the first Greek mathematician and philosopher who introduced the basic
concepts of geometry to the Greeks. He founded his school in Miletus, teaching mathematics and astronomy. In
his younger days, he used his almost mystical ability to foresee the future to corner the olive oil market. Thales
foresaw a good season for olives one year and bought every local olive press he could find. With the wealth he
gained from this venture, he was able to finance his school. Thales was so honored and revered for his wisdom
and knowledge that he became known as one of the Seven Wise Men of Greece.
Functions → Polynomial Functions → Polynomial Operations → Polynomial Division →
Synthetic Division
Dividing two polynomials using long division can be tedious. It is also easy to make mistakes. Therefore, we
usually like to try and find a shortcut method for dividing polynomials. One such shortcut can be used when the
divisor (the denominator) takes the form x – k. We call this shortcut synthetic division. This is not so much a
formula as a technique for solving certain types of problems. This is an important distinction. You will need to
be able to apply the technique, not just plug the problem into a specific formula.
We give the details of synthetic division for a cubic polynomial. To apply this to higher-degree polynomials,
we simply continue the pattern established below.
Synthetic Division for a Cubic Polynomial
To divide ax3 + bx2 + cx + d by x – k, use the following pattern:
Vertical pattern: Add terms
Diagonal pattern: Multiply by k
The resulting expression gives us:
=
Functions
Synthetic Division – 123
Notes:
Synthetic division only works for divisors of the form x – k.
[remember that x + k = x – (–k)].
You cannot use synthetic division to divide a polynomial by a quadratic such as x2 + 5.
Synthetic division for higher degree polynomials follows a similar pattern as the one given above.
If we have any missing terms, then we use a zero for a placeholder just as we do for regular long
division of polynomials.
Let's look at an example:
Use synthetic division to divide 3x3 – 17x2 + 15x – 25 by x – 5.
solution
Set up an array like the one given below. Apply the pattern for synthetic division given above.
Now we can rewrite our quotient (and remainder) in standard form:
3
2
3x − 17x + 15x − 25
=
x−5
3x2 – 2x + 5
answer
Note that in this example, we have a remainder of 0. To double-check our answer, we can multiply our quotient
3x2 – 2x + 5 by the divisor x – 5 and see if we obtain the dividend:
(3x2 – 2x + 5)(x – 5)
124 – Synthetic Division
=
3x3 – 15x2 – 2x2 + 10x + 5x – 25
multiply two polynomials
using Distributive Property
=
3x3 – 17x2 + 15x – 25
add like terms
check
Functions
Advanced
To find solutions of a polynomial equation using the Factor Theorem.
No definitions on this page.
Marin Mersenne (1588-1648) was a French number theorist and good friend of René Descartes. His chief role in
mathematics was not developing his own theories, but in spreading the information developed by other
mathematicians. It was through Mersenne that both Descartes and Galileo achieved recognition throughout
Europe. He did do his own work and attempted to find a rule which would allow him to calculate large prime
numbers. Unfortunately his ideas were flawed, but his work stimulated other mathematicians to tackle the
problem of finding large prime numbers, which today are called "Mersenne Numbers".
Functions → Polynomial Functions → Polynomial Operations → Polynomial Division →
Synthetic Division → Factor Theorem
The Remainder Theorem tells us that when we divide a polynomial f(x) by a divisor x – k, then the remainder is
equal to that function evaluated at x = k. But what happens when the remainder is 0? As it turns out, a
remainder of 0 indicates that the divisor x – k is a factor of the polynomial f(x). This result is known as the
Factor Theorem:
The Factor Theorem
A polynomial f(x) has a factor (x – k) if and only if f(k) = 0.
This is a simple statement that has a few surprising ramifications, so let's take a moment to see just why the
Factor Theorem is true.
We can use the Division Algorithm with the factor (x – k) to write down f(x), which is a polynomial of degree n,
as:
f(x)
=
(x – k)q(x) + r(x)
Division Algorithm
By the Remainder Theorem, we know that r(x) = r = f(k), which means that we can now write f(x) as:
f(x)
=
(x – k)q(x) + f(k)
Remainder Theorem
where q(x) is a polynomial of degree n – 1. If f(k) = 0, then we have:
f(x)
=
(x – k)q(x)
From this we can tell that (x – k) is a factor of f(x). Conversely, if (x – k) is a factor of f(x), division of f(x)
by (x – k) yields us a remainder of 0. Therefore, by the Remainder Theorem, we have f(k) = 0.
Now that we have demonstrated the principle, it only makes sense to look at it from an applied standpoint.
Functions
Factor Theorem – 125
Use synthetic division to show that x is a solution of the following third-degree polynomial equation. Use the
result to factor the polynomial completely. List all real zeros of the function.
Polynomial Equation
x3 – 7x + 6 = 0
Value of x
x = 2
solution
First, we divide the polynomial x3 – 7x + 6 by x – 2, using synthetic division.
2
1
1
0
–7
6
2
4
–6
2
–3
0
Since we end up with a remainder of 0, we can apply the Factor Theorem to conclude that (x – 2) is a factor of
the polynomial f(x) = x3 – 7x + 6. From the Division Algorithm, we can write f(x) as:
x3 – 7x + 6
=
(x – 2)(x2 + 2x – 3) + 0
Division Algorithm
Now, if we substitute in x = 2, then we end up with one factor becoming zero, which subsequently turns the
whole expression to 0, and therefore we can say that x = 2 is indeed a solution to the polynomial equation:
x3 – 7x + 6
=
0
We are also asked to factor the polynomial completely. We can apply standard factoring techniques to the
quadratic expression (x2 + 2x – 3) to obtain:
x3 – 7x + 6
=
(x – 2)(x – 1)(x + 3)
factor quadratic
answer
Now that we have factored our expression completely, we can see that the roots of the polynomial equation we
were given are:
x
=
2
x
=
1
x
=
–3
126 – Factor Theorem
answer
Functions
Advanced
To evaluate a function for a given value using synthetic division instead of direct substitution.
No definitions on this page.
Pythagoras of Samos (c.580-500 B.C.), after whom the Pythagorean Theorem is named, established
a school in southern Italy to teach philosophy and mathematics. Students at his school lived by very
strict rules. They were conscientious vegetarians, took a vow of silence for the first five years of their
membership, and were forbidden to keep any written records. Pythagoras opened up his school to
both men and women, women even being allowed to teach. Members disdained from taking credit
for individual discoveries, instead attributing them to Pythagoras or to the group.
Functions → Polynomial Functions → Polynomial Operations → Polynomial Division →
Synthetic Division → Remainder Theorem
The remainder obtained from synthetic division has an important interpretation. This is given in the Remainder
Theorem:
The Remainder Theorem
Let f(x) be a polynomial of positive degree n. Then for any number k,
f(x)
=
q(x)(x – k) + f(k)
remainder theorem
where q(x) is a polynomial of degree n – 1.
The Remainder Theorem has the interesting effect of allowing us to evaluate a polynomial function. To
evaluate the polynomial function f(x) when x = k, divide f(x) by x – k. The remainder will be f(k).
To see why this is true, consider the Division Algorithm, which allows us to write f(x) as:
f(x)
=
(x – k)q(x) + r(x)
Division Algorithm
By the Division Algorithm, we know that either r(x) = 0 or the degree of r(x) is less than the degree of x – k.
This second piece of information tells us that r(x) must be a constant. That is, r(x) = r. If we evaluate f(x) at
x = k, we get:
f(k)
f(k)
=
(k – k)q(k) + r
evaluate f(x) at x = k
=
0q(k) + r
simplify
=
r
Remainder Theorem
Let's see an example of how this works.
Functions
Remainder Theorem – 127
Express the function in the form f(x) = (x – k)q(x) + r for the given value of k and demonstrate that f(k) = r.
Function
f(x)
15x4 + 10x3 – 6x2 + 14
=
Value of k
2
k = −
3
solution
In order to get the function in the form f(x) = (x – k)q(x) + r, we need to divide f(x) by x – k using synthetic
division.
−
2
15
3
10
–6
0
–10
0
4
0
–6
4
15
14
−
8
3
Using the bottom row as the coefficients of the terms in q(x), we have
q(x)
15x3 – 6x + 4
=
and
r
34
=
remainder
3
Thus we can write our function f(x) as:
f(x)
=
(15x3 − 6x + 4) +
⎟
3⎠
3
⎛x +
⎜
⎝
2⎞
34
answer
In order to demonstrate that f(k) = r, we simply substitute in x = k into our original function:
f ⎛⎜ −
2⎞
⎟
⎝ 3⎠
⎟
⎝ 3⎠
2
⎛ 2⎞
⎛ 2⎞
⎟ + 10 ⎜ − ⎟ − 6 ⎜ ⎟ + 14
⎝ 3⎠
⎝ 3⎠
⎝ 3⎠
=
15⎛⎜
16 ⎞
simplify
=
15⎛⎜
16 ⎞
convert all fractions to LCD
8
4
+ 10⎛⎜ − ⎞⎟ − 6⎛⎜ ⎞⎟ + 14
⎟
⎝ 81 ⎠
⎝ 27 ⎠ ⎝ 9 ⎠
=
2⎞
3
15 ⎛⎜ −
=
f ⎛⎜ −
2⎞
4
=
=
128 – Remainder Theorem
24
36
81
− 10⎛⎜ ⎞⎟ − 6⎛⎜ ⎞⎟ + 14⎛⎜ ⎞⎟
⎟
⎝ 81 ⎠
⎝ 81 ⎠ ⎝ 81 ⎠
⎝ 81 ⎠
240
81
918
81
34
3
−
240
81
−
216
81
+
1134
81
substitute in x = k
simplify
simplify
reduce to a fraction in lowest terms
answer
Functions
Advanced
To determine the number of real zeros for a given polynomial function.
No definitions on this page.
René Descartes (1590-1650) was more than just a mathematician. In his younger years he served as a soldierof-fortune in various armies throughout Europe, including the Bavarian Army in present-day Germany. It was
during this time, on November 10, 1619, he fell into a fevered delirium and had a mystical experience that
changed his life. Supposedly he experienced three vivid dreams in succession. As a result, he became
convinced it was his divine mission to found a new philosophical system in which everything in the sciences is
connected by a chain of mathematical logic. Furthermore, physics was nothing but geometry. The really strange
bit is that modern physics relies more and more on geometry to explain what goes on at the quantum level.
Functions → Polynomial Functions → Real Zeros
For any polynomial function f of degree n, the following statements are true. Remember that the zeros of a
function are the x-values for which the function equals zero.
1.
The graph of f has, at most, n – 1 turning points. Turning points are points at which the graph changes
from increasing to decreasing and vice-versa.
2.
The function f has, at most, n real zeros. This statement comes from the Fundamental Theorem of
Algebra.
Finding the zeros of a polynomial function is one of the most important applications of algebra. The zeros
represent solutions to polynomial equations. It is important to realize the connection between an algebraic
approach and a graphical approach to a problem. Given a graph, we can often use that graph to help us
determine the location of any zeros of a function. On the other hand, we can use algebraic techniques to
determine the shape of a graph based on the solutions to an equation.
In general, we can make the following statements about the real zeros of a polynomial function.
Real Zeros of Polynomial Functions
If f(x) is a polynomial function and a is a real number,
the following statements are equivalent:
Note:
1.
x = a is a zero of the function f.
2.
x = a is a solution of the polynomial equation f(x) = 0.
3.
(x – a) is a factor of the polynomial f(x).
4.
(a, 0) is an x-intercept of the graph of f.
Finding real zeros of polynomial functions is closely related to factoring and finding x-intercepts.
Let's take a look at an example of how we can use the statements above to find the solutions to a polynomial
equation.
Functions
Real Zeros – 129
Find all the real zeros of the following polynomial function.
f(t)
t3 – 5t2 + 6t
=
solution
First, we can factor out a t from all three of the terms. Then we can factor the remaining quadratic equation.
Then we apply Statement 3 from above to determine the real zeros of the polynomial function.
f(t)
=
t3 – 5t2 + 6t
original function
=
t(t2 – 5t + 6)
factor out common monomial factor t
=
t(t – 2)(t – 3)
factor remaining quadratic
Now that we have reduced f(t) to its factored form, we want values of t which would make f(t) = 0. According
to our factors, t = 0, t = 2 or t = 3. Thus the real zeros are:
t
=
0
t
=
2
t
=
3
answer
Here is a graph of the function plotted with MathCAD 2000:
The corresponding t-intercepts to t = 0, t = 2, and t = 3 are (0, 0), (2, 0), and (3, 0). This figure also has two
turning points. Since this is a third-degree polynomial it can have at most two turning points.
The Intermediate Value Theorem
We can use the values of a function at two distinct points to determine if a function has a zero between those
two points. The Intermediate Value Theorem is an existence theorem in that it tells us such zeros exist, but not
how to go about finding them. If (a, f(a)) and (b, f(b)) are two points on the graph of a polynomial function
such that f(a) f(b), then for any number d between f(a) and f(b) there must be a number c between a and b such
that f(c) = d. When we are looking for zeros of a function, then we want f(c) = 0.
In even more formal terms, this is written as:
130 – Real Zeros
Functions
Intermediate Value Theorem
Let a and b be real numbers such that a < b. If f is a polynomial function such that f(a) ≠ f(b),
then in the interval [a, b], f takes on every value between f(a) and f(b).
We can use this theorem to help us find zeros as follows. If we can find a number x = a such that f(a) > 0, and
another number x = b such that f(b) < 0, then the function must cross the x-axis at least once somewhere
between x = a and x = b. The point(s) where the function crosses the x-axis are the zeros for that function.
Let's see how this works.
Find at least one interval of length 1 for the following function in which the given function is guaranteed to
contain a zero.
f(x)
=
x3 – 3x2 + 3
solution
First, let's start with a few trial numbers. A good range of numbers to start with is positive and negative
numbers near zero. Then we evaluate the function f at those numbers. When f(x) switches from negative to
positive or from positive to negative, we know that we have crossed the x-axis. Therefore, the zero will be
contained in that interval.
x
f(x)
–2
–17
–1
–1
0
3
1
1
2
–1
3
3
Because f(–1) is negative and f(0) is positive, you can conclude from the Intermediate Value Theorem that there
must be a zero somewhere in the interval [–1, 0]. Similarly, because f(1) is positive and f(2) is negative, there is
another zero in the interval [1, 2]. Finally, a third zero is found in the interval [2, 3]. If we look at the graph of
this function, we can see that it will give us the same information:
answer
Functions
Real Zeros – 131
132 – Real Zeros
Functions
Advanced
To find the number of real (and imaginary) roots of a polynomial function.
positive real zeros – positive real x-values for which the function equals zero.
negative real zeros – negative real x-values for which the function equals zero.
variation in sign – two consecutive coefficients in a polynomial function having opposite signs.
In 1649, René Descartes (1590-1650) traveled to Sweden to tutor 20-year old Queen Christina. She preferred to
start her days at 5 am. The combination of extremely cold winters and early rising resulted in Descartes
contracting a fatal case of pneumonia. He died early spring of 1650. In 1666, his remains were exhumed and
returned to France. The French ambassador received permission to remove Descartes's right forefinger from the
corpse. Descartes's skull is said to have been removed by a guard and sold several times. Supposedly,
Descartes's skull is currently on display at the Museé de l'Homme in the Palais de Chaillot in France.
Functions → Polynomial Functions → Real Zeros → Descartes's Rule of Signs
From the previous thought, we know that any nth-degree polynomial can have at most n real zeros. Of
course, many polynomials do not have that many real zeros. For example, f(x) = x2 + 1 has no real zeros and
f(x) = x3 + 1 has only one real zero. However, no polynomial can have more than n real zeros. We can use
the following theorem, called Descartes's Rule of Signs to shed more light on the zeros of a polynomial
function.
Descartes's Rule of Signs
Let f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 be a polynomial with real coefficients and a0 ≠ 0.
1.
The number of positive real zeros of f is either equal to the number of variations in signs of f(x)
or less than that number by an even integer.
This means that for every change of sign (from +an to –an – 1 to + an – 2, etc), you will have a
zero for each change in sign or two. If you do not have a zero for each change in sign, then you
will have two less than the maximum number of zeros. If you do not have that many, then you
will have four less than the maximum, etc… These are all for positive real zeros.
2.
The number of negative real zeros of f is either equal to number of variations of sign of f(–x) or
less than that number by an even integer.
The same rule that applies to the positive real zeros applies to the negative zeros as well.
For example,
If you have 4 changes in sign, then you will have either 4 zeros, 2 zeros, or no zeros. You will never have 3
zeros or 1 zero.
If you have 5 changes in sign, you can have 5 zeros, 3 zeros or 1 zero. You will always have at least one zero
for an odd number of changes in sign. You will never have an even number of zeros if you have an odd number
of changes in sign.
Functions
Descartes's Rule of Signs – 133
Notes: Variation in sign means that two consecutive coefficients have opposite signs.
Where there is only one variation in sign, Descartes’s Rule of Signs guarantees the existence of exactly
one positive (or negative) real root.
Descartes's Rule of Signs does not tell us what the zeros are. It simply states how many zeros we will
find, as well as whether they will be positive or negative. We need to use other methods to actually
find the zeros.
We must note the qualifier that the roots given by Descartes's Rule of Signs are all real. According to the
Fundamental Theorem of Algebra, the degree of the polynomial tells us the total number of roots. Therefore, if
Descartes's Rule of Signs yields us three real roots (positive, negative, or both) for a 5th degree polynomial,
then we must conclude that the remaining 2 roots are imaginary.
Use Descartes's Rule of Signs to determine how many real solutions the following function has as well as if
they are positive or negative.
h(x)
=
3x4 – 4x + 7
given function
solution
Our given function has coefficients that are positive, negative, and positive again. This translates to 2 variations
in sign.
Descartes's Rule of Signs tells us that the number of positive real solutions is equal to the number of changes in
sign or less than that by an even number. Thus, we can have either 2 positive real roots or (2 – 2) = 0 positive
real roots.
In order to determine our negative real roots, we have to replace "x" with "–x" in the function and simplify,
noting any variations in sign again:
h(–x)
=
3(–x)4 – 4(–x) + 7
replace "x" with "–x"
=
3x4 + 4x + 7
simplify
All the coefficients are positive. Thus, we conclude from Descartes's Rule of Signs that there are no negative
real roots, since there are no variations in sign.
Thus we have at most, 2 positive real roots. answer
Since we have just found how many real roots there are, we can use the Fundamental Theorem of Algebra to
determine that since we have a 4th-degree polynomial, we must obtain 4 solutions. We have already stated that
at most 2 of these solutions are real. Therefore, we must conclude that there are at least 2 complex solutions.
(Complex solutions always come in pairs, so we can have either two complex solutions or four complex
solutions, but not three complex solution.)
Use Descartes's Rule of Signs to determine how many real solutions the following function has as well as if
they are positive or negative.
f(x)
=
2x3 + x2 + 3x + 3
given function
solution
In our original function, we can see that the coefficients are all positive. Since this is the case, we can conclude
from Descartes's Rule of Signs that there are no positive real roots.
To find the negative real roots, we substitute "–x" in for "x" in our original function and simplify:
134 – Descartes's Rule of Signs
Functions
f(–x)
=
2(–x)3 + (–x)2 + 3(–x) + 3
replace "x" with "–x"
=
–2x3 + x2 –3x + 3
simplify
Now we have a negative coefficient, followed by a positive coefficient, followed by a negative coefficient,
followed last by a positive constant. There are three variations in sign (– to + to – to +). We can then use
Descartes's Rule of Signs to conclude that there are either 3 negative real roots or (3 – 2) = 1 negative real root.
If we later find out that there is only 1 negative real root, then, by elimination, we must conclude that the
remaining two roots are complex.
Thus we have at most 3 negative real roots. answer
Functions
Descartes's Rule of Signs – 135
136 – Descartes's Rule of Signs
Functions
Advanced
To find rational zeros of polynomial functions.
No definitions on this page.
Although Albert Einstein (1879-1955) is most famous for his theories on relativity, he won his Nobel Prize for his
work on the photoelectric effect. When light rays strike metal, electrons are released from the metal to fly freely
through the air. This is the "photoelectric effect". The number of electrons released is not a function of the
intensity or brightness of the light. Einstein demonstrated that the number of electrons released is a function of
the color of the light. Color is determined by the wavelength of light. A shorter wavelength corresponds to higher
energy. Energy is delivered via packets known as quanta. The more quanta of energy that strike the metal, the
higher the probability that an electron will be released. His paper on the photoelectric effect was published in
1905, but he didn't get the Nobel Prize until 1921.
Functions → Polynomial Functions → Real Zeros → Rational Zero Test
For polynomials with integer coefficients, the zeros of that polynomial may be found by looking at the factors
of the constant term and the factors of the leading coefficient. All possible zeros can be expressed as a ratio of
one factor of the constant term over one factor of the leading coefficient. However, this just gives us the
possible zeros. We have to test each and every zero in the actual polynomial to see if works. We call this
procedure the Rational Zero Test, and explain it below:
The Rational Zero Test
If the polynomial f(x) = anxn + an-1xn–1 + . . . + a2x2 + a1x + a0 has integer coefficients,
every rational zero of f has the form
Rational zero
=
where p and q have no common factors other than 1, and
p
=
a factor of the constant term a0
q
=
a factor of the leading coefficient an
Possible rational zeros are formed by listing all rational numbers whose numerators
are coefficients of the constant term (a0) and whose denominators are factors of the
leading coefficient (an).
Possible rational zeros
=
factors of constant term
factors of leading coefficient
In essence, we divide each of the factors of the constant term by each of the factors of the leading coefficient.
Note that we will need to look at both positive and negative factors. The factors of 6, for example, are ±1, ±2,
±3, ±6 because we can have:
Functions
Rational Zero Test – 137
(1)(6)
=
6
and
(–1)(–6)
=
6
and
(2)(3)
=
6
and
(–2)(–3)
=
6
Dividing the factors will give us a list of possible zeros. Any possible zeros that crop up more than once are
disregarded. For example, if we have 6 as the constant term and 2 as the leading coefficient, we end up with:
possible rational zeros
±1, ±2, ±3, ±6
±1, ±2
=
factors of 6
factors of 2
After dividing each factor of 6 by each factor of 2, we obtain a list of:
±1, ±1/2, ±2, ±2/2, ±3, ±3/2, ±6, ±6/2
We can reduce some of the fractions to get:
±1, ±1/2, ±2, ±1, ±3, ±3/2, ±6, ±3
Since the factors ±1 and ±3 appear more than once, we really only need to list them once:
±1, ±1/2, ±2, ±3, ±3/2, ±6
possible rational zeros of a polynomial with
leading coefficient 2 and constant term 6
Once the list of possible zeros is formed, trial and error methods are needed to determine, which, if any, the
actual zeros of the polynomial are. Note that this does not tell us how many zeros we have. We can use
Descartes Rule of Signs to tell us how many real zeros we have then apply the Rational Zero Test to determine
which of the possible zeros we come up with will work. Once we have found as many zeros (both positive and
negative) as are allowed by the degree of the polynomial, we can stop.
Find the rational zeros of
f(x)
=
x3 + x2 + 1
solution
Here we have a leading coefficient of 1 and a constant of 1. The only possible zeros we can have, from the
Rational Zero Test are ±1. However, when we test this, we will find that neither 1 nor –1 will make our
function equal to zero:
f(1)
f(–1)
=
1 3 + 12 + 1
substitute in x = 1
=
3
simplify
=
(–1)3 + (–1)2 + 1
substitute in x = –1
=
1
simplify
Thus, we can conclude from the Rational Zero Test that this polynomial function does not have any rational
zeros. However, if we graph this function, we can see that it does have a real zero (between –2 and –1):
138 – Rational Zero Test
Functions
answer
Let's look at another example where the leading coefficient is 1 but the constant term is not 1:
Find the rational zeros of:
f(x)
=
x3 – 7x – 6
solution
Here the leading coefficient is 1 and the constant term is 6, so by the Rational Zero Test, we have the following
possible rational zeros:
±6, ±3, ±2, ±1
±1
factors of 6 (constant term)
factors of 1 (leading coefficient)
Which we can simplify to:
±6, ±3, ±2, ±1
possible rational zeros
Note that we have a total of eight possible zeros: –6, –3, –2, –1, 1, 2, 3, 6. We have to check each of them in
our polynomial function and see which ones will make our function equal to zero.
Testing these possible zeros, we find the following three rational solutions:
x
=
–1
x
=
–2
x
=
–3
These are the only three that work. answer
Of course, many polynomials do not have a convenient leading coefficient of 1. When that happens, we
dramatically increase the number of possible zeros. We can use a number of methods to shorten the search.
First, if we happen to have an expensive calculator, we can probably use its built-in polynomial solving
functions. The HP 48G, for example, allows us to simply input the coefficients of the polynomial and the
calculator finds the roots for us. Not all calculators are quite so sophisticated. However, a graphing calculator
can also be of use. By looking at the graph of a polynomial function, in particular those points where it crosses
the x-axis, we can narrow our focus to those possible roots that are likely to be in the region where the graph
crosses the x-axis. Finally, if we do not have a polynomial-solving or graphing calculator, we can resort to
synthetic division.
Synthetic division let's us know if one of our potential zeros is an actual zero if we divide the polynomial by
x – k, where k is one of our possible roots. If we obtain a zero remainder, then we know that we have a root.
As always, we only need to continue until we obtain n roots, where n is the degree of the polynomial.
Let's see an example of a more complicated polynomial function and apply the Rational Zero Test.
Functions
Rational Zero Test – 139
Find all the rational zeros of the following function
f(x)
4x3 – 12x2 – x + 15
=
solution
The leading coefficient is 4 and the constant term is 15, so we have the following possible
zeros:
±1, ±3, ±5, ±15
±1, ±2, ±4
factors of 15 (constant term)
factors of 4 (leading coefficient)
Written out, this gives us:
±1, ±1/2, ±1/4, ±3, ±3/2, ±3/4, ±5, ±5/2, ±5/4, ±15, ±15/2, ±15/4
for a total of 24 possible zeros. This would be extremely tedious to try and plug each one into our function to
see if it works. Let's graph the function and see if we can eliminate some zeros. Since this is a cubic
polynomial, we expect to have at most three solutions, but not all of them need be rational. Here is the graph:
Based on this graph, it looks as though we have a zero located at x = –1. Let's use synthetic division to check it
out.
–1
4
4
–12
–1
15
–4
16
–15
–16
15
0
Since we have a remainder of 0, we can conclude that x = –1 is a rational solution of f(x). Another possible
solution is located at x = 3/2. Again, we use synthetic division to check it out:
3
2
4
4
140 – Rational Zero Test
–12
–1
15
6
–9
–15
–6
–10
0
Functions
Again, we end up with a remainder of 0, so we conclude that x = 3/2 is a rational zero of the function f(x).
Finally, we see that there is a third (and final) possible zero located at x = 5/2. Let's try it and see:
5
4
2
4
–12
–1
15
10
–5
–15
–2
–6
0
Once again, lo and behold, we end up with a remainder of 0! We have found three rational zeros:
x
=
x
=
x
=
–1
3
2
5
2
and
and
answer
Since our polynomial is cubic, we know that we have found all three of the possible solutions.
Functions
Rational Zero Test – 141
142 – Rational Zero Test
Functions
Advanced
To add, subtract, multiply and divide complex numbers.
complex number – any number, real or imaginary, of the form a + bi, where a and b are real numbers and
imaginary number – a number which is the square root of a negative real number.
complex conjugate – the complex conjugate of a + bi is a – bi, and vice versa. The product of complex
conjugates is a real number.
Girolamo Cardano (1501-1576) was more than just an Italian algebraist and physician. He was the illegitimate
son of a noted lawyer, a compulsive gambler, a popular astrologer, a former prisoner of the Inquisition (largely
due to his astrology practice), and the father of a convicted murderer. Despite the many sordid details of his life,
he still made significant contributions to mathematics and medicine. Cardano published over 200 papers on
human medicine, physics, natural science, philosophy, and music. Cardano was one of the first to recognize the
existence of the square root of negative numbers, otherwise known as imaginary numbers, although he did not
quite know how to deal with them.
Functions → Polynomial Functions → Complex Numbers
For certain quadratic equations, we are unable to find any real solutions to the equation. The classic example of
such an equation is:
x2 + 1
=
0
equation with no real solutions
We cannot factor this equation. If we subtract 1 from both sides, we end up with:
x2
=
–1
There is no real number x which we can square to give us –1. However, mathematicians refused to let this stop
them from finding a solution to the equation. Since they were unable to find a real solution, they decided that
there could only be imaginary solutions (sometimes called fictitious solutions). The imaginary solutions
became an entire new system of numbers, the basis of which is the solution to the equation given above. They
defined this new unit, called i, as:
i
=
−1
imaginary unit
Thus, i2 = –1. Any imaginary number is written as a multiple of the imaginary unit above. For example, if we
have the equation:
x2 + 4
=
0
another equation with no real solution
Then the solutions of this equation are:
Functions
Complex Numbers – 143
x2
=
x
=
−4
take square root of both sides
=
4( −1)
factor out a 4
=
4⋅ −1
use a property of radicals to separate
–4
subtract 4 from both sides
=
2 −1
simplify first radical
=
2i
substitute in i = (–1)
We do not have to work with imaginary numbers by themselves. We can combine real numbers with imaginary
numbers. The result is what is known as a complex number. All complex numbers can be written in the
standard form, a + bi.
Definition of Complex Numbers
For real numbers a and b, the number
a + bi
is a complex number written in standard form.
If a = 0 and b ≠ 0, then the complex number bi is an imaginary number.
If b = 0 and a ≠ 0, then the complex number a is a real number.
The above definition leads to an interesting conclusion. Although the set of real numbers and the set of
imaginary numbers are independent of each other, both are subsets of the set of complex numbers. That is, the
set of complex numbers contains all the real numbers AND all the imaginary numbers, as shown in the diagram
below:
Equality of Complex Numbers
Two complex numbers a + bi and c + di, written in standard form, are equal to each other
a + bi
=
c + di
equality of two complex numbers
if and only if a = c and b = d.
Operations with Complex Numbers
Naturally, it is not enough to know what a complex number is. We also have to know how to use them. Adding
and subtracting complex numbers is fairly straightforward. We add (or subtract) the real parts together. We
then add (or subtract) the complex parts together.
144 – Complex Numbers
Functions
Addition and Subtraction of Complex Numbers
If a = bi and c + di are two complex numbers written in standard form,
then their sum and difference are defined as follows:
Sum
(a + bi) + (c + di)
=
(a + c) + (b + d)i
adding complex numbers
Difference
(a + bi) – (c + di)
=
(a – c) + (b – d)i
subtracting complex numbers
Simplify.
a.)
(4 – i) + (7 – 3i)
b.)
(5 + 5i) + 3i
c.)
(3 – 2i) + (4 + 2i)
a.)
(4 – i) + (7 – 3i)
solution
b.)
c.)
Note:
(5 + 5i) + 3i
=
4 – i + 7 – 3i
remove parentheses
=
4 + 7 – i – 3i
group like terms
=
(4 + 7) – (1 + 3) i
put inside parentheses
=
11 – 4i
simplify into standard form
answer
=
5 + 5i + 3i
remove parenetheses
=
5 + (5 + 3)i
factor out an i-term
=
5 + 8i
simplify into standard form
answer
(3 – 2i) + (4 + 2i)
=
3 – 2i + 4 + 2i
remove parentheses
=
3 + 4 – 2i + 2i
group like terms
=
7
simplify
answer
In example (c), we ended up with a real number for the answer. Remember that a real number can be
written in complex form as a + 0b.
Multiplying two complex numbers involves applying the Distributive Property as we would do when
multiplying two polynomials. In the case of complex numbers, we treat i as we would treat any variable, such
as x.
Functions
Complex Numbers – 145
Simplify.
d.)
(i)(4 – 3i)
e.)
(2 – i)(5 – 3i)
f.)
(3 + 2i)(3 – 2i)
d.)
(i)(4 – 3i)
solution
e.)
f.)
Note:
=
4i – 3i2
Distributive Property
=
4i – 3(–1)
remember that i2 = –1
=
3 + 4i
rewrite in standard form
answer
(2 – i)(5 – 3i)
(3 + 2i)(3 – 2i)
=
10 – 6i – 5i + 3i2
FOIL method
=
10 – 11i + 3i2
simplify
=
10 – 11i + 3(–1)
remember that i2 = –1
=
10 – 3 – 11i
group like terms
=
7 – 11i
simplify in standard form
answer
=
9 – 6i + 6i – 4i2
FOIL method
=
9 – 4(–1)
remember that i2 = –1
6i terms cancel each other
=
9+4
simplify
=
13
simplify
answer
Example (f) shows us that we can also obtain a real number by multiplying two complex numbers.
This phenomenon is explained in more detail below.
Complex Conjugates and Division
Whenever we see any type of rational expression or function, we like to try and simplify the denominator
because that makes it a whole lot easier to work with. Complex rational expressions are no exception. In
general, we like to try to convert the denominator to a real number. Example (f) from above showed us how
this is possible. We can obtain a real number from complex numbers by multiplying a complex number by its
complex conjugate. A pair of complex numbers of the form a + bi and a – bi are called complex conjugates.
When we multiply these two numbers together, we end up with:
(a + bi)(a – bi)
=
a2 – abi + abi – b2i2
=
a2 – b2(–1)
remember that i2 = –1
=
a2 + b2
real number result
If we want to find the quotient of a + bi and c + di where c and d are both not equal to zero, then multiply the
numerator and the denominator by the complex conjugate of the denominator. We are essentially multiplying
the top and bottom of the rational expression by a very special form of "1":
146 – Complex Numbers
Functions
a + bi
a + bi ⎛ c − di ⎞
⎜
=
c + di
⎟
c + di ⎝ c − di ⎠
( ac + bd ) + ( bc − ad ) i
=
2
c +d
2
multiply numerator and denominator by complex conjugate
multiply numerator and denominator
Now we have a complex number in the numerator and a real number in the denominator.
Simplify.
4
2+ i
solution
4
2+ i
=
=
=
=
Note:
Functions
4
2+ i
⋅ ⎛⎜
⎟
⎝2−i⎠
8 − 4⋅ i
4+ 1
8 − 4⋅ i
5
8
5
−
4
5
2−i⎞
i
multiply top and bottom by complex conjugate of denominator
expand numerator and denominator
simplify denominator
rewrite in standard form
answer
From this example, we can see that converting the denominator to a real number allows us to write the
complex number in standard form.
Complex Numbers – 147
148 – Complex Numbers
Functions
Advanced
To review the definition of rational functions.
rational function – a function defined by a simplified rational expression.
domain – the set of values which are allowable substitutions for the independent variable.
asymptote – a line approached by the graph of a function.
Thales of Miletus (c.640-546 B.C.) was an extremely well-traveled mathematician. He went to Egypt to either
pursue business opportunities or to study Egyptian science and philosophy or both. It is in Egypt where he
learned the basic principles of geometry. According to one legend, Thales was observing the Great Pyramid of
Cheops one day when Pharaoh and his entourage were passing by. Pharaoh had heard of Thales's reputation
as a wise man and asked Thales to calculate the height of the pyramid. One way he might have accomplished
this is with the "magical" Egyptian shadow-stick, which, when placed in the ground allows one to establish a
relationship between the height of the stick's shadow and the height of the object to be measured. A more
plausible method is that Thales measured the pyramid's shadow at the same time when his own shadow
appeared to be the same length as himself, thus assuring the Great Pyramid's shadow accurately represented its
true height.
Functions → Rational Functions
A rational function, contrary to popular belief, is not a function that makes sense. The name comes from the
fact that a rational function consists of a ratio of two other functions. Consider the expression:
5
4
This is the ratio of 5 to 4, also written as 5:4. In the same way, if we have a function p(x) = 3x and wish to
express this function in relation to some other function q(x) = 4x2 + 1, then we can write it in exactly the same
form as above:
3x
2
4x + 1
ratio of p(x) = 3x to q(x) = 4x2 + 1
In more general terms:
Definition of Rational Function
A rational function can be written as:
f(x)
=
q(x) ≠ 0
where p(x) and q(x) are both polynomials and q(x) is not equal to zero.
For the time being, we will assume that p(x) and q(x) have no common factors. (If they do, then we can apply
the principles of polynomial division in order to simplify the polynomial f(x)).
The domain of a rational function of x includes all real numbers except those values which would cause the
denominator q(x) to return a value of zero. In our rational expression above, we can use all real numbers
because there are no real numbers that will cause the denominator q(x) = 4x2 + 1 to equal zero.
Functions
Rational Functions – 149
In this section of the Algebra Brain, we will examine rational functions in some detail, paying particular
attention to the asymptotes of rational functions. We will also look at the notion of partial fractions, whereby
we write rational expressions as the sum of two or more simpler rational expressions.
150 – Rational Functions
Functions
Advanced
To find the asymptotes of a rational function.
asymptote – a line approached by the graph of a function such that the graph never touches or crosses the line.
vertical asymptote – a vertical line approached by the graph of a function such that the graph never touches or
crosses the vertical line.
horizontal asymptote – a horizontal line approached by the graph of a function such that the graph never
touches or crosses the horizontal line.
The Pythagoreans shared a mystical belief that the order inherent in the universe could be revealed in numbers.
From this belief, they developed representations of numbers: 1 was the point, 2 the line, 3 the surface, and 4 the
solid. They even assigned moral qualities to numbers. 4, for example, represented justice, while 10 (a.k.a. the
tetractys) indicated the sum of all nature because it was the sum of 1 + 2 + 3 + 4 (the point, line, plane, and
solid).
Functions → Rational Functions → Asymptotes
Rational functions do not have the same sort of graphs as other polynomial functions. Remember that a rational
function is defined as one polynomial function divided by another polynomial function. Let's discuss the
behavior of the simplest rational function. This will give us the tools needed to make some generalizations
about their behavior.
Consider the function:
f(x)
=
1
x
rational function
The domain of f(x) is all real values of x such that the denominator of f(x) does not equal zero. In this case, if
x = 0, then f(x) becomes undefined and therefore x = 0 is excluded from the domain of f(x). Let's take a look
at what happens to f(x) at values near x = 0.
x
f(x)
–1
–1
–0.5
–2
x
f(x)
0←
←
1
1
–0.25
–4
0.5
2
–0.1
–10
0.25
4
–0.01
–100
0.1
10
0.01
100
–0.001
–1000
→ 0
→–
0.001
1000
We can graph these values of x and f(x) to obtain the following:
Functions
Asymptotes – 151
Note:
As x approaches 0 from the left, f(x) decreases without bound (goes to – ). As x approaches 0 from
the right, f(x) increases without bound (goes to + ).
Also note that the graph of f(x) never crosses either the x- or y-axes. Because of this fact, we call the
x- and y-axes the asymptotes of the graph of f(x). In other words, the line x = 0 is the vertical
asymptote of the graph of f and the line y = 0 is the horizontal asymptote of the graph of f. This
means that the values of f(x) = 1/x approach zero as x increases or decreases without bound (as x
moves toward
).
The formal definition of vertical and horizontal asymptotes of a function is as follows:
Definition of Vertical and Horizontal Asymptotes
The line x = a is a vertical asymptote of the graph of f if
f(x) →
as x
or f(x) → –
a, either from the right or from the left.
The line y = b is a horizontal asymptote of the graph of f if
f(x) → b
as x →
or x → – .
Eventually (as x → or x → – ) the distance between the graph of f and the asymptote must approach zero.
However, by definition, the graph never actually touches the asymptote. You will learn more about this
astonishing behavior when you study calculus. Now that we know the definition of asymptotes, we need some
method by which we can go about finding them for a particular rational function.
152 – Asymptotes
Functions
Asymptotes of a Rational Function
Let f be the rational function given by:
f(x)
=
=
where p(x) and q(x) have no common factors.
1.
The graph of f has vertical asymptotes at the zeros of q(x).
2.
The graph of f has one or no horizontal asymptotes determined as follows:
a.
If n < m, the graph of f has the x-axis (y = 0) as a horizontal asymptote.
b.
If n = m, the graph of f has the line
y
=
as a horizontal asymptote
c.
If n > m, the graph of f has no horizontal asymptotes.
The theorem basically boils down to comparing the degree of the polynomial in the numerator to the degree of
the polynomial in the denominator. In the example we used to introduce asymptotes, the degree of p(x) = 0 and
the degree of q(x) = 1, which means n < m, therefore we used part (2a) of our theorem on asymptotes to
determine what the horizontal asymptote was (y = 0). Similarly, the only time f(x) = 1/x has a zero is when x is
infinitely large, which means that the vertical asymptote is x = 0. Let's look at an example.
Find the domain of the function and any vertical or horizontal asymptotes.
f(x)
2+ x
=
2−x
solution
First of all, the domain of this function includes all values of x except those values that cause the function to be
undefined. In the case of rational functions, they are undefined whenever the denominator is zero. We want to
find values of x which will make q(x) = x – 2 = 0. Those values of x are excluded from our domain. In this
example, if x = 2, then q(x) = 0 and f(x) becomes undefined. Therefore, the domain of f(x) is:
all real numbers except x = 2
domain of f(x)
answer
We use the numbers excluded from the domain of f(x) to find our vertical asymptotes. In other words, vertical
asymptotes are located at values of x that make the denominator of f(x) go to zero. We have already found such
a number for this function. Therefore, the vertical asymptote for f(x) is:
x
Functions
=
2
vertical asymptote
answer
Asymptotes – 153
Horizontal asymptotes are found by comparing the degrees of the polynomials in numerator and denominator.
Both happen to be of degree "1" in this example. Since they are equal, then according to our definition above, a
horizontal asymptote is located at:
y
=
y
=
y
=
a1
b1
1
−1
–1
horizontal asymptote when p(x) and q(x) have same degree
substitute in a1 = 1 and b1 = –1, leading coefficients of p(x) and q(x), respectively
simplify
answer
The graph of this function looks like:
Applications of Asymptotes
Believe it or not, asymptotes have an important role in real-world applications. Of particular interest is costbenefit analysis. A particular process, whether it is manufacturing or something else, often has a function
associated with it that determines how much it costs to obtain a certain percentage benefit. For example, it may
cost a power plant only $10,000 to remove 10% of the pollutants it emits, but it may cost them $100,000 to
remove 30% of the pollutants and $1,000,000 to remove 70% of the pollutants. As the percentage of pollutants
removed increases, the cost also increases. Removing 100% of the pollutants it emits would bankrupt the
power plant. It just does not have the capacity to remove every single piece of unclean material from its
emissions. This is important information to know because laws determine how much the power plant needs to
remove in order to be compliant with the law. The bean-counters in Accounting then have to decide where to
find the extra money they need when the law starts tightening its emission controls.
Another interesting example of rational functions is population control. When a population of
animals is introduced into an environment, it increases dramatically at first, but tapers off as
the resources in the environment decrease. The population can only grow if there are
sufficient resources.
154 – Asymptotes
Functions
Advanced
To use a set of guidelines to sketch rational functions.
slant asymptote – an asymptote to a rational function that occurs when the degree of the numerator is greater
than the degree of the denominator.
Pythagoras (c.580-500 B.C.) was a noted astronomer. He was the first to declare the spherical nature of the
Earth. He also deduced that the sun, moon, and planets each rotate around their own axes. Pythagoras noted
that celestial bodies orbit around a central point which he mistakenly identified as the Earth. Later Pythagoreans
deduced there was another central point not on the Earth, but never identified this point as the sun. Finally,
Pythagoras established the period of the Earth's rotation: 24 hours.
Functions → Rational Functions → Sketching Rational Functions
Rational functions can be difficult to graph if we don't know what we are doing. Fortunately, there are several
guidelines we can use to help us sketch the graph of a rational function.
Guidelines for Graphing Rational Functions
Let f(x) = p(x)/q(x), where p(x) and q(x) are polynomials with no common factors.
Note:
1.
Find and plot the y-intercept (if any) by evaluating f(0).
2.
Find the zeros of the numerator (if any) by solving the equation p(x) = 0. Then plot the
corresponding x-intercepts.
3.
Find the zeros of the denominator (if any) by solving the equation q(x) = 0. Then sketch the
corresponding vertical asymptotes.
4.
Find and sketch the horizontal asymptote (if any) by using the rule for finding the horizontal
asymptote of a rational function.
5.
Plot at least one point between and one point beyond each x-intercept and vertical asymptote.
6.
Use smooth curves to complete the graph between and beyond the vertical asymptote.
Testing for symmetry can also be very useful. For example, the graph of f(x) = 1/x is symmetrical with
respect to the origin and the graph of g(x) = 1/ x2 is symmetrical with respect to the y-axis.
It is usually not enough to simply give you some guidelines. It is also helpful to see how these guidelines can
be applied.
Sketch the graph of the rational function. As sketching aids, check for intercepts, symmetry, vertical and
horizontal asymptotes.
g(x)
Functions
=
x
2
x −9
Sketching Rational Functions – 155
solution
Let's go down the list of guidelines given above.
1.
The y-intercept occurs at g(0):
g(0)
=
=
0
2
0 −9
let x = 0
0
Thus, the y-intercept is located at (0, 0).
2.
The x-intercepts occur when y = 0, which only happens when p(x) = 0. In this case, p(x) = x, so that
implies that p(0) = 0 and we get:
x
=
0
x-coordinate of x-intercept
Thus, the x-intercept is also located at (0, 0) in this particular example.
3.
Vertical asymptotes occur when q(x) = 0. So we solve the equation x2 – 9 = 0:
x2 – 9
=
0
q(x) = 0
(x – 3)(x + 3)
=
0
factor
x
=
3
one solution
vertical asymptote
=
–3
other solution
vertical asymptote
or
x
Sketch these asymptotes on your graph.
4.
To find horizontal asymptotes, we compare the degree of the numerator to the degree of the
denominator. Here we have p(x) = x, which is degree 1. We also have q(x) = x2 – 9, which is degree 2.
Since 1 < 2, we can conclude that the horizontal asymptote is located at y = 0. This is just the x-axis,
so it is probably already part of your graph and we don't need to sketch it.
5.
We next plot one point on each side of the left hand vertical asymptote x = –3. We plot one point
on each side of the x-intercept x = 0. We plot one point on each side of the right hand vertical
asymptote x = 3. We should end up with 6 points total.
x
g(x)
6.
–4
–0.571
–3
undefined
–2
0.4
–1
0.125
0
0
1
–0.125
2
–0.4
3
undefined
4
0.571
Finally, we connect our points with smooth curves. Remember that the graph can
never cross an asymptote! We show you the graph twice, once with all the bells and
whistles and once without all the fancy sketching tricks we used.
156 – Sketching Rational Functions
Functions
answer
If we remove the asymptotes (remember that they are not actually part of the graph) we have:
Zooming out on the graph, we can get an even clearer picture of what it looks like:
Functions
Sketching Rational Functions – 157
Slant Asymptotes
Not all rational functions have just horizontal or vertical asymptotes. If the degree of the polynomial in the
numerator is exactly one more than the degree of the polynomial in the denominator, then the rational function
has a slant asymptote. The exact equation of the slant asymptote is found by long division. For instance, if we
divide x2 – x by x + 1, we end up with:
2
f(x)
x −x
=
divide two polynomials
numerator is one degree higher than polynomial
x+1
x−2+
=
2
x+1
The slant asymptote is the quotient part of the resulting expression, not the remainder part. In
the above example, the slant asymptote would be:
y
=
x–2
quotient part
Sketch the graph of the function.
2
f(x)
=
x −x
x+1
solution
To sketch the graph of this function we use all the same tricks and tools that we did before.
1.
The y-intercept is located at (0, 0) because f(0) = 0.
2.
The x-intercepts can be found by factoring the numerator and solving the equation p(x) = 0. In this
case factoring the equation produces:
x2 – x
=
0
p(x) = 0
x(x – 1)
=
0
factor
x
or
x
=
0
x-coordinate of one intercept
=
1
x-coordinate of other intercept
The x-intercepts are located at (0, 0) and (0, 1).
3.
The vertical asymptotes are located wherever the denominator q(x) = 0. In this case, that happens to
occur when x = –1, so this is also the one and only vertical asymptote for this function. Sketch the
vertical asymptote.
4.
The horizontal asymptote is found by comparing the degree of the numerator to the degree of the
denominator. Here the degree of p(x) is 2 and the degree of q(x) is 1. Since 2 > 1, we can conclude
that this rational function has no horizontal asymptote. However, since the degree of p(x) is exactly
one more than the degree of q(x), it does have a slant asymptote, which we have already determined
exists at y = x – 2. Sketch this slant asymptote.
5.
Now we need a few points. We plot points on either side of each asymptote and on either side of each
x-intercept.
x
f(x)
–3
–6
–2
–6
–1
undefined
158 – Sketching Rational Functions
–0.5
1.5
0
0
0.5
–0.167
1
0
2
0.667
3
1.5
Functions
6.
Finally, we sketch the graph of this rational function, using the asymptotes to help us:
answer
Functions
Sketching Rational Functions – 159
160 – Sketching Rational Functions
Functions
Advanced
To decompose a rational function into its component parts.
partial fraction decomposition – the method by which we transform one rational function into the sum of two or
more simpler rational functions.
S. I. Ramanujan (1887-1920) was an Indian number theorist and one of the most unusual mathematicians of all
time. Before he had completed high school, he came across G. S. Carr's A Synopsis of Elementary Results in
Pure and Applied Mathematics, a compilation of five thousand formulae with few proofs. Ramanujan decided to
prove the theorems in Carr's compilation. Since Ramanujan was unfamiliar with Western mathematical tools, he
quickly fashioned his own and often wandered into strange mathematical realms of his own creation. Had
Ramanujan not caught the attention of mathematics professors in England, we might never have uncovered his
Notebooks, theorems of his own which baffle mathematicians to this very day.
Functions → Rational Functions → Partial Fractions
Sometimes we are faced with a rational function problem that seems too complicated for us. For example, if we
have the function:
f(x)
=
−5x + 15
2
x + 3x − 4
rational function
how do we go about simplifying this expression? As it turns out, this expression is nothing more than:
f(x)
=
2
x−1
−
7
x+4
equivalent rational function
We have a difficult time seeing the direct connection. We are about to show you how to transform the first
rational function above into the second rational function. The method is known as partial fraction
decomposition.
Before we demonstrate this method with algebraic expressions, let's first use a concrete example with some real
numbers. The method is exactly the same regardless if we use real numbers or algebraic expressions.
Recall that when you simplify the expression:
1
3
+
1
2
you find the least common denominator, which in this case is 6. Then you can rewrite the fraction as:
1
3
+
1
2
=
=
2
6
+
3
6
5
6
Now what if you had simply been given the fraction 5/6 and had been told to break it down into its parts? How
would you go about solving it?
Functions
Partial Fractions – 161
The answer lies with partial fraction decomposition. In a sense, it is the reverse of simplification of fractions.
To "decompose" the fraction 5/6, we first need to write the denominator in its factored form:
5
6
5
=
factor denominator
( 2) ( 3)
Now we can write the fraction as the sum of two fractions, each having in its denominator one of the two
factors:
5
6
=
2
+
3
the black squares represent unknown quantities.
In order to fill in the blanks, let's assign the numerators values of A and B respectively:
5
6
A
=
2
+
B
3
Next, we want to get rid of the 6 on the left, so we multiply both sides by 6:
5
=
5
=
6A
2
+
6B
multiply both sides by 6
3
3A + 2B
simplify fractions on right
Now to find A and B, we have to do a little bit of guesswork since we have 1 equation and 2 unknowns. In
general, you will end up with n equations with n unknowns. Then you can solve the system of linear equations
using standard algebraic techniques.
If we let A = 1, then we have:
5
=
3(1) + 2B
let A = 1
2
=
2B
subtract 3 from both sides
1
=
B
divide both sides by 2
We have determined that A and B are both 1. Therefore, we can now write our original fraction 5/6 in its
decomposed form:
5
6
=
1
2
+
1
3
partial fraction decomposition
Now we can do the same sort of tricks to transform
−5x + 15
2
x + 3x − 4
into
2
x−1
−
7
x+4
partial fraction decomposition
solution
First, we factor the denominator, like so:
−5x + 15
2
x + 3x − 4
162 – Partial Fractions
=
−5x + 15
( x − 1) ( x + 4)
factor denominator
Functions
Next, we let the expression equal the sum of two fractions, each having in its denominators one of the two
factors. Again, we use A and B to represent the unknown quantities of the numerator.
−5x + 15
A
=
( x − 1) ( x + 4)
x−1
B
+
x+4
rewrite fraction as sum of two fractions
partial fraction decomposition
Then we multiply both sides of the equation by the denominator on the left:
−5x + 15
( x − 1) ( x + 4)
⎛ A + B ⎞ ( x − 1) ( x + 4)
⎜ −1
⎟
x + 4⎠
⎝x
multiply both sides
by (x – 1)(x + 4)
A( x − 1) ( x + 4)
Distributive Property on
right hand side
simplify left hand side
( x − 1) ( x + 4)
=
–5x + 15
=
–5x + 15
=
A(x + 4) + B(x – 1)
cancel common factors in
each rational expression
–5x + 15
=
Ax + 4A + Bx – B
Distributive Property
x−1
+
B( x − 1) ( x + 4)
x+4
Now at this point it looks like we have 1 equation with 3 unknowns because we still have no idea what A and B
are. However, for both sides to be equal the coefficients of x on both sides have to be the same and the constant
terms must also add up to the same number. Therefore, the sum of A and B is equal to –5 and the quantity 4A –
B is equal to 15. In other words, we can rewrite the last step above as:
–5x + 15
=
(A + B)x + 4A – B
rewrite last step from above
We next break this down into a system of two linear equations with two unknowns:
A+B
and
4A – B
=
–5
coefficients of x
=
15
constant terms
We can solve this system by substitution:
4A – 15
=
B
solve second linear equation for B
A + (4A – 15)
=
–5
substitute in B = 4A – 15 into first linear equation
5A
=
10
combine like terms and add –15 to both sides
A
=
2
divide both sides by 5
2+B
=
–5
substitute in A = 2 into first linear equation
B
=
–7
subtract 2 from both sides
Once we have found A and B, we can plug these into our partial fraction decomposition:
−5x + 15
2
x + 3x − 4
=
2
x−1
−
7
x+4
answer
It really works! Although it does take a great deal of effort, you will be glad you learned about how to do this
when you take calculus.
In general, there are five basic situations that occur with partial fraction decomposition. The denominator of a
rational function can have distinct linear factors, distinct quadratic factors, repeated linear factors, repeated
quadratic factors, or any combination of the previous four situations.
Functions
Partial Fractions – 163
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
and
(ax2 + bx + c)n
where (ax2 + bx + c) is irreducible.
3.
Linear factors: For each factor of the form (px + q)m, the partial fraction decomposition
must include the following sum of m fractions:
4.
Quadratic factors: For each factor of the form (ax2 + bx + c)m, the partial fraction decomposition
must include the following sum of n fractions:
Instead of providing an example of each possible situation here, we have decided to give each situation its own
thought. When decomposing fractions, you will usually end up with a "basic equation" that contains the
numerator of the original fraction on the left and an expression formed from the factors of the denominator of
the original fraction on the right. Solving this "basic equation" varies depending on if we have linear or
quadratic factors. Here are some guidelines for solving these "basic equations":
Guidelines for Solving the Basic Equations
Linear Factors
1.
Substitute the roots of the distinct linear factors into the basic
equation
2.
For repeated linear factors use the coefficients determined in step 1 to
rewrite the basic equation. Then substitute other convenient values
of x and solve for the remaining coefficients (for example, let x = 0).
Quadratic Factors
1.
Expand the basic equation.
2.
Group terms according to powers of x.
3.
Equate the coefficients of like powers to obtain a system of linear
equations involving A, B, C, and so on.
4.
Solve the system of linear equations.
164 – Partial Fractions
Functions
Advanced
To decompose a rational function that has only distinct linear factors in the denominator.
No definitions on this page.
Leonardo Pisano Fibonacci (1170?-1250?), an Italian number theorist, is most famous for his recursive
sequence of numbers: 1, 1, 2, 3, 5, 8, 13, 21, etc. This sequence stems from a problem involving rabbits. A pair
of rabbits are kept in an enclosure and allowed to breed, producing offspring at the rate of one pair of rabbits per
month, beginning with the second month. How many rabbits will exist at the end of the year, assuming none of
the rabbits die? Each number in the Fibonacci Sequence corresponds to the number of pairs of rabbits at the
end of each month. That is, in January and February you would have 1 pair of rabbits, in March you have 2
pairs, in April you have 3 pairs, and so on.
Functions → Rational Functions → Partial Fractions → Distinct Linear Factors
In the previous thought, we introduced you to the notion of partial fractions. Now we are going to give an indepth example of one of the five possible cases that occur with rational functions. In this example, we will be
looking at the case where the denominator of the rational function consists of distinct linear factors. By
"distinct", we mean that each linear factor only occurs one time in the denominator of the rational function.
We also reproduce the method for decomposing fractions into their component parts (we leave out the part
about quadratic factors in the denominator because it doesn't apply here).
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
and
(ax2 + bx + c)n
where (ax2 + bx + c) is irreducible.
3.
Functions
Linear factors: For each factor of the form (px + q)m, the partial fraction decomposition
must include the following sum of m fractions:
Distinct Linear Factors – 165
Decompose the following rational function into its partial fraction decomposition.
2
f(x)
=
x + 3x + 6
3
x − 2x − x + 2
solution
According to the rules of partial fraction decomposition given above, the first thing we need to do is see if we
can divide the numerator by the denominator. That only happens when the degree of p(x) is greater than or
equal to the degree of q(x). In this case the degree of p(x) is 2 and the degree of q(x) is 3, so we cannot use
polynomial long division.
Next, we factor the denominator if possible. After examining the denominator for awhile, we notice that we can
apply the FOIL method in reverse:
2
x + 3x + 6
3
x − 2x − x + 2
2
=
x + 3x + 6
FOIL method in reverse
(x2 − 1)( x − 2)
2
=
x + 3x + 6
factor difference of two squares
completely factored denominator
( x − 1) ( x + 1) ( x − 2)
Once we have completely factored the denominator, we can begin the process of decomposing our rational
function. First, we set up the basic equation, where the rational function is set equal to the sum of its parts:
2
x + 3x + 6
( x − 1) ( x + 1) ( x − 2)
=
A
x−1
+
B
x+1
+
C
x−2
Basic Equation
Next, we multiply both sides of the basic equation by the denominator on the left and simplify the right hand
side of the equation.
2
x + 3x + 6
( x − 1) ( x + 1) ( x − 2)
( x − 1) ( x + 1) ( x − 2)
=
⎛ A + B + C ⎞ ( x − 1) ( x + 1) ( x − 2)
⎜ x − 1 x + 1 x − 2⎟
⎝
⎠
multiply both sides by denominator on left
x2 + 3x + 6
=
A(x + 1)(x – 2) + B(x – 1)(x – 2) + C(x – 1)(x + 1)
simplify both sides
x2 + 3x + 6
=
A(x2 – x – 2) + B(x2 – 3x + 2) + C(x2 – 1)
FOIL method on right hand side
x2 + 3x + 6
=
Ax2 – Ax – 2A + Bx2 – 3Bx + 2B + Cx2 – C
Distributive Property
x2 + 3x + 6
=
Ax2 + Bx2 + Cx2 – Ax – 3Bx – 2A + 2B – C
regroup terms according to degree of x
x2 + 3x + 6
=
(A + B + C)x2 – (A + 3B)x – 2A + 2B – C
factor out an x2 and an x
At this point, we take a moment to step back and see what we have accomplished. In order for both sides to be
equal, the coefficients of the polynomial on the left have to equal the coefficients of the polynomial on the right.
For instance, on the left, the coefficient of the x2-term is 1, so the quantity (A + B + C) has to be equal to 1.
Using this line of reasoning, we set up a system of linear equations where the coefficients of the left hand side
have to be equal to the sums of A, B and C on the right hand side:
166 – Distinct Linear Factors
Functions
1
=
A+B+C
coefficients of x2
3
=
–(A + 3B)
coefficients of x
7
=
–2A + 2B – C
constant terms
We use substitution to solve this, although we could use a variety of methods, including a method involving a
matrix, but we won't do that to you here.
Solve the second linear equation for A:
3
=
–(A + 3B)
second linear equation
coefficients of x
3
=
–A – 3B
Distributive Property
A
=
–3 – 3B
solve for A
Now substitute this value for A into the remaining two linear equations and solve:
1
=
(–3 – 3B) + B + C
let A = 3 – 3B in first linear equation
coefficients of x2
1
=
–3 – 2B + C
simplify
2 + 2B
=
C
solve for C
Substitute A = 3 – 3B and C = –2 + 2B into the third linear equation
6
=
–2(–3 – 3B) + 2B – (2 + 2B)
substitute in A = –3 – 3B and C = 2 + 2B
constant terms
6
=
6 + 6B + 2B – 2 – 2B
Distributive Property
2
=
6B
simplify
=
B
divide both sides by 6
1
3
Now that we know B, we can back-substitute this into the second linear equation.
3 – 3B
−3 − 3⎛⎜
1⎞
⎟
⎝ 3⎠
–4
=
A
second linear equation
=
A
substitute in B = 1/3
=
A
simplify
So B = 1/3 and A = –4. Now we can find C by substituting these values into the first or third linear equations.
For the sake of argument, we will substitute our values for A and B into the first linear equation:
Functions
Distinct Linear Factors – 167
1
=
A+B+C
1
=
−4 +
=
C
8
3
1
3
+C
first linear equation
coefficients of x2
substitute in A = –4 and B = 1/3
solve for C
Therefore, we have:
A
=
B
=
C
=
–4
1
3
and
and
8
3
so our Basic Equation looks like:
2
x + 3x + 6
( x − 1) ( x + 1) ( x − 2)
=
=
168 – Distinct Linear Factors
A
x−1
−4
x− 1
+
B
x+1
+
1
+
3
x+ 1
C
x−2
8
+
3
x− 2
Basic Equation
partial fraction decomposition
answer
Functions
Advanced
To decompose a rational function that has only distinct quadratic factors in the denominator.
No definitions on this page.
Pierre de Fermat (1601-1655) was a French number theorist. He is regarded by some mathematicians as the
inventor of differential calculus, having set out the fundamentals of differential calculus in a 1636 paper. Among
his more famous contributions to math and science is Fermat\s Principle of optics, wherein he states that light
travels by the path of shortest duration. Fermat's son published much of Fermat's work after his death, including
a theorem scribbled in a margin that had no corresponding proof. The solution to Fermat's Last Theorem
puzzled mathematicians for almost 300 years until Andrew J. Wiles developed the first acceptable proof in 1944.
Functions → Rational Functions → Partial Fractions → Distinct Quadratic Factors
If we do not have any linear factors in the denominator of a rational function, then we must have an irreducible
quadratic expression in the denominator. An irreducible quadratic expression is one that has no real linear
factors. The most common example of an irreducible quadratic is:
x2 + 1
irreducible quadratic
This expression can only be factored by using complex numbers. There are no real numbers that will make this
expression equal zero.
If we have distinct irreducible quadratic factors in our denominator, we can still use the method of partial
fractions to break the rational function up into the sum of its parts. However, the numerator of each of the
component bits is a linear expression in x (or some other independent variable), whereas with linear factors in
the denominator, we had a partial fraction decomposition involving constant terms in the numerators. An
example of what we are talking about would probably be helpful right about now:
A
x+1
+
Ax + B
2
x +1
B
x−1
+
partial fraction decomposition
distinct linear factors
constant terms in numerators
Cx + D
2
x +4
partial fraction decomposition
distinct irreducible quadratic factors
linear expressions in numerators
Note the differences between the two partial fraction decompositions.
For your convenience, we once again provide you with the method of partial fraction decomposition, this time
omitting the part about linear factors.
Functions
Distinct Quadratic Factors – 169
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
(ax2 + bx + c)n
and
where (ax2 + bx + c) is irreducible.
3.
(the part about linear factors goes here)
4.
Quadratic factors: For each factor of the form (ax2 + bx + c)m, the partial fraction decomposition
must include the following sum of n fractions:
Let's see an example of how this works for distinct irreducible quadratic factors.
Distinct Quadratic Factors
Write the partial fraction decomposition for the following rational expression.
3x
4
2
x + 5x + 4
solution
Before we do anything else, we first need to make sure that this is a "proper" fraction. A proper rational
expression is one where the degree of the numerator is less than the degree of the denominator. If it is not, then
we have to perform long division in order to transform the rational expression into a polynomial plus a "proper"
rational expression. However, in this case we don't need to worry ourselves about that, as the degree of the
numerator, 1, is clearly less than the degree of the denominator, 4.
Now we need to see if we can factor the denominator. By carefully looking at the denominator, we realize that
it can indeed be factored:
3x
4
=
2
x + 5x + 4
3x
(x2 + 1)(x2 + 4)
factor the denominator
Note that each factor in the denominator is an irreducible quadratic. Now we use the method of partial fraction
decomposition in order to write the basic equation for this rational expression:
(x
3x
2
)(
2
)
+1 x +4
170 – Distinct Quadratic Factors
=
Ax + B
2
x +1
+
Cx + D
2
x +4
Basic Equation
Functions
We next need to multiply both sides of the Basic Equation by the denominator on the left:
3x
=
⎛ Ax + B + Cx + D ⎞ ( x2 + 1) ( x2 + 4)
⎜ 2
⎟
2
x +4 ⎠
⎝ x +1
multiply both sides by (x2 + 1)(x2 + 4)
3x
=
(Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
Distributive Property
cancel common factors in
numerator and denominator
3x
=
Ax3 + 4Ax + Bx2 + 4B + Cx3 + Cx + Dx2 + D
multiply out all binomials
3x
=
Ax3 + Cx3 + Bx2 + Dx2 + 4Ax + Cx + 4B + D
rearrange terms in decreasing order
of powers of x
3x
=
(A + C)x3 + (B + D)x2 + (4A + C)x + 4B + D
factor out all x-terms
In order for both sides to be equal, the coefficients of all x-terms have to be equal. Since there are no x3-terms
on the left hand side, the sum (A + D), which is the coefficient of x3 on the right hand side, must also be equal to
zero. We compare the remaining coefficients of the x-terms and construct a system 4 linear equations with 4
unknowns (A, B, C, and D).
0
=
A+C
coefficient of x3
0
=
B+D
coefficient of x2
3
=
4A + C
coefficient of x
0
=
4B + D
constant term
We have several choices with which we can begin solving this system. Notice that both the second and fourth
equations involve D and both equal 0. By toying with these two equations, we can discover:
0
=
B+D
second linear equation
B
=
–D
solve for D
0
=
4B + D
fourth linear equation
0
=
4(–D) + D
substitute in B = –D
0
=
–3D
simplify
0
=
D
divide both sides by –3
which implies that B = D = 0. Thus, B = 0 and D = 0. Next, we solve the first and third equations.
Functions
0
=
A+C
first linear equation
A
=
–C
solve for A
3
=
4A + C
third linear equation
3
=
4(–C) + C
substitute in A = –C
3
=
–3C
simplify
–1
=
C
divide both sides by –3
Distinct Quadratic Factors – 171
which also gives us A = –C = –(–1) = 1. Thus, A = 1 and C = –1. Altogether, we have:
A
=
1
and
B
=
0
and
C
=
–1
and
D
=
0
We plug these values for A, B, C and D into our Basic Equation and get:
(x
3x
)
=
(x2 + 1)(x2 + 4)
=
2
)(
2
+1 x +4
3x
Ax + B
2
+
x +1
x
2
x +1
Cx + D
2
x +4
−
x
2
x +4
Basic Equation
substitute in A = 1, B = 0, C = –1, D = 0
partial fraction decomposition
answer
As we can see from this example, partial fraction decomposition is a nasty, convoluted, exhausting process.
Many people would rather have their teeth extracted without any anesthetic rather than decompose rational
functions. The only time we actually want to perform partial fraction decomposition is in calculus when we
have to integrate unpleasant-looking rational functions. Then partial fraction decomposition is a positive boon.
172 – Distinct Quadratic Factors
Functions
Advanced
To decompose a rational function that contains repeated linear factors in the denominator.
No definitions on this page.
Leonardo Pisano Fibonacci (1170?-1250?) is credited with bringing the Hindu-Arabic number system to the rest
of Western Europe. He studied the Hindu-Arabic system while in Africa and quickly saw the advantages of using
Hindu-Arabic numbers for calculation as opposed to the cumbersome Roman numerals in use at the time. His
book Liber abaci (Book of Calculations), published in 1202, was intended to introduce Hindu-Arabic numbers and
show how they could be used in business transactions.
Functions → Rational Functions → Partial Fractions → Repeated Linear Factors
Using the method of partial fraction decomposition is very much the same when we have repeated linear factors
as when we have distinct linear factors. There is only one main difference. If we have a denominator in our
rational expression that can be factored into repeated linear factors, then its partial fraction decomposition takes
the form:
A
( x − c)
+
B
( x − c)
2
C
+
( x − c)
3
+ ....
partial fraction decomposition
repeated linear factors
c is an integer
Note that we specified that c must be an integer. Partial fractions will still work if we have "messy" real
numbers for c (such as radicals or fractions), but it gets horribly, horribly complicated when c is not a "nice"
number, like 2.
Again, we provide you with the partial fraction decomposition method for reference:
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
and
(ax2 + bx + c)n
where (ax2 + bx + c) is irreducible.
3.
Functions
Linear factors: For each factor of the form (px + q)m, the partial fraction decomposition
must include the following sum of m fractions:
Repeated Linear Factors – 173
Note:
We omit the decomposition into partial fractions when we have quadratic factors because that doesn't
apply to this case.
Repeated Linear Factors
Write the partial fraction decomposition for the rational expression. Check your result algebraically.
2
4x + 3
( x − 5)
3
solution
As always, the first thing we need to do is check to make sure this is a proper fraction. Since the degree of the
numerator is 2 and the degree of the denominator is 3 and 2 < 3, then we conclude that this is indeed a proper
fraction and we can proceed to factor the denominator.
The denominator can easily be factored:
2
2
4x + 3
( x − 5)
=
3
4x + 3
factor the denominator
( x − 5) ( x − 5) ( x − 5)
Once we have factored the denominator, we are ready to build our Basic Equation, using the guidelines outlined
above:
2
4x + 3
( x − 5) ( x − 5) ( x − 5)
=
A
x−5
B
+
( x − 5)
2
+
C
( x − 5)
3
Basic Equation
As we do every time we apply partial fraction decomposition, we multiply both sides by the denominator on the
left:
4x2 + 3
=
⎡ A + B + C ⎤ ( x − 5) ( x − 5) ( x − 5)
⎢x − 5
2
3⎥
( x − 5)
( x − 5) ⎦
⎣
multiply both sides by
(x – 1)(x – 1)(x – 1)
4x2 + 3
=
A(x – 5)(x – 5) + B(x – 5) + C
Distributive Property
cancel common factors
in numerator and denominator
4x2 + 3
=
A(x2 – 10x + 25) + Bx – 5B + C
start multiplying out everything
4x2 + 3
=
Ax2 – 10Ax + 25A + Bx – 5B + C
Distributive Property
4x2 + 3
=
Ax2 + (B – 10A)x + 25A – 5B + C
rearrange terms
At this point, we can make the brilliant leap of logic that A = 4, since the coefficients of x2 on each side have to
be equal in order for the equation to be valid. We also note that the quantity (B – 10A) = 0 since there are no xterms on the left hand side. Finally, we can conclude that 25A – 5B + C = 3, which is the constant term on both
sides. In short, we have the following system of three linear equations with three unknowns:
4
=
A
coefficient of x2
0
=
B – 10A
coefficient of x
3
=
25A – 5B + C
constant term
174 – Repeated Linear Factors
Functions
Since we already know A, we can substitute it into the second equation to find B. Once we know A and B, we
substitute them into the third linear equation and solve for C:
0
40
=
B – 10A
second linear equation
=
B – 10(4)
substitute in A = 4
=
B
add 40 to both sides
Next we substitute A = 4 and B = 40 into the third linear equation:
3
=
25A – 5B + C
third linear equation
3
=
25(4) – 5(40) + C
substitute in A = 4 and B = 40
3
=
100 – 200 + C
simplify
103
=
C
add 100 to both sides to solve for C
Now that we know:
A
=
4
and
B
=
40
and
C
=
103
we can plug these values into our Basic Equation to complete the partial fraction decomposition of our original
rational expression:
A
2
4x + 3
( x − 5) ( x − 5) ( x − 5)
=
x−5
4
2
4x + 3
( x − 5) ( x − 5) ( x − 5)
=
x−5
B
+
( x − 5)
2
40
+
( x − 5)
2
C
+
( x − 5)
3
Basic Equation
3
partial fraction decomposition
answer
103
+
( x − 5)
In order to check our partial fraction decomposition, we multiply each numerator on the right by the missing
factors in its denominator. We then simplify the resulting expression and compare it to the numerator on the
left. If they match, then the partial fraction decomposition is correct.
4
x−5
+
40
( x − 5)
2
+
103
( x − 5)
3
=
=
4( x − 5) ( x − 5) + 40( x − 5) + 103
( x − 5) ( x − 5) ( x − 5)
(2
)
4 x − 10x + 25 + 40x − 200 + 103
( x − 5) ( x − 5) ( x − 5)
2
=
4x − 40x + 100 + 40x − 200 + 103
( x − 5) ( x − 5) ( x − 5)
2
=
Functions
4x + 3
( x − 5) ( x − 5) ( x − 5)
rewrite partial fraction
decomposition in terms of
LCD
multiply out polynomial
in numerator
expand polynomial in
numerator
simplify numerator
check
Repeated Linear Factors – 175
176 – Repeated Linear Factors
Functions
Advanced
To decompose a rational function that contains repeated quadratic factors in the denominator.
No definitions on this page.
Benoit B. Mandelbrot (1924- ) is most famous for his psychedelic Mandelbrot Set, an application of fractal
geometry. The basic idea behind fractal geometry is that a very simple set of instructions can produce a highly
complex situation, such as a picture or even a living organism. Furthermore, it is infinitely complex in that
attempting to zoom in on a particular feature yields just as much complexity as the whole situation. Mandelbrot's
theories have led to advances in engineering, chemistry, biology, and physics.
Functions → Rational Functions → Partial Fractions → Repeated Quadratic Factors
In related thoughts, we looked at examples of partial fraction decomposition involving distinct linear factors,
distinct quadratic factors, and repeated linear factors. Now we will look at an example where the denominator
of a rational expression has repeated quadratic factors. The principles of how to solve this type of problem
remain the same as in all the other examples; we just add one slight wrinkle.
If we have repeated quadratic factors in our denominator, then the partial fraction decomposition takes the form:
B 1⋅ x + C 1
2
ax + bx + c
+
B 2⋅ x + C 2
(ax
2
+ bx + c
)
2
+ .... +
Bn ⋅ x + C n
(ax
2
+ bx + c
)
n
partial fraction decomposition
repeated quadratic factors
In other words, if the denominator has the factor (ax + bx + c) repeated 3 times, then we end up with the sum of
3 fractions, which take the form given above. As always, we once again give you the summary of the method
of partial fraction decomposition (omitting the part about repeated linear factors, because it just doesn't apply
here).
Functions
Repeated Quadratic Factors – 177
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
(ax2 + bx + c)n
and
where (ax2 + bx + c) is irreducible.
3.
(the part about linear factors goes here)
4.
Quadratic factors: For each factor of the form (ax2 + bx + c)m, the partial fraction decomposition
must include the following sum of n fractions:
Repeated Quadratic Factors
Write the partial fraction decomposition of the following rational expression. Check your result algebraically.
2
2x + x + 8
4
2
x + 8x + 16
solution
First, we check to make sure that we don't have to perform long division. Since the degree of the numerator is 2
and the degree of the denominator is 4 and 2 < 4, we don't have to perform long division to get this into a proper
rational expression. It is already proper.
Our next step is to factor the denominator if possible. After inspecting the denominator, we can see that we can
factor it as follows:
2
2x + x + 8
4
2
x + 8x + 16
2
2x + x + 8
=
(x2 + 4)(x2 + 4)
factor the denominator
Notice that we have a repeated quadratic factor. Next, we build the Basic Equation, according to the guidelines
we established at the top of the page:
Ax + B
2
2x + x + 8
(x2 + 4)(x2 + 4)
=
178 – Repeated Quadratic Factors
2
x +4
+
Cx + D
(x2 + 4)2
Basic Equation
Functions
Now we continue solving the problem as we do for any partial fraction decomposition problem. We multiply
both sides by the denominator on the left and cancel common factors in the numerators and denominators on the
right:
2x2 + x + 8
=
⎡ Ax + B + Cx + D ⎤ ( x2 + 4) ( x2 + 4)
⎢ x2 + 4
(x2 + 4)2 ⎥⎦
⎣
multiply both sides by
(x2 + 4)(x2 + 4)
2x2 + x + 8
=
(Ax + B)(x2 + 4) + Cx + D
cancel common factors in
numerators and denominators
2x2 + x + 8
=
Ax3 + 4Ax + Bx2 + 4B + Cx + D
multiply out all polynomials
2x2 + x + 8
=
Ax3 + Bx2 + 4Ax + Cx + 4B + D
rearrange terms in decreasing
order of powers of x
2x2 + x + 8
=
Ax3 + Bx2 + (4A + C)x + 4B + D
factor out an x-term
Now we can compare the coefficients of all the x-terms on both sides of this equation. We have 0 x3-terms on
the left and A x3-terms on the right, so this implies that A = 0. We have 2 x2-terms on the left and B x2-terms on
the right, so this implies that B = 2. We can use A = 0 and B = 2 to find the remaining 2 unknown quantities.
We set up a system of two linear equations:
1
=
4A + C
coefficient of x
8
=
4B + D
constant terms
We already said that A = 0 and B = 2, so we can substitute these into the linear equations and solve for the
unknown quantities:
1
=
4(0) + C
substitute in A = 0
1
=
C
simplify
8
=
4(2) + D
substitute in B = 2
8
=
8+D
simplify
0
=
D
subtract 8 from both sides
Thus, we have:
A
=
0
and
B
=
2
and
C
=
1
and
D
=
0
Finally, we substitute our values for A, B, C and D into our Basic Equation to complete the partial fraction
decomposition:
Functions
Repeated Quadratic Factors – 179
Ax + B
2
2x + x + 8
(x2 + 4)(x2 + 4)
=
2
2
(x2 + 4)(x2 + 4)
(x2 + 4)2
x +4
2
2x + x + 8
Cx + D
+
=
+
2
x +4
x
(x2 + 4)2
Basic Equation
substitute in A = 0, B = 2, C = 1 and D = 0
partial fraction decomposition
answer
To check our answer, we try to reverse the process we just completed. We simplify the partial fraction
decomposition so that it matches our original rational expression.
2
2
x +4
+
x
(x2 + 4)2
=
=
(2 )
x
+
2
(x2 + 4) (x2 + 4)2
2x +4
(2 )
(x2 + 4)2
2x +4 +x
2
=
180 – Repeated Quadratic Factors
2x + 8 + x
(x2 + 4)2
rewrite expression in terms of LCD
combine fractions with like denominators
expand polynomial in numerator
matches our original rational expression
check
Functions
Advanced
To decompose a rational function into its component parts using the method of partial fractions.
No definitions on this page.
Leonardo Pisano Fibonacci (1170?-1250?) was so famous in his time he even attracted the attention of royalty.
In 1225, Holy Roman Emperor Frederick II traveled to Pisa to hold a mathematical tournament to test
Fibonacci\'s skills. Fibonacci was the only competitor to correctly answer all three of the tournament questions.
Here is the third question, posed as a riddle:
Three men owned, respectively, a half, a third, and a sixth of an unknown quantity of money. Each man took an
unspecified amount, leaving none left. Then each man returned a half, a third, and a sixth of what he had first
taken. The returned money was divided into equal thirds and redistributed to the men. This resulted in each
man acquiring his fair share. How much money did each man own?
Functions → Rational Functions → Partial Fractions → Mixed Factors
Before you look at this thought, you should go and review the other thoughts under Partial Fractions. In this
thought, we will combine all the techniques we used in the other thought to decompose rational expressions into
their partial fraction components.
If we have a mixture of linear and quadratic factors in the denominator, then we apply the method of partial
fractions separately for each of those factors. For instance, if we have x(x2 + 1) in the denominator, then the
partial fraction decomposition takes the form:
A
x
+
Bx + C
2
x +1
partial fraction decomposition
mixed linear and quadratic factors
Let's look at an example of how it works in practice. Again, we give you the method of partial fraction
decomposition for your convenience.
Functions
Mixed Factors – 181
Decomposition of N(x) / D(x) into Partial Fractions
1.
Divide if improper: If N(x) / D(x) is an improper fraction, divide the numerator into the
denominator to obtain:
=
(polynomial)
+
where N1(x) is the remainder from the division of N(x) by D(x).
Then apply Steps 2, 3, and 4 (below) to the proper rational expression N1(x) / D(x).
2.
Factor denominator: Completely factor the denominator into the form:
(px + q)m
(ax2 + bx + c)n
and
where (ax2 + bx + c) is irreducible.
3.
Linear factors: For each factor of the form (px + q)m, the partial fraction decomposition
must include the following sum of m fractions:
4.
Quadratic factors: For each factor of the form (ax2 + bx + c)m, the partial fraction decomposition
must include the following sum of n fractions:
Mixed Linear and Quadratic Factors
Write the partial fraction decomposition for the rational expression. Check the result algebraically.
x
(
2
)
( 2x − 1) 2x + 1
solution
The degree of the numerator is 1 and the degree of the denominator is 3. Since 1 < 3, we do not have to
perform long division on this rational expression and can proceed with factoring the denominator.
x
(
2
)
( 2x − 1) 2x + 1
=
x
(
)
2
( 2x − 1) 2x + 1
denominator is already factored
Note that we have a mixture of linear and irreducible quadratic factors in the denominator. We now build a
Basic Equation using the guidelines outlined above for each linear and quadratic factor. The Basic Equation
turns out to look like:
x
(
2
)
( 2x − 1) 2x + 1
182 – Mixed Factors
=
A
2x − 1
+
Bx + C
2
2x + 1
Basic Equation
Functions
As always, our next step is to multiply both sides of the Basic Equation by the denominator on the left and
cancel common factors in the numerators and denominators on the right.
x
=
⎛ A + Bx + C ⎞ ( 2x − 1) ( 2x2 + 1)
⎜ 2x − 1
⎟
2
2x + 1 ⎠
⎝
multiply both sides by (2x – 1)(2x2 + 1)
x
=
A(2x2 + 1) + (Bx + C)(2x – 1)
cancel common factors in numerators and
denominators
x
=
2Ax2 + A + (2Bx2 – Bx + 2Cx – C)
start expanding polynomials
x
=
2Ax2 + 2Bx2 – Bx + 2C + A – C
rearrange terms in order of decreasing
powers of x
x
=
(2A + 2B)x2 + (–B + 2C)x + A – C
factor out x-terms
Once again, we compare coefficients on both sides of the equation. We set up a system of three linear equations
with three unknowns:
0
=
2A + 2B
coefficient of x2
1
=
–B + 2C
coefficient of x
0
=
A–C
constant term
If we solve the third equation for A, we get:
0
=
A–C
third linear equation
C
=
A
solve for C
If we solve the first linear equation for A, we get:
0
=
2A + 2B
first linear equation
0
=
A+B
divide both sides by 2
–B
=
A
subtract B from both sides
We substitute A = –B and C = A into the second equation to find a value for A:
1
=
–B + 2C
second linear equation
1
=
A + 2A
substitute in A = –B and C = A
1
=
3A
simplify
=
A
divide both sides by 3
1
3
Once we have found A, we back-substitute into the first and third equations to find out that B and C are:
Functions
Mixed Factors – 183
A
=
B
=
C
=
1
and
3
−
1
and
3
1
3
Now that we have found values for A, B and C, we can plug these values into our Basic Equation:
x
(
)
2
( 2x − 1) 2x + 1
2
)
=
)
=
( 2x − 1) 2x + 1
x
(
2x − 1
Bx + C
+
−1
2
( 2x − 1) 2x + 1
3
2⋅ x − 1
Basic Equation
2
2x + 1
1
x
(
A
=
3
+
1
⋅x +
3
substitute in values for A, B and C
2
2x + 1
1
3( 2x − 1)
−
x−1
(
rewrite rational expression
partial fraction decomposition
answer
)
2
3 2x + 1
In order to check our answer, we rewrite all rational expressions in terms of the Least Common Denominator
(LCD):
1
3( 2x − 1)
−
(
x−1
2
)
3 2x + 1
=
(
)
2
1 2x + 1
(
)
2
( x − 1) ( 2x − 1)
−
(
2
=
2x + 1 − ( x − 1) ( 2x − 1)
(
=
(
2
=
=
184 – Mixed Factors
)
2x + 1 − 2x − 3x + 1
(
)
2
3( 2x − 1) 2x + 1
2
=
)
2
3( 2x − 1) 2x + 1
2
2
)
3( 2x − 1) 2x + 1
3( 2x − 1) 2x + 1
rewrite rational
expressions
in terms of LCD
combine fractions with like
denominators
expand polynomial in
numerator
2
2x + 1 − 2x + 3x − 1
(
)
2
3( 2x − 1) 2x + 1
3x
(
)
2
3⋅ ( 2⋅ x − 1) ⋅ 2⋅ x + 1
x
(
2
)
( 2x − 1) 2x + 1
simplify numerator
simplify numerator
cancel like factors in
numerators and
denominators
check
Functions
There are two main categories of functions studied in algebra. By now you should already be familiar with
algebraic functions, which include polynomial functions and rational functions. In this section of the Algebra
Brain, we will study the second main category: transcendental functions.
An algebraic function is one that can be expressed as a finite number of sums, differences, multiples, quotients,
and radicals involving xn. A transcendental function is one that does not include any of the above operations.
Common transcendental functions include logarithmic functions, exponential functions, trigonometric functions
(including inverse trigonometric functions) and hyperbolic functions. In the Algebra Brain, we will only
concentrate on exponential and logarithmic functions.
An exponential function is one in which the independent variable becomes the exponent.
Logarithmic functions are used to solve exponential functions for the independent variable. A logarithm is the
inverse of an exponent just as subtraction is the inverse of addition.
Functions
Transcendental Functions – 185
186 – Transcendental Functions
Functions
Advanced
To define the exponential function and see how it can be used to model real world situations.
exponential function – a function with the independent variable as an exponent. A function with an equation of
the form y = abx.
base – the number b in the expression bn.
Gottfried Wilhelm Leibnitz (1646-1716), one of the fathers of calculus, first wrote of the binary numbers (base 2)
in his treatise De Progressione Dyadica. He also corresponded with a Jesuit missionary in China, Pere Joachim
Bouvet. Through Bouvet, Leibnitz realized a direct connection between the hexagrams of the Chinese I Ching
and his own binary system, as shown below:
Functions → Transcendental Functions → Exponential Functions
Recall that a polynomial function usually involves powers of x, such as x4, x3, and x2. Here we have a
variable number raised to some integer power. Now we are going to flip things around and raise an integer
number to some variable power. For example, we might need to deal with 2x. Note that 2x ≠ x2 unless x
happens to equal 2.
A function that involves real numbers raised to variable powers is called an exponential function, because the
variable is now the exponent. The real number is called the base of the exponential function and is read "base
a", where a is the real number, such as 2.
Definition of Exponential Function
The exponential function f with base a is denoted by
f(x)
bax
=
exponential function
where b ≠ 0, a > 0, a ≠ 1, and x is any real number.
Note:
The base a = 1 cannot be used in an exponential function because then we have:
f(x)
=
1x
=
1
which is a constant function and not an exponential function.
In order to be able to work effectively with exponential functions, you will need to be very familiar with the
properties of exponents. It might be a good time at this point for you to review how exponents work before
continuing with this discussion.
Evaluating exponential functions can be tricky. It is relatively simply to evaluate 4x = 64 when x = 3 or 4x = 2
when x = 1/2. However, to evaluate 4x for any real number x, it is necessary to include all the irrational numbers
as well, such as √2. For most applications involving irrational exponents, we have to work with approximate
values that we obtain by using a calculator.
Functions
Exponential Functions – 187
Let's look at the behavior of exponential functions. Their graphs are quite a bit different from any other type of
function you may have studied so far. However, the graph of an exponential function is distinctive enough to
be immediately recognizable as such once you know what to look for.
In the same coordinate plane, sketch the graph of each function.
a.)
f(x)
=
2x
b.)
g(x)
=
4x
and
solution
We plot points using standard plotting techniques. We assign integer values for x, then evaluate f(x) and g(x) at
those points, listing them in a table and then plotting them in the plane.
x
f(x)
g(x)
–2
1
4
1
16
–1
1
2
1
4
0
1
2
3
1
2
4
8
1
4
16
64
Note that the function g is increasing much more rapidly than the function f. This will easily show up on our
graph, which is shown below:
answer
For values of x > 1, the function g increases faster than f. For values of 0 < x < 1, the function g decreases faster
than f. This is true in general of exponential functions. If we have two exponential expressions ax and bx such
that a > b, then ax > bx for values of x > 1 and ax < bx for values of x < 1.
In the same coordinate plane, sketch the graph of each function.
a.)
F(x)
=
2–x
b.)
G(x)
=
4–x
188 – Exponential Functions
and
Functions
solution
Again we are not restricting x to any particular domain, allowing for x to be any real number. We list some
values for each function based on convenient integer values for x. We could just as easily let x be irrational.
x
–3
–2
–1
0
F(x)
8
4
2
1
G(x)
64
16
4
1
1
1
2
1
4
2
1
4
1
16
Note that this table is basically the mirror image of the table in the previous example. Also note that both
graphs are decreasing. The graph of G(x) = 4–x is decreasing more rapidly than the graph of F(x) = 2–x. The
graph of these functions looks like:
answer
The graphs of F and G are the mirror images of f and g. That is, F is a reflection of f (in the y-axis).
Notice that each of these graphs has only one y-intercept. They also have a horizontal asymptote located
at y = 0. They are also continuous for all real values of x. We can summarize the behavior of the graphs
of y = ax, a > 1 and y = a–x, a > 1 as follows.
Graph of y = ax, a > 1
· Domain: (– , )
· Range: (0, )
· Intercept: (0, 1)
· Increasing
· x-axis is horizontal asymptote
(ax → 0 as x → – )
· Continuous
Graph of y = a–x, a > 1
· Domain: (– , )
· Range: (0, )
· Intercept: (0, 1)
· Decreasing
· x-axis is horizontal asymptote
(a–x → 0 as x → )
· Continuous
Finally, we need to see what happens when we transform the function f(x) = ax so some new function of the
form:
g(x)
=
b ± ax + c
Let's see if we can’t guess what will happen to the function. If we add or subtract b from the function f, all we
are doing is moving the y-intercept of the function. So if we have the function h(x) = 2x – 1, the graph of this
function will look identical to the function f(x) = 2x given in the first example except that we move the entire
function 1 unit down the y-axis so that its new y-intercept is (0, 0).
If we have k(x) = 2x + 1, then we are also changing the y-intercept of the function. The new function k will have
its y-intercept located at the point (0, 2) when x = 0, because k(0) = 20 + 1 = 21 = 2. The function k(x) = 2x + 1 will
increase faster than f(x) = 2x.
Functions
Exponential Functions – 189
If we have j(x) = –2x, this is equivalent to –f(x), which means that the graph of j(x) can be obtained by reflecting
f(x) in the x-axis.
Finally, if we have m(x) = 2–x, this is equivalent to f(–x), which means that the graph of m(x) can be obtained by
reflecting f(x) in the y-axis.
Each of these four situations is shown below.
f(x) = 2x
h(x) = 2x – 1
f(x) = 2x
k(x) = 2x + 1
f(x) = 2x
j(x) = –2x
f(x) = 2x
m(x) = 2–x
190 – Exponential Functions
Functions
Advanced
To learn the definition of logarithmic functions.
common logarithmic function – logarithmic function that uses base 10.
John Napier (1550-1617), the Scottish mathematician who invented logarithms, also created an early calculating
device in the form of a chess board. He expressed his numbers in normal base ten notation, but the means of
calculation relied on base two. Essentially, he used the fact that the sum or product of any whole number can be
written as the sum or product of powers of 2. For instance, the number 17 is equal to 24 + 20 = 16 + 1 = 17.
Functions → Transcendental Functions → Logarithmic Functions
What are logarithms and what is a logarithmic function? We can define a logarithm in several different ways.
One way is graphically. To understand the graphical interpretation of the logarithm, you will first need to
understand inverse functions and the exponential function. A logarithmic function can be defined as the inverse
of the exponential function. That is, the domain of an exponential function is the range of a logarithmic function
and the range of an exponential function is the domain of a logarithmic function. The main reason why
exponential functions all have an inverse is because they pass the Horizontal Line Test. In other words, any
horizontal line drawn across the graph of the exponential function only crosses the exponential function once.
The formal definition of a logarithmic function is as follows:
Definition of Logarithmic Function
For x > 0 and 0 < a ≠ 1,
y
=
loga x
if and only if
x
=
ay
The function given by
f(x)
=
loga x
is called the logarithmic function with base a.
Note:
The equations y = loga x and x = ay are equivalent. The first equation is in logarithmic form and the
second equation is in exponential form. This will become very useful knowledge when we begin
solving exponential and logarithmic equations.
A logarithm is an exponent, plain and simple. The expression loga x is the exponent the base a must be raised to
obtain x. For example, log 2 16 = 4 because 2 must be raised to the fourth power (24) to get 16.
A logarithm can have any base we need. Common bases include base 2 (binary), base 8 (octal), and base 16
(hexadecimal). The most common base by far is base 10 (decimal). Since it is so common, we don't even write
log10 x, we simply use log x. A logarithmic function with base 10 is called the common logarithmic function.
Logarithmic functions have a number of properties that we can exploit when dealing with them.
Functions
Logarithmic Functions – 191
Properties of Logarithms
1.)
loga 1 = 0 because a0 = 1
2.)
loga a = 1 because a1 = a
3.)
loga ax = x because ax = ax
4.)
If loga x = loga y, then x = y
Graphs of Logarithmic Functions
What does the graph of a logarithmic function look like? We have already stated that logarithmic functions are
inverses of exponential functions. By definition, the graphs of inverse functions are reflections of each other in
the line y = x.
In the same coordinate plane, sketch the graph of each function.
a.)
f(x)
=
3x
b.)
g(x)
=
log3 x
solution
a.)
For the function f(x) = 3x, we can construct a table of values as follows:
x
f(x)
–2
1
9
–1
1
3
0
1
2
3
1
3
9
27
We can then plot these points and connect them with a smooth curve to obtain the red curve on the
graph shown below.
b.)
Since g(x) = log3 x is the inverse of f(x) = 3x, the graph of g(x) is the reflection of f(x) about the line
y = x, as shown by the blue curve on the graph shown below.
answer
Compare the logarithmic curve (blue) to the exponential curve (red) above. The exponential function has one yintercept and one horizontal asymptote located at y = 0. By contrast, the logarithmic function as one x-intercept
and one vertical asymptote located at x = 0. This behavior is typical for logarithmic functions, although we can
192 – Logarithmic Functions
Functions
move them around the Cartesian plane in much the same fashion as we move exponential functions. The basic
characteristics of logarithmic graphs are summarized below:
Graph of y = loga x, a > 1
· domain: (0, )
· range: (– , )
· intercept: (1, 0)
· increasing
· y-axis is a vertical asymptote
(loga x → – as x → 0–)
· continuous
· reflection of the graph of y = ax
about the line y = x
Find the domain, vertical asymptote and x-intercept of the following logarithmic function. Then sketch its
graph.
f(x)
=
–log6 (x + 2)
solution
First of all, remember that a logarithm of any kind is just an exponent. The negative sign tells us that this is a
negative exponent. The graph of an exponential function with a negative exponent is reflected about the y-axis.
Since a logarithmic function is the inverse of an exponential function, it stands to reason that a negative
logarithmic function is reflected about the x-axis. Therefore, the graph of this function looks like the upsidedown image of the logarithmic function given above.
Next, we need to find the domain of this function. Logarithmic functions are only defined when the argument
of the function is greater than zero. In this case, the argument of f(x) is (x + 2). In order for (x + 2) = 0, x has
to be equal to –2. Therefore, the domain of this function is all values of x such that x > –2.
answer
The vertical asymptote also occurs at the point where the argument of the function goes to zero. We already
stated that f(–2) = log6 (–2 + 2) = log6 (0) = undefined, so the vertical asymptote for this function is the line
x = –2. answer
The x-intercept occurs at the point f(x) = 0. Therefore, we need to solve the equation f(x) = 0:
–log6 (x + 2)
=
0
set f(x) = 0
x+2
=
1
equivalence of logarithm and exponent:
60 = (x + 2) = 1
x
=
–1
subtract 2 from both sides
Thus, the x-intercept is located at (–1, 0).
The graph of this function looks like:
Functions
Logarithmic Functions – 193
answer
194 – Logarithmic Functions
Functions
Advanced
To review the properties of logarithms, which are closely related to the properties of exponents.
No definitions on this page.
Logarithms are a result of John Napier's (1550-1617) interest in astronomy. Astronomy relies heavily on tedious
calculation of trigonometric functions. Napier invented logarithms to speed the calculations. Henry Biggs worked
with Napier and constructed a set of standardized logarithmic tables. These tables were used until computers
and calculators became powerful enough to render the tables obsolete (around the mid to latter part of the 20th
century).
Functions → Transcendental Functions → Logarithmic Functions → Properties of Logarithms
Logarithms are exponents. As such, it should come as no surprise that the logarithm operator has several
properties associated with it. Many of these properties are similar to those exhibited by exponents. Again, this
is because a logarithm is exactly the same as an exponent.
Here is the list of basic properties, which are also found in the thought Logarithmic Functions:
Properties of Logarithms
1.)
loga 1 = 0 because a0 = 1
2.)
loga a = 1 because a1 = a
3.)
loga ax = x because ax = ax
4.)
If loga x = loga y, then x = y
Change of Base
In addition to these basic properties, we often find it useful or necessary to change the base of a logarithm to
something more practical. For instance, many scientific calculators cannot evaluate log5 12 because they are
not equipped to handle base 5 numbers. However, they can evaluate logs of base 10 (common logs). So if we
can somehow transform log5 to log10, we can go ahead and use our calculator to evaluate the expression log5 12
once it has been converted from base 5 to base 10. This is known as "change-of-base". The formula for
changing from one base to another base is given by:
Change-of-Base Formula
Let a, b, and x be positive real numbers such that a 1 and b 1.
Then loga x is given by:
loga x
Note:
Functions
=
Change-of-Base Formula
One way to think of the change-of-base formula is that logarithms to base a are simply constant
multiples of logarithms to base b. The constant multiplier is 1 / (logb a).
Properties of Logarithms – 195
Evaluate the logarithm using the change-of-base formula. Round the answer to three decimal places.
a.)
log3 7
b.)
log9 0.4
c.)
log15 1250
solution
Although we are free to change the base of each logarithm to any base we want, a convenient base is 10, since
most calculators can deal with log base 10 rather easily. Thus, we will use the change-of-base formula to
convert each logarithm to base 10.
a.)
log3 7
log10 7
=
log10 3
0.845
≈
b.)
log9 0.4
log 10 0.4
=
log 10 9
−0.398
0.954
≈
log15 1250
simplify, rounding to three decimal places
answer
1.771
≈
c.)
use a calculator to evaluate each logarithm
round to three decimal places
0.477
≈
Change-of-Base Formula
let b = 10, x = 7 and a = 3
–0.417
=
≈
≈
Change-of-Base Formula
let b = 10, x = 0.4, and a = 9
use a calculator to evaluate each logarithm
round to three decimal places
simplify, rounding to three decimal places
answer
log 10 1250
log 10 15
3.097
1.176
2.634
Change-of-Base Formula
let b = 10, x = 1250, and a = 3
use a calculator to evaluate each logarithm
round to three decimal places
simplify, rounding to three decimal places
answer
Simplifying Logarithms
Many times we are given a logarithm that is unnecessarily complicated. There are a number of properties we
can use to transform a complicated logarithm into something that is much simpler for us to work with. These
are collectively given as some more properties of logarithms, below:
196 – Properties of Logarithms
Functions
More Properties of Logarithms
Let a be a positive number such that a 1. Let n be a real number.
If u and v are positive real numbers, then the following properties are true.
1.)
loga (uv)
2.)
loga un
3.)
Note:
=
loga u + loga v
Product Property of Logarithms
=
loga u – loga v
Quotient Property of Logarithms
=
n loga u
Power Property of Logarithms
There is no general property that can be used to rewrite loga (u
that loga (u v) = loga u + loga v.
v). More specifically, it is not true
In order to see why the first property is true, consider the following. Let
x
=
loga u
y
=
loga v
and
Then their corresponding exponential forms are given by:
ax
=
u
exponential form
ay
=
v
exponential form
If we multiply u and v, we get:
uv
=
a xa y
multiply u and v together
=
ax + y
add exponents of like bases
The corresponding logarithmic form of uv is:
loga (uv)
=
x+y
logarithm form of uv
=
loga u + loga v
substitute in x = loga u and y = loga v
proof
We can prove the other two properties with a similar approach.
Use the properties of logarithms to write the logarithm as the sum, difference, and/or constant multiples of
logarithms. Assume all variables are positive.
Functions
a.)
log8 x3
b.)
log xyz
c.)
log ⎛⎜
⎞
⎟
⎝ x + 1⎠
x
2
Properties of Logarithms – 197
solution
a.)
Here we clearly need to apply the Power Property of Logarithms to move the exponent of x3 out in
front of the logarithm:
log8 x3
b.)
=
Power Property of Logarithms
answer
Since we have a product of three variables as the argument of our logarithm, the Product Property of
Logarithms tells us that we can write this as the sum of three logarithms, one for each variable. Note
that we do not specify the base of the log, so it is implied that the base is 10.
log xyz
c.)
3 log8 x
=
Product Property of Logarithms
answer
log x + log y + log z
This one is tricky. First, we use the Quotient Property of Logarithms to separate the rational
expression inside the parentheses into the difference of two logarithms. Then we note that the
denominator of the rational expression is the equivalent of (x2 + 1)1/2, which means that we next need
to apply the Product Property of Logarithms to the second term in our difference:
log ⎛⎜
⎞
⎟
⎝ x + 1⎠
x
2
=
(
)
2
(2
log ( x) − log x + 1
=
log ( x) −
1
2
(2
1
)2
=
198 – Properties of Logarithms
Quotient Property of Logarithms
log ( x) − log x + 1
convert square root to fractional exponent
)
log x + 1
Product Property of Logarithms
answer
Functions
Advanced
To learn about base e and some of its applications.
natural logarithm – logarithms using the base e.
David Hilbert's (1862-1943) reputation as a mathematician was founded on his research into invariant theory. An
invariant is an expression that remains the same under various transformations. For instance, multiplying a
linear equation by 2 does not change the solution of the equation. Most research on invariants up until that time
required massive amounts of calculation. Hilbert took a path that did not require explicit calculation. Other
mathematicians who studied invariant theory were taken aback. One described Hilbert's approach as "not
mathematics, but theology."
Functions → Transcendental Functions → Logarithmic Functions → Natural Base e
Natural Exponential Function
One of the most useful and interesting numbers in mathematics is the number e, where
e
≈
2.718281828….
Along with π, the number e is one of the most famous irrational numbers.
The number e has a number of applications, particularly when we use it as a base in exponential functions.
The function f(x) = ex is called the natural exponential function. The graph of this function is shown below.
Notice its similarity to other exponential functions.
What makes the natural exponential function different from all other exponential functions is that it is the only
function whose slope at any point along the curve is equal to the function evaluated at that point. For instance,
if we draw a line so that it is tangent to the curve at the point (1, e), the slope of the tangent line is e. If we draw
the tangent line at the point (2, e2), the slope of the tangent line is e2. You will learn much more about the
behavior of e and the slope of curves in calculus.
Functions
Natural Base e – 199
Since the number e is a base used in exponential functions, all the properties of exponential functions apply
equally to the natural exponential function. For example:
exe2
=
e2 + x
add exponents of like bases
(ex)2
=
e2x
multiply exponents
One particular application that may be of interest (pardon the pun) is compound interest on a bank account. It
can be shown that the expression
n
⎛1 + 1 ⎞
⎜
⎟
n⎠
⎝
approaches the number e as n approaches infinity. Compound interest is defined in terms of this expression.
For interest that is compounded continuously, n → , and thus, we replaces the expression above with e.
Natural Logarithmic Function
In addition to its usefulness in exponential functions, e is so useful that we use it as one of the two most
common bases for logarithms (base 10 is the other base we usually use). We even give the logarithm of base e
a special symbol:
ln x
logarithm of base e of x
The symbol "ln" stands for natural logarithm. The symbol "ln x" is read as "el en of x".
More formally, we can define the natural logarithmic function as:
The Natural Logarithmic Function
The function defined by
f(x)
=
loge x
=
ln x
x>0
is called the natural logarithmic function.
All of the properties that apply to common logarithms apply equally to natural logarithms. The number e is just
another base, like base 2, base 6, or base 10, so it only stands to reason that it would obey those properties.
Properties of Natural Logarithms
1.)
ln 1 = 0 because e0 = 1
2.)
ln e = 1 because e1 = e
3.)
ln ex = x because ex = ex
4.)
If ln x = ln y then x = y
The domain and the range of the natural logarithmic function are the same as for the common logarithmic
function. The argument of the natural logarithm can never be less than or equal to zero. The graph of the
natural logarithmic function is shown below. Notice its similarity to the common logarithmic function.
200 – Natural Base e
Functions
If you compare this to the graph of the natural exponential function given above, you can see that the domain of
the natural exponential function is the range of the natural logarithmic function. Since this is the case, we can
conclude that the natural exponential function and the natural logarithmic function are inverses of each other.
Shown below are both functions on the same graph. Notice that they are reflections of each other about the line
y = x, just like the normal exponential function and the normal logarithmic function:
Functions
Natural Base e – 201
202 – Natural Base e
Functions
Advanced
To solve exponential and logarithmic equations.
inverse properties – one property that can be used to undo another property, and vice versa.
In 2002, the executives and auditors at Enron, WorldCom, Arthur Andersen and a host of other companies were
commended by the Ig Nobel committee (a spoof of the actual Nobel committee) "for adapting the mathematical
concept of imaginary numbers for use in the business world."
Functions → Transcendental Functions → Exponential Functions → Exponential / Logarithmic Eq
So far in this section of the Algebra Brain, we have looked at the definitions and graphs of exponential and
logarithmic functions. Now it is time to put the two together and start solving equations involving exponents
and logarithms.
Let's start simple. Suppose we have the equation 2x = 64. How do we go about solving this equation for x?
One of the properties of exponents states that ax = ay if and only if x = y. We can solve 2x = 64 by rewriting the
equation in the form 2x = 26. This implies that x = 6, which is indeed the case.
Most of the time, the problem is not so simple and we have to rely on other means to solve the equation. What
if we had 2x = 9? This is not quite so easily solved using the property we used before. However, we can take
advantage of the fact that exponential and logarithmic functions are inverses of each other. Thus, if we are
presented with an exponential equation, we can use a logarithm to solve it. If we have a logarithmic equation,
we can exponentiate to solve it. The following are the inverse properties of logarithmic and exponential
functions:
Inverse Properties of Logarithmic and Exponential Functions
base a
loga ax = x
1.)
base e
ln ex = x
eln x = x
2.)
If we use the above properties to solve 2x = 9, we can take the log2 of both sides to obtain:
2x
=
9
original equation
logz 2x
=
log2 9
take log2 of both sides
x
=
log2 9
log2 2x = x because 2x = 2x
Using the Change-of-Base Formula and a calculator, we can approximate a value for x:
x
=
=
≈
log2 9
log( 9)
log( 2)
3.17
value for x in log2 form
Change-of-Base Formula
approximate logarithms using a calculator
As we can see, this means that 23.17 ≈ 9, which makes sense since 9 is between 8 (23) and 16 (24).
Functions
Exponential / Logarithmic Equations – 203
Here are a few guidelines to use when solving exponential and logarithmic equations:
Guidelines for Solving Exponential and Logarithmic Equations
1.)
To solve an exponential equation:
First isolate the exponential expression, then take the logarithm of both sides and solve for the
variable.
2.)
To solve a logarithmic equation:
Rewrite the equation in exponential form and solve for the variable.
As always, it is nice to see these guidelines in action, so here are a couple of examples.
Solve the exponential equation algebraically. Round the result to three decimal places.
7 – 2e3x
=
1
solution
Since we have an exponential equation, we apply the first guideline given above. We start by rearranging the
equation so that the exponential term (2e3x) is all by itself. Then we proceed to take the natural logarithm of
both sides (since we have base e). This will allow us to solve for the variable by applying one of the inverse
properties given above.
7 – 2e3x
=
1
given equation
6
=
2e3x
subtract 1 from both sides
add 2e3x to both sides
3
=
e3x
divide both sides by 2
ln 3
=
ln e3x
take natural log of both sides
ln 3
=
3x
since ln ex = x, it stands to reason that ln e3x = 3x
=
x
divide both sides by 3
≈
x
use a calculator to evaluate the left hand side
round to three decimal places
answer
ln⋅ 3
3
0.366
Solve the logarithmic equation algebraically. Round the result to three decimal places.
ln x + ln (x – 2)
=
1
solution
This is somewhat tricky. First of all, we recognize that we can, in fact, combine the logarithms into a single
logarithm. Then we exponentiate both sides. This will allow us to use the second inverse property of logarithms
to solve for the variable.
204 – Exponential / Logarithmic Equations
Functions
ln x + ln (x – 2)
=
1
given equation
ln (x(x – 2))
=
1
sum of logarithms of same base is equal to logarithm
of the product of the arguments of the logarithms
ln (x2 – 2x)
=
1
simplify expression inside parentheses
(2 )
=
e1
exponentiate both sides
x2 – 2x
=
e
since eln x = x, it makes sense that eln (argument) = (argument)
x2 – 2x + 1
=
e+1
add 1 to both sides to complete the square on the left
(x – 1)2
=
e+1
factor left hand side
x–1
=
take square root of both sides
x
=
add 1 to both sides
x
≈
e
ln x − 2x
2.928
round to three decimal places
answer
Note that we discarded the possible negative solution for x. This is because if x is –0.928, we have
ln (–0.928 – 2) in our original equation, which will give us ln (–2.928). We can never take the
logarithm of a negative number, so we can disregard any negative values of x.
Functions
Exponential / Logarithmic Equations – 205
206 – Exponential / Logarithmic Equations
Functions
Exponential and logarithmic functions have a wide range of applications in the real world. Exponential
functions are used to model exponential growth and decay, particularly population growth and radioactive
decay.
We can also use an exponential function to compile statistical data. The famous "bell curve" is one such device.
A group of test scores, for example, will often be displayed as a bell curve, where the peak of the curve
represents the median scores on the exam. High scores are on the right side of the curve, and low scores are on
the left side of the curve.
Although an exponential growth model can be used to predict population growth under certain conditions for
certain organisms, a much more realistic model is the logistics growth model. An environment can only support
so much growth. After a while, the rate of growth tapers off as the resources of the environment diminish. The
logistics growth model reflects this more realistic trend of population growth.
Finally, we have logarithmic models, which use logarithms to give us information about a situation. Often a
logarithmic model is more convenient because the numbers involved are really large or really small. Thus, we
only talk about the exponent, which is a more manageable number. Situations that involve logarithms include
the pH scale for measuring hydrogen ion concentrations in solution, the decibel scale for measuring relative
sound intensity, and the Richter scale, used to measure the magnitude of earthquakes.
Thus, we have four basic models where logarithmic and exponential functions come in handy:
1.)
Model
Exponential Growth / Decay:
• If b > 0, exponential growth
• If b < 0, exponential decay
Basic Form of Equation
y = aebx
− ( x− b)
2.)
Gaussian:
y
=
3.)
Logistics Growth:
y
=
ae
2
c
1
− ( x− c)
1 + be
4.)
Logarithmic:
d
y
=
a + b ln x
y
=
a + b log10 x
or
The graphs of the basic forms of these models are shown below:
Exponential Growth Model
Functions
Exponential Decay Model
Exponential and Logarithmic Models – 207
Gaussian Model
Logistics Growth Model
Note: The logistics growth model contains
two horizontal asymptotes.
Logarithmic Model (base e)
Logarithmic Model (base 10)
In the thoughts below this one, we will look at each of the models in more detail.
208 – Exponential and Logarithmic Models
Functions
Advanced
To demonstrate how to use exponential functions to model real world situations.
No definitions on this page.
Omar Khayyam (c.1048-1131) was a Persian mathematician, poet, astronomer, and philosopher. His greatest
mathematical achievement was the application of Greek geometric methods to solve cubic equations. Greek
algebra used geometry to solve algebraic problems involving quadratic equations. Khayyam used similar
techniques to solve cubic equations, using the conic sections.
Functions → Transcendental Functions → Exponential Functions → Exponential / Logarithmic Eq →
Logarithmic / Exponential Models → Exponential Growth and Decay
Exponential growth and decay is one of the more common applications of exponential functions. The basic
equation for any type of growth/decay problem is given by:
y
aebx
=
exponential growth / decay equation
If b > 0, then we have an exponential growth problem. If b < 0, then we have an exponential decay problem.
Exponential Growth
Many types of organisms experience exponential growth in the initial stages of their population. Bacteria are a
prime example, but this also applies to viruses, parasites, even larger organisms, such as frogs and snakes, if
they have enough food to eat.
Exponential growth equations often take the form:
y(t)
=
Ce
kt
exponential growth equation
where C is the initial amount of whatever is growing, k is a constant of proportionality and t is time.
A certain colony of genetically engineered bacteria doubles its population every 3 hours. Suppose we start out
with a mere 250 bacteria. How long will it take to for the colony to reach 100,000 members?
solution
First, we always write down what we are given in any kind of word problem. Here, we know that we start with
250 bacteria. After 3 hours, this number has doubled to 500. Thus we have:
C
=
250
number of bacteria we started with
y(3)
=
500
number of bacteria after 3 hours
t
=
3
length of time needed to double colony size
k
=
?
constant of proportionality we don't know
In order to find out how long it will take for the colony to reach 100,000 members, we first need to find k, the
constant of proportionality:
Functions
Exponential Growth and Decay – 209
y(t)
=
Cekt
exponential growth equation
500
=
250e3k
substitute in y(3) = 500, C = 250, and t = 3
2
=
e3k
divide both sides by 250
ln 2
=
ln e3k
take natural log of both sides
ln 2
=
3k
Power Property of Logarithms and ln e = 1
=
k
divide both sides by 3
≈
k
approximate k to three decimal places
ln( 2)
3
0.231
Now that we have an approximate value for k, we can go ahead and use the exponential growth equation again,
this time letting y(t) = 100,000, C = 250, and k ≈ 0.231. This will let us solve the equation for t.
y(t)
=
Cekt
exponential growth equation
100,000
=
250e0.231 t
substitute in y(t) = 100,000, C = 250 and k = 0.231
400
=
e0.231 t
divide both sides by 250
ln 400
=
0.231 t
take natural log of both sides
=
t
divide both sides by 0.231
≈
t
round t to three decimal places
answer
ln( 400)
0.231
25.937
After about 26 hours, our colony of 250 bacteria will have increased to 100,000 members.
Exponential Decay
This is essentially the opposite of exponential growth. Here we start out with some initial amount of material
and over time the amount decreases at an exponential rate. The basic equation for exponential decay looks
almost identical to the equation for exponential growth with one slight difference:
y(t)
=
Ce
− kt
exponential decay equation
Note that we have added a negative sign in front of the k. This tells us immediately that we are dealing with
exponential decay. The rest of the terms have the exact same meaning as they do for exponential growth. C is
our initial amount, y(t) is the amount we have left after time t, and k is the constant of proportionality. The
minus sign in front of the k tells us that the initial amount is decreasing over time.
Radium-226 (226Ra) has a half-life of 1620 years. This means that if we have a 1-kg sample of radium right
now, in 1620 years, the amount of radium we have left will be around 0.5 kg, one-half the mass of our original
sample.
Suppose we have a 10-g sample of 226Ra now. How much will we have left 10 years from now? 20 years?
10,000 years?
210 – Exponential Growth and Decay
Functions
solution
Once again, we have to determine the constant of proportionality, k. We start be letting C = 10 (our original
amount), y(1620) = 5 (one-half our original amount) , and let t = 1620, the half life of radium.
5
=
10e–k (1620)
exponential decay equation with
y(1620) = 5, C = 10, t = 1620
0.5
=
e–1620 k
divide both sides by 10
ln 0.5
=
–1620 k
take natural log of both sides
=
k
divide both sides by –1620
≈
k
approximate value for k
−ln( 0.5)
1620
0.0004279
Now that we know k, we can start using k to help us figure out how much radon we have after 10 years, 20
years, and 10,000 years.
y(10)
=
10e–0.0004279 (10)
substitute in k = 0.0004279, C = 10, and t = 10
y(10)
≈
9.957
evaluate with a calculator
answer
After 10 years, our 10-g sample has only decreased by about 43 milligrams. Now let's see how much we have
left after 20 years:
y(20)
=
10e–0.0004279 (20)
substitute in t = 20
y(20)
≈
9.915
evaluate with a calculator
answer
After 20 years, our 10-g sample has only decreased by 85 milligrams. Finally, let's see how much we have left
after 10,000 years. We can expect that it should be reduced by a significant amount since half of it disappears
after only 1620 years.
y(10000)
=
10e–0.0004279 (10000)
substitute in t = 10000
y(10000)
≈
0.139
evaluate with a calculator
answer
We have only 1.4% of our original sample left after 10,000 years.
Functions
Exponential Growth and Decay – 211
212 – Exponential Growth and Decay
Functions
Advanced
To learn about one of the basic exponential functions used to model data.
normal distribution – the continuous function that the binomial distribution approaches as n, the number of
trials, increases without bound.
bell-shaped curve – The graph of a normal distribution.
Nicolaus Mercator (1619?-1687) is a lesser-known, but still important mathematician. In 1668, he published his
greatest work, Logarithmicotechnia>, wherein he computes a table of logarithms. He also described an equation
that now bears his name. It is a series expansion that computes the area under a hyperbola. Besides his
mathematical skills, he was an inventor, designing a pendulum watch named after Christiaan Huygens, creator of
the pendulum clock.
Functions → Transcendental Functions → Exponential Functions → Exponential / Logarithmic Eq →
Logarithmic / Exponential Models → Gaussian
One of the most useful models for looking at data is the Gaussian model, named after that paragon of
mathematics, Carl Friedrich Gauss.
In general, Gaussian models take the form:
− ( x− c)
y
=
2
c
ae
Gaussian model
This type of model is commonly used in probability and statistics to represent populations that are normally
distributed . For standard normal populations, the model takes the form:
2
−x
y
=
1
σ 2π
e
2
standard normal distribution model
2σ
where σ is the standard deviation (σ is the lowercase of the Greek letter sigma). When we graph the Gaussian
model, we end up with a bell-shaped curve .
Bell-shaped curves are useful in that they can tell us the percentage of a certain event occurring. Let's look at an
example.
A certain game requires us to roll two six-sided dice (abbreviated as 2d6) and add the two numbers together. If
we roll "1" and "5", the total result is "6". The smallest number we can roll is 2 and the largest number we can
roll is 12. The number we roll on 2d6 roughly follows a normal distribution given by:
− ( x− 7)
y
=
0.1445e
15.2
2
2 ≤ x ≤ 12
where x is the number we roll on 2d6. Sketch the graph of this function. From the graph, estimate the average
number we can roll on 2d6.
Functions
Gaussian Model – 213
solution
Using an automatic graphing utility, we can graph the function below:
The peak of the bell-curve corresponds to the mean, or average. In this case, the average number rolled on 2d6
is 7. answer
What the graph illustrates is that we can expect about 50% of our rolls to total 7 or more and about 50% of the
rolls to total 7 or less. We will explore probabilities in much more detail in another section of the Algebra
Brain.
214 – Gaussian Model
Functions
Advanced
To model real-world situations using logarithmic functions.
No definitions on this page.
John Napier (1550-1617) was rumored to be a magician as well as a mathematician. Supposedly he possessed
supernatural powers and kept a black rooster as a supernatural familiar.
Functions → Transcendental Functions → Exponential Functions → Exponential / Logarithmic Eq →
Logarithmic / Exponential Models → Logarithmic
The logarithmic model is used in a variety of situations. In general, the logarithmic model
takes the form:
y
=
a + b ln x
base e
y
=
a + b log x
base 10
Which one we use depends largely on the situation. Base 10 is very commonly used to
measure the magnitude of earthquakes, the intensity of sound, or the hydrogen ion
concentration in aqueous solution (called pH).
The level of sound β (measured in dB), with an intensity of I (measured in watts per square
centimeter W / cm2) is given by:
β (I)
=
10log 10⎛⎜
I
⎞
⎟
⎝ I0 ⎠
level of sound in decibels (dB)
where I0 is an intensity of 10–16 W / cm2, the faintest sound that can be heard by the human
ear.
If the loudest car stereo on earth can produce 165 dB level sounds, what is the intensity of the
sound wave it produces?
Functions
Logarithmic Model – 215
solution
We let β(I) = 165 and solve the equation decibel equation for I:
165
=
10log 10⎛⎜
16.5
=
log 10⎛⎜
16.5
=
log I – log 10–16
Quotient Property of Logarithms
16.5
=
log I – (–16)
simplify second logarithm using Properties of Logarithms
16.5
=
log I + 16
simplify
0.5
=
log I
subtract 16 from both sides
100.5
=
10log I
exponentiate both sides
100.5
=
I
simplify
3.162
≈
I
approximate value of I to three decimal places
answer
I
− 16
⎝ 10
I
− 16
⎝ 10
⎞
⎟
⎠
⎞
⎟
⎠
substitute in β(I) = 165 and I0 = 10–16
divide both sides by 10
Thus, the intensity of the sound wave produced by the car stereo is about 3.162 W / cm2. This
is a considerable amount of energy per unit area. It is a blast loud enough to pop balloons,
shatter glass, and even make your hair stand on end. It will also cause you to go permanently
deaf if you do not take precautions.
Let's compare this car stereo (featured on Ripley's Believe it or Not!) to a commercial jetliner,
which produces a noise level of about 140 dB when it takes off.
How much more intense is the sound produced by the car stereo above than that of a commercial jetliner?
solution
First, we need to find the intensity of the sound produced by the jetliner. Since we already stated that a jetliner
has a sound level of about 140 dB, we solve for the intensity exactly the same as we did above:
140
=
10log 10⎛⎜
14
=
log 10⎛⎜
14
=
log I – log 10–16
Quotient Rule of Logarithms
14
=
log I + 16
simplify second log with Properties of Logarithms
–2
=
log I
subtract 16 from both sides
10–2
=
I
exponentiate both sides
0.01
=
I
simplify
I
− 16
⎝ 10
I
− 16
⎝ 10
216 – Logarithmic Model
⎞
⎟
⎠
⎞
⎟
⎠
substitute in β(I) = 140 and I0 = 10–16
divide both sides by 10
Functions
The sound that the jetliner produces has an intensity of about 0.01 W / cm2. If we divide the
car stereo sound by the sound of the jetliner, we will find how much louder the stereo is than
the jetliner:
3.162
0.01
=
316.2
divide intensity of stereo by intensity of jetliner
Thus, the car stereo is 316 times louder than the jetliner.
Functions
answer
Logarithmic Model – 217
218 – Logarithmic Model
Functions
Advanced
To see a more realistic model of population growth.
logistics curve – a growth curve described by the equation below:
sigmoidal curve – another name for a logistics growth curve.
Omar Khayyam (c.1048-1131), in attempting to prove Euclid's Fifth Postulate, inadvertently discovered nonEuclidean geometries. The Fifth Postulate states that, given a line a, only one line parallel to a can be drawn
through a point not on a. Khayyam tried to prove the postulate by constructing a quadrilateral with two equal
sides that are perpendicular to the base. By definition, the quadrilateral's upper angles must be equal. Khayyam
hypothesized three possible scenarios: the angles are right angles, acute angles, or obtuse angles. He rejected
the last two hypotheses because the Fifth Postulate says that converging lines must intersect. Although he
dismissed the last two hypotheses, other mathematicians later speculated about non-Euclidean spaces where
converging lines don't meet.
Functions → Transcendental Functions → Exponential Functions → Exponential / Logarithmic Eq →
Logarithmic / Exponential Models → Logistics Growth
We have already looked at one model of population growth: exponential growth. This model is only accurate
under extremely favorable conditions, when the population has abundant resources that it can use to expand
itself. This model is generally only accurate within a controlled environment, such as a laboratory. The real
world is a much harsher place. In the real world, an environment cannot support population growth indefinitely.
Eventually an area will run out of food or water. To ensure the survival of the species, a population of animals
automatically adjusts its growth rate so that it does not exceed the maximum capacity of the environment
(humans tend to be the exception to this rule).
The population initially grows exponentially just as it did before, but after awhile the growth rate declines. The
model used to describe this growth pattern is known as the logistics curve given by the function:
a
y
− ( x− c)
=
1 + be
logistics growth equation
d
where y is the population and x is the time. A graph of this curve (called a sigmoidal curve) is shown below:
Functions
Logistics Growth – 219
A conservation organization releases 200 animals of an endangered species into a game preserve. The
organization believes that the preserve has a carrying capacity of 2000 animals and that the growth of the herd
will be modeled by the logistics curve.
p(t)
=
2000
1 + 5e
− 0.1643 t
where t is measured in months.
a)
Graph the function. Use the graph to determine the horizontal asymptotes and interpret the meaning of
the larger p-value in the context of the problem.
b)
Estimate the population after 6 months
c)
After how many months will the population be 1600?
solution
a)
The graph of the function looks as follows:
From the graph, it can be seen that there are horizontal asymptotes at y = 2000 and y = 0. These
represent the boundaries for which the environment is viable. The larger p-value represents the
maximum number of animals who can live in this environment. After the population reaches 2000, the
environment is no longer able to support the population.
b)
The graph shows that after 6 months, the population has jumped from 200 to around 600 animals.
220 – Logarithmic Model
Functions
c)
Use the logistics growth formula to find out when the population reaches 1600.
2000
substitute known quantity for p
750
=
750(1 + 5e–0.1643 t)
=
2000
multiply both sides by (1 + 5e–0.1643 t)
1 + 5e–0.1643 t
=
1.25
divide both sides by 750
5e–0.1643 t
=
0.25
subtract 1 from both sides
e–0.1643 t
=
0.05
divide both sides by 5
ln e–0.1643 t
=
ln 0.05
take ln of both sides
–0.1643t
=
ln 0.05
simplify using properties of ln
t
=
t
≈
1 + 5e
− 0.1643 t
ln( 0.05)
−0.1643
18.23 months
solve for t
evaluate to two decimal places
The population will reach 1600 around 18 months and 7 days from the initial time.
Functions
answer
Logistics Growth – 221
222 – Logarithmic Model
Functions
In this section of the Algebra Brain, we will examine in detail several procedures we can do with functions.
Arithmetic Combinations – Adding, subtracting, multiplying and dividing two functions by one another.
This creates a new function whose domain consists of all values common to the domains of f and g.
Composition of Functions – Using one function as the argument for another function. For example, if we
have two functions f(x) = 2x and g(x) = x2, then f(g(x)) = f(x2) = 2(x2) = 2x2.
Reflections in Coordinate Axes – Reflecting the graph of a function about the x- or y-axis. Very handy when
trying to draw certain functions. If we know one half of the function, then we can use this property to
draw the other half of the function.
Vertical and Horizontal Shifts – Moving the graph of a function around in the coordinate plane. This is
extremely important when analyzing conic sections.
Functions
Translations and Combinations – 223
224 – Translations and Combinations
Functions
Advanced
To learn how to add, subtract, multiply and divide two or more functions.
arithmetic combination – the sum, difference, product, or quotient of two or more functions.
Nicole d'Oresme (c.1320-1382) was a French algebraist, physicist, philosopher, and theologian. Among his
contributions to mathematics is the realization that exponents can be, and often are, negative integers or even
irrational numbers. He used a graphical approach in his mathematical analysis which became a cornerstone for
the development of analytic geometry in the 17th century by René Descartes. d'Oresme even understood basic
properties of exponents, such as xm + xn = xm+n.
Functions → Translations and Combinations → Arithmetic Combinations
We can combine two functions by the operations of addition, subtraction, multiplication, and division. The
result is a new function that is the sum/difference/product/quotient of the original two functions.
For instance, if we have the functions
f(x)
=
3x2 – 2
g(x)
=
2x + 1
and
then the sum, difference, product, and quotient of f and g are as follows:
f(x) + g(x)
f(x) – g(x)
f(x)g(x)
f ( x)
g ( x)
=
(3x2 – 2) + (2x + 1)
add two functions together
=
3x2 + 2x – 1
simplify
sum of functions f and g
=
(3x2 – 2) – (2x + 1)
subtract one function from another function
=
3x2 – 2 – 2x – 1
Distributive Property
=
3x2 – 2x – 3
simplify
difference of functions f and g
=
(3x2 – 2)(2x + 1)
multiply two functions together
=
6x3 + 3x2 – 4x – 2
FOIL method
product of functions f and g
2
=
3x − 2
2x + 1
x ≠ –1/2
divide one function by another function
As always, we have to keep the domains of our given functions in mind when we combine them. When we
divided f by g, for instance, we had to note that the value x = –1/2 is excluded from the domain because
otherwise we have a division by zero, which is a big no-no in algebra. In other words, g(x) ≠ 0 when we divide
two functions f and g. In more general terms, the domain of the arithmetic combination of functions consists
of all those terms that are common to both f and g. If the domain of f is [–5, 3] and the domain of g is [–1, 7],
Functions
Arithmetic Combinations – 225
then the domain of any combination of f and g will be [–1, 3], because the numbers inside this range are
common to both f and g.
Sum, Difference, Product, and Quotient of Functions
Let f and g be two functions with overlapping domains. Then, for all x
common to both domains, the sum, difference, product, and quotient
of f and g are defined as follows.
Note:
1.)
Sum:
(f + g)(x)
=
f(x) + g(x)
2.)
Difference:
(f – g)(x)
=
f(x) – g(x)
3.)
Product:
(fg)(x)
=
f(x) · g(x)
4.)
Quotient:
=
g(x) ≠ 0
While here we only list the sum/difference/product/quotient of two functions, these properties easily
extend to three or more functions. That is, we can have
m(x)
=
f ( x) + g ( x) −
p ( x) q ( x)
arithmetic combination of 5 functions
k ( x)
Find (a) (f + g)(x), (b) (f – g)(x), (c) (fg)(x), and (d) (f / g)(x). What is the domain of f / g?
f(x)
=
x2 + 5
g(x)
=
1–x
solution
a.)
b.)
c.)
d.)
f(x) + g(x)
f(x) – g(x)
f(x)g(x)
f ( x)
g ( x)
=
(x2 + 5) + (1 – x)
add the two functions together
=
x2 – x + 6
write in standard form
answer
=
(x2 + 5) – (1 – x)
subtract g from f
=
x2 + x – 4
rewrite in standard form
answer
=
(x2 + 5)(1 – x)
multiply f and g together
=
x2 – x3 + 5 – 5x
FOIL method
=
–x3 + x2 – 5x + 5
rewrite in standard form
2
=
x +5
x−1
divide f by g
answer
The domain of f / g includes only those values of x such that g(x) ≠ 0. In this case, g(x) = x – 1, so that means
we exclude x = 1 from our domain. Therefore, our domain for f / g is all real numbers except x = 1.
answer
226 – Arithmetic Combinations
Functions
Advanced
To learn how to take a function of a function.
composition of functions – the function that results from first applying one function, then another; denoted by
the symbol o.
S. I. Ramanujan (1887-1920) was an Indian mathematician of great genius. He was also a devoutly religious
man who was reluctant to travel because of certain religious restrictions. Ramanujan attributed his mathematical
gifts to his household deity, the goddess Namagiri. He combined his passion for mathematics with his devout
faith, once commenting to a friend, "An equation for me has no meaning unless it expresses a thought of God."
Functions → Translations and Combinations → Composition of Functions
Suppose we have the functions
f(x)
=
x2 + 1
g(x)
=
x+1
and
One way to combine these functions is to take a function of a function. That is, the function g(x) becomes the
argument for the function f(x). We write this as:
f(g(x))
take a function of a function
When we substitute in g(x) = x + 1 inside the parentheses, we get:
f(x + 1)
=
(x + 1)2 + 1
substitute in g(x) = x + 1
=
x2 + 2x + 1 + 1
expand binomial
=
x2 + 2x + 2
simplify
We call this ability to take a function of a function composition of functions, and denote it as f o g.
Definition of Composition of Two Functions
The composition of the function f with the function g is given by:
(f o g)(x)
=
f(g(x))
composition of f with g
Similarly, the composition of the function g with the function f is given by:
(g o f)(x)
=
g(f(x))
composition of g with f
The domain of (f o g) is the set of all x in the domain of g such that g(x) is in the domain of f.
Note:
Composition of function is not a commutative operation. That is,
f(g(x))
Functions
≠
g(f(x))
composition is not commutative!
Composition of Functions – 227
Also, we can extend this definition for more than two functions. Thus, we can find (f o g o h)(x) as:
(f o g o h)(x)
=
f(g(h(x)))
composition of three functions
In such cases, we usually start with the inner function first and work our way to the outer function.
Given f(x) = 2x – 3 and g(x) = x + 6, find:
a.)
(f o g)(2)
b.)
(g o f)(2)
solution
a.)
To compose f with g, we find f(g(x)) and evaluate it for x = 2:
(f o g)(2)
b.)
=
f(g(2))
definition of composition of functions
=
f(2 + 6)
substitute in g(2) = 2 + 6
=
2(8) – 3
let f(2 + 6) = 2(2 + 6) – 3
=
13
simplify
answer
Now we reverse the order of the composition and evaluate it for x = 2:
(g o f)(2)
=
g(f(2))
definition of composition of functions
=
g(2(2) – 3)
substitute in f(2) = 2(2) – 3
=
g(1)
simplify inside parentheses
=
1+6
evaluate g(1)
=
7
simplify
answer
Here is a classic example of the fact that (f o g)(x) ≠ (g o f)(x), thus proving the point that this is not a
commutative operation.
228 – Composition of Functions
Functions
Advanced
To learn how to reflect a graph of a function in the coordinate plane.
No definitions on this page.
The original discoverer of conic sections was Manaechmus, a fourth century B.C. Greek geometer. He used
simple properties of the parabola and the rectangular hyperbola to solve the problem of the mean proportionals.
It is unknown exactly how he proved that the curves were conics. Fifty years after his time, a considerable
amount of work had been done on the conics. At that time, the parabola, the ellipse, and the hyperbola were
known, respectively, as the sections of a right-angled, acute-angled, and obtuse-angled cone. Later it was
shown that all three conics could be derived from the same cone.
Functions → Translations and Combinations → Reflections in Coordinate Axes
One of the most common types of ways of transforming a function is to reflect the function about either the xor the y-axis. This means that the function on one side of the axis is a mirror image of the function on the other
side of the axis.
There are two ways of transforming a function so that it has a reflection. To reflect a function about the x-axis,
we let our new function h(x) = –f(x). Suppose we let f(x) = x2, then h(x) = –x2. When we graph f(x) and h(x) in
the same coordinate plane, we obtain the following graph:
To reflect the function f about the y-axis, we let h(x) = f(–x). That is, we replace x in the function f with –x.
Suppose we have f(x) = x3, then h(x) = f(–x) = (–x)3. As a result, when x = –1, we have h(–1) = (–(–1))3 = (1)3 = 1.
That is, when x is a negative number, we obtain a positive result. If we graph f(x) = x3 and h(x) = (–x)3 on the same
coordinate plane, we obtain the following graph:
Functions
Reflections in Coordinate Axes – 229
To summarize, we can use the following guidelines for reflecting functions along the axes:
Reflections in the Coordinate Axes
Reflections in the coordinate axes of the graph of y = f(x) are represented as follows:
1.)
Reflections in the x-axis: h(x) = –f(x)
2.)
Reflections in the y-axis: h(x) = f(–x)
230 – Reflections in Coordinate Axes
Functions
Advanced
To learn how to move the graph of a function around in the Cartesian plane.
No definitions on this page.
The Greek word arithmetike—derived from the Greek word arithmos, meaning "number"—does not refer to
calculation. Instead it denotes theoretical, speculative thinking about numbers. Nichomachus of Gerasa (c.60c.100 A.D.) discusses numbers in his work Introduction to the Arithmetic. He introduces various kinds of
numbers, such as odd, even, prime, and perfect. He considered them to have human-like qualities, like
goodness. One theorem he developed involves grouping positive odd integers like so: 1; 3 + 5; 7 + 9 + 11; 13 +
15 + 17 + 19; etc. Nichomachus discovered the sum of each grouping is equal to the cubed integers: 13 = 1; 23 =
3 + 5; 33 = 7 + 9 + 11; 43 = 13 + 15 + 17 + 19; and so forth.
Functions → Translations and Combinations → Vertical and Horizontal Shifts
Suppose we have the simple function of f(x) = x. This is a diagonal line of slope 1 that passes through the
origin. If we want to move the line up along the y-axis by 1 unit, then we simply add 1 to the overall function to
obtain a new function:
f(x)
=
x
original function
g(x)
=
x+1
add 1 to our original function
=
f(x) + 1
substitute in x = f(x)
The net result is that we have shifted the graph of f(x) upward by 1 unit. If we wanted to move the graph of f(x)
downward by 1 unit, then logic would suggest that we subtract 1 from f(x):
h(x)
=
f(x) – 1
subtract 1 from f(x)
=
x–1
substitute in f(x) = x
This behavior holds true even if we have more complicated functions, such as f(x) = x3 – 3x2 + 4x – 1.
In order to move the graph of this function up 1 unit, we add 1 to the overall function:
g(x)
=
f(x) + 1
add 1 to f(x)
=
(x3 – 3x2 + 4x – 1) + 1
substitute in f(x) = x3 – 3x2 + 4x – 1
=
x3 – 3x2 + 4x
simplify
Now, what if we want to shift the graph to the right or to the left? How do we go about doing that? Suppose
we have k(x) = x2. This is just a parabola centered on the origin. To move the graph to the right by 1 unit, we
replace x with (x – 1) to obtain a new function:
j(x)
Functions
=
k(x – 1)
substitute in x – 1 for x
=
(x – 1)2
graph of k shifted 1 unit to the right
Vertical and Horizontal Shifts – 231
In order to move the graph of k(x) = x2 to the left by 1 unit, we reverse the sign, replacing x with (x + 1):
q(x)
=
k(x + 1)
substitute in x + 1 for x
=
(x + 1)2
graph of k shifted 1 unit to the left
We have a few rules we can use to tell us how the graph of a function f is shifted.
Vertical and Horizontal Shifts
Let c be a positive real number. Vertical and horizontal shifts in the
graph of y = f(x) are represented as follows:
1.)
Vertical shift c units upward:
h(x) = f(x) + c
2.)
Vertical shift c units downward:
h(x) = f(x) – c
3.)
Horizontal shift c units to the right: h(x) = f(x – c)
4.)
Horizontal shift c units to the left: h(x) = f(x + c)
We can combine these guidelines to move graphs of functions in both the horizontal and vertical directions.
Sketch on the same set of coordinate axes a graph of f for c = –2, 0, 2.
f(x)
=
(x – c)3 + c
solution
Notice that f contains the terms (x – c), which means that we shift the graphs 3 units to the right or left
(depending on the value of c). Also notice that we add c to this term, which means we move the graph up or
down, again depending on the value of c. Thus, we have three functions:
f1(x)
f2(x)
f3(x)
=
(x – (–2))3 + (–2)
substitute in c = –2
=
(x + 2)3 – 2
simplify
function is shifted 2 units left and 2 units down
=
(x – 0)3 + 0
substitute in c = 0
=
x3
simplify
function is not shifted at all, centered on the origin
=
(x – 2)3 + 2
substitute in c = 2
function is shifted 2 units right and 2 units up
The graphs of these functions look like:
232 – Vertical and Horizontal Shifts
Functions
answer
Notice that while we move the function around the Cartesian plane, we do not change its overall shape. The
group of function represented on the graph above is called a family of functions. You will study families of
functions in much more detail when you study integration in calculus.
Functions
Vertical and Horizontal Shifts – 233
234 – Vertical and Horizontal Shifts
Functions
Advanced
To define inverse functions.
inverse function – the functions f and g are inverses of each other if f(g(x)) = x in the domain of g and g(f(x)) = x
for all x in the domain of f.
George Berkeley (1685-1753)—Irish philosopher, Anglican cleric, and mathematician—is known for his rejection
of contemporary mathematics for being unclear and logically deficient. He is most famous for attacking Isaac
Newton's infinitesimal calculus. He accepted the fact that Newton's calculations provided true results, but he
believed the reasoning upon which those calculations were based was flawed. This critique inspired other
mathematicians to develop a logical clarification of calculus.
Functions → Inverse Functions
Recall that one way we can represent a function is as a set of ordered pairs. Suppose we have the function
f(x) = x + 3 for values of x from the set A = {1, 2, 3, 4}. This function maps onto the set B = {4, 5, 6, 7}.
We can write this as follows:
f(x)
=
x+3
{(1, 4), (2, 5), (3, 6), (4, 7)}
Now, if we interchange the first and second coordinate of each of the ordered pairs, we form the inverse
function of f, which we denote by f –1 (read as "f-inverse"; this should not be confused as being an exponent).
In essence we have a function mapped from the set B to the set A, which we write as follows:
f –1
=
x–3
{(4, 1), (5, 2), (6, 3), (7, 4)}
Notice that the domain of f (the first coordinate) is equal to the range for f –1, and vice versa, as shown below:
Also notice that the functions f and f –1 have the effect of "undoing" each other. That is, we can use f(x) to
"undo" f –1(x) and vice versa. Suppose we form the composition of f and f –1. Then we have:
f(f–1(x))
Functions
=
f(x – 3)
composition of functions
=
(x – 3) + 3
substitute in (x – 3) for x in the function f(x)
=
x
simplify
Inverse Functions – 235
Similarly, if we form the composition of f –1 and f, we obtain:
f –1(f(x))
=
f –1(x + 3)
composition of functions
=
(x + 3) – 3
substitute in (x + 3) for x in the function f –1(x)
=
x
simplify
In general, for two functions to be inverses, both compositions of the functions must return a value of x. More
specifically, we can define an inverse function as follows:
Definition of the Inverse of a Function
Let f and g be two functions such that
f(g(x))
=
x
for every x in the domain of g
g(f(x))
=
x
for every x in the domain of f
and
Under these conditions, the function g is the inverse of the function f.
The function g is denoted by f –1 (read "f-inverse"). Thus:
f(f –1(x))
=
x
f –1(f(x))
=
x
and
The domain of f must be equal to the range of f –1 and
the range of f must be equal to the domain of f –1.
Note:
Do not be confused by the use of –1 to denote the inverse of the function f. Whenever we write f –1, we
are always referring to the inverse function, not the reciprocal of f(x).
Show that f and g are inverses of each other.
f(x)
=
1 – x3
g(x)
=
3
1−x
solution
To demonstrate that they are inverses, we form the compositions f(g(x)) and g(f(x)) and see if they both equal
the identity function x:
f(g(x))
( 3 1 − x)
composition of f with g
=
f
=
1−
=
1 – (1 – x)
simplify radical/exponent
=
1–1+x
Distributive Property
=
x
check
236 – Inverse Functions
( 3 1 − x) 3
substitute in for x
Functions
Now we form the composition of g with f:
g(f(x))
=
g(1 – x3)
=
3
=
3
=
3
=
x
(
composition of g with f
1− 1−x
1−1+ x
x
3
3
3
)
substitute in for x
Distributive Property
simplify under the radical sign
simplify
check
Since we obtain the identity function x when we form both compositions, we conclude that these functions are
inverses of each other.
answer
Functions
Inverse Functions – 237
238 – Inverse Functions
Functions
Advanced
To find the inverse of a function.
No definitions on this page.
The curve known as the "Witch of Agnesi" is named after Marie Gaëtann Agnesi (1718-1798), an Italian
algebraist, geometer, linguist, and philosopher. The "Witch" part of the name derives from a confusion of terms.
The Italian word versiera (to turn) somehow became associated with the Italian word avversiera, which means
"devil's wife", or "witch". The curve itself is the graph of the Cartesian equation y(x2 + a2) = a3, where a is a
constant. Oddly, the graph resembles the point hat of a witch, possibly another reason for its name
Functions → Inverse Functions → Finding Inverse Functions
In the previous thought, we introduced you to the idea that some functions have an inverse. We also showed
you how to demonstrate that two functions are inverses of each other. However, we did not give you any
means for finding the inverse of a function if it has one.
For simple functions, we can find inverses by inspection. For instance, if we have f(x) = x + 3, then the obvious
inverse of this function is f –1(x) = x – 3. Below we list some guidelines for finding the inverse of a given
function. The key step in these guidelines is Step 2—interchanging the roles of x and y. The reason why we do
this is because of the fact that inverse functions have ordered pairs with the coordinates reversed—i.e. the
domain of a function f is the range of the function f –1.
Guidelines for Finding the Inverse of a Function
Note:
1.)
In the equation for f(x), replace f(x) with y.
2.)
Interchange the roles of x and y.
3.)
If the new equation does not represent y as a function of x, the function f does not have an
inverse function. If the new equation does represent y as a function of x, solve the new
equation for y.
4.)
Replace y with f –1(x).
5.)
Verify that f and f –1 are inverses of each other by showing that the domain of f is equal to the
range of f –1, the range of f is equal to the domain of f –1, and f(f –1(x)) = x = f –1(f(x)).
In Step 3, above, it is possible that a function has no inverse. For instance, the function f(x) = x2 has no
inverse function for all values of x in the domain of f. However, if we restrict the domain of f to only
positive values of x, then the function f(x) = x2 does have an inverse. Knowing the domain and ranges
of our given functions is extremely important in determining if they have inverses or not.
Find the inverse.
f(x)
Functions
=
2x + 3
7
Finding the Inverse – 239
solution
We use the guidelines given above.
1.)
2.)
First, we replace f(x) with y:
f(x)
=
y
=
2x + 3
2y + 3
=
interchange x and y
7
This new equation does represent y as a function of x, the function f does have an inverse function, so
we solve the equation in Step 2 for y:
2y + 3
x
=
7x
=
2y + 3
multiply both sides by 7
7x – 3
=
2y
subtract 3 from both sides
=
y
divide both sides by 2
2
equation from Step 2
7
Now we replace y with f –1(x).
f –1(x)
5.)
replace f(x) with y
7
7x − 3
4.)
given equation
7
Next, we interchange x and y:
x
3.)
2x + 3
7x − 3
=
replace y with f –1(x)
answer
2
Finally, we notice that both f and f –1 have domains and ranges that consist of the entire set of real
numbers. That is, the range of f is the set of all real numbers, and the domain of f –1 is the set of all real
numbers. We also need to double check that f(f –1(x)) = x = f –1(f(x)).
f(f –1(x))
=
f ⎛⎜
7x − 3 ⎞
2⎛⎜
7x − 3 ⎞
⎝
⎝
=
⎟
⎠
2
⎟+3
⎠
2
composition of f with f –1
evaluate f(f –1(x))
7
( 7x − 3) + 3
=
7
7x
=
=
7
x
240 – Finding the Inverse
cancel common factors
combine like terms in numerators
cancel common factors
check
Functions
We also find f –1(f(x)):
f –1(f(x))
f
=
− 1⎛
⎜
⎝
7⎛⎜
⎝
=
2x + 3 ⎞
⎟
⎠
7
7
2x + 3 ⎞
⎟−3
⎠
7
composition of f –1 with f
evaluate f –1(f(x))
2
( 2x + 3) − 3
=
2
2x
=
=
cancel common factors
combine like terms in numerators
2
cancel common factors
check
x
Since both compositions give us the identity function x, we feel confident that we have indeed found
the inverse of the function f.
Find the inverse of the function g.
g(x)
=
x3 + 2
solution
Here we will move a bit more quickly than on the last example. First, we substitute in y for g(x), then we switch
the variables x and y. We solve the equation for y, and then substitute in g–1(x) for y. We will leave the process
of double-checking the inverse up to you to do on your own.
g(x)
=
x3 + 2
given equation
y
=
x3 + 2
substitute in y = g(x)
x
=
y3 + 2
switch the variables x and y
x–2
=
y3
subtract 2 from both sides
3
x−2
=
y
take cube root of both sides
3
x−2
=
g–1(x)
replace y with g–1(x)
answer
Notice that the domain of g(x) is the range of g–1(x), namely both are the set of all real numbers. Similarly, the
range of the function g(x) is the domain of g–1(x) and both are again the set of all real numbers.
Functions
Finding the Inverse – 241
242 – Finding the Inverse
Functions
Advanced
To determine if a function has an inverse by applying the Horizontal Line Test.
No definitions on this page.
al-Battani (858?-929), also known by his Latinized names of Albatenius, Albategnius, and Albategni, was a
prominent Arab astronomer and trigonometer. His innovative use of trigonometric methods to perform
astronomical calculations led to much greater accuracy in his measurements. He determined the length of the
solar year to be 365 days, 5 hours, 48 minutes, and 24 seconds, very close to the value we use today. He also
applied trigonometric techniques to astrology, as well as astronomy. Although scientists today have little or no
use for astrology, to the ancient Arabs, astrology was an integral part of everyday life. They firmly believed the
positions of the stars and planets had a direct effect on people's lives.
Functions → Inverse Functions → Horizontal Line Test
The graphs of inverse functions are related to one another. This should not come as too big a surprise.
Remember that one definition of inverse functions involves swapping the order of their coordinate points. For
example, if the function f contains the point (a, b), then its inverse f –1 contains the point (b, a) and vice versa.
As a result, the graph of f –1 is a reflection of the graph of f about the line y = x, as indicated by the diagram
below:
In the thought Finding the Inverse, we told you that some functions do not have an inverse. The classic
example of a function that does not have an inverse is the function f(x) = x2. The reason why this function has
no inverse has to do with the basic definition of a function. If we let x be any real number, then we will have
elements from the set of the domain mapping onto the same element in the set of the range. For example, if we
let x1 = 2 and x2 = –2, then f(x1) = 4 and f(x2) = 4. In order to have an inverse, we reverse the order of the
coordinates. But then we have one element of the domain of f –1 pointing to two different elements in its range,
which is not allowed according to the basic definition of a function. However, if we restrict the domain of f to
include only nonnegative values of x, then the function does have an inverse.
We can use a geometric means to find the inverse of a function. The reflective property of the graphs of inverse
functions gives us a nice test we can utilize to determine whether or not a function has an inverse. We call this
remarkably simple test the Horizontal Line Test.
Functions
Horizontal Line Test – 243
Horizontal Line Test
A function f has an inverse function if and only if no horizontal line
intersects the graph of f at more than one point.
One of the disadvantages of this test is that there is no way for you do determine what the inverse actually is. It
is an existence theorem because it tells us only if an inverse function exists or not.
Use a graphing utility to graph the function and use the Horizontal Line Test to determine whether the function
has an inverse.
g(x)
=
−2x 16 − x
2
solution
We use MathCAD 2000® to draw our graphs, but you are free to use whatever tool you prefer. The graph of
g(x) is shown below. The blue line represents the application of the Horizontal Line Test.
Since the line clearly intersects the graph at more than one point, we conclude that this function fails the
Horizontal Line Test and hence has no inverse.
answer
244 – Horizontal Line Test
Functions
Unit 4
Quadratics
Quadratics – 245
246 – Quadratics
General Information about Quadratic Equations
Besides linear equations, the other type of equation that shows up most often in algebra are the quadratic
equations. In most general terms, a quadratic equation in x is one that takes the form
ax2 + bx + c
=
0
quadratic equation in x
where a, b, and c are all real numbers and a ≠ 0. If a = 0, then the first term goes to zero and we are left with a
linear equation. This equation is also known as a second-degree polynomial equation in x.
While we have stated that the general form has one side equal to zero. If we allowed one side to be another
variable, such as y, then we have what is known as a quadratic function in x because the value for y is entirely
dependent on the value for x:
y
=
ax2 + bx + c
quadratic function
On a graph, this function can be described by a parabolic curve. Note that only the variable x is raised to the
second power.
We will study quadratic equations in detail, learning the four basic methods of solving a quadratic equation:
1.
Factoring
2.
Completing the Square
3.
Extracting Square Roots
4.
The Quadratic Formula (the granddaddy of all algebraic formulas)
Generally, we try to factor first. If that fails, then we try to complete the square. If we can neither factor nor
complete the square, then we apply the Quadratic Formula. Extracting square roots is only reserved for special
occasions.
Conic Sections
If we raise the variable y to the second power as well as x, then we end up with what is called a conic section. A
conic section is what happens when we intersect a cone with a plane. We will examine these in intense detail in
the appropriate section of the Algebra Brain. There are four basic conic sections:
1.
Parabola
2.
Ellipse
3.
Circle (yes, that's right, a circle)
4.
Hyperbola (a what?)
The parabola is a fairly common shape. Throw a baseball across a field and it will travel in a parabolic path.
An ellipse is also a fairly common shape. The planets all travel in elliptical orbits. The circle is by far one of
the most recognizable shapes. Coins are circular, compact discs are circular, pens can be circular in crosssection, pipes, wires, and all kinds of other things are made to be circular. By far the most obscure conic
section is the hyperbola. However, it does play an important role in the motions of comets and asteroids, as well
as in long-distance communication.
Quadratics – 247
248 – Quadratics
Advanced
To learn the basic definition of a quadratic equation and some applications of how they are used in
the real world.
No definitions on this page.
The ancient Chinese had a simple process for finding the square root of a number. It was derived from the
geometrical division of a square into smaller areas. The offshoot of this method was a solution to the quadratic
equation of the form x2 + bx = c, when b and c are positive. This is all explained in detail in the Chinese work Jin
zhang shuan shu (Nine Chapters on the Mathematical Art).
Quadratics → Quadratic Equations
Next to linear equations, quadratic equations are among the major emphasis areas of algebra. Indeed, much of
algebra was developed specifically to help solve problems dealing with quadratic equations. In particular,
quadratic equations have a wide range of applications in physics, architecture, business, finance, and even
warfare (although usually that involves physics).
We can define a quadratic equation in standard form as follows:
Definition of a Quadratic Equation
A quadratic equation in x is an equation that can be written in the standard form
ax2 + bx + c
=
0
quadratic equation in standard form
where a, b, and c are all real numbers and a ≠ 0.
This is also known as a second-degree polynomial equation in x.
Under this area of the Algebra Brain, we will look at four main methods for solving quadratic equations. Each
method only works if we have a quadratic equation in standard form! If we have a quadratic equation that is not
in standard form, then we need to somehow transform it using equivalent equations so that it becomes a
quadratic equation in standard form. For example, if we have the equation
4x2 – 5x + 3
=
–2x
example of quadratic equation not in standard form
then we would add 2x to both sides, thus the right hand side becomes equal to 0 and all the terms are then on the
left hand side:
4x2 – 5x + 3
=
–2x
4x2 – 5x + 3 + 2x
=
–2x + 2x
add 2x to both sides
4x2 – 3x + 3
=
0
combine like terms
quadratic equation in standard form
Once we have the equation in standard form, we can then apply a variety of techniques to solve the equation for
x. In this particular example, we would probably head straight for the Quadratic Formula as this equation does
not factor easily, nor does it really lend itself to completing the square, nor is it in an acceptable form for
extracting square roots. As it turns out, this equation does not contain any real roots. We can solve it using the
method of completing the square, but this method is actually more tedious than the Quadratic Formula in this
instance. Thus, after applying the Quadratic Formula, we discover that the two roots of this equation are:
Quadratics
Quadratic Equations – 249
x
=
x
=
3
8
3
8
+
⎛ 39 ⎞ i
⎜
⎟
⎝ 8 ⎠
and
−
⎛ 39 ⎞
⎜
⎟i
⎝ 8 ⎠
solutions to the quadratic equation
One of the most important skills required for solving quadratic equations is the ability to recognize which
technique to use. As with any mathematical skill, it only comes through practice, practice, and more practice.
Let's look at a real life example of how quadratic equations work.
On a particular university campus, there is a square courtyard covered in lush green grass. Despite the fact that
there are "DON'T WALK ON THE GRASS" signs posted all around the grass, students still insist on walking
on the grass in order to cut down the time it takes to get to their classes. The path they generally follow across
the grass covers a distance of 47 feet. The path they travel is shown in the diagram below:
How many feet does a student save by walking across the lawn instead of walking on the sidewalk?
solution
Based on the figure above, we let x be the length of the shorter path around the grass and then we use a ruler on
our diagram to discover that the longer path is twice the length of the shorter path, which we can represent with
2x. Since we have a right triangle, we apply the Pythagorean Theorem to get:
x2 + (2x)2
=
472
Pythagorean Theorem
x2 + 4x2
=
2209
simplify
5x2
=
2209
combine like terms
x2
=
441.8
divide both sides by 5
x
=
x
≈
250 – Quadratic Equations
441.8
21.02
extract positive square root
(see thought on Extracting Square Roots)
approximate value of x to nearest hundreth
Quadratics
The total distance on the sidewalk is:
x + 2x
=
3x
total length of path around grass
=
3( 441.8)
substitute in for x
≈
63.06 feet
simplify, rounding to nearest hundreth
Cutting across the grass saves the student having to walk about 63.06 – 47 = 16.06 feet. answer
Note:
Eventually the university became so tired of students traipsing across their nice green grass that they
finally put a sidewalk down where the students were actually walking, instead of where the university
wanted the students to walk.
Quadratics
Quadratic Equations – 251
252 – Quadratic Equations
Quadratics
Advanced
To solve quadratic equations by factoring.
No definitions on this page.
The Englishman Thomas Harriot (1560-1621) is credited with developing much of the early work in solving
quadratic equations by factoring and applying the Zero Product Property, although others may have applied the
method before him. Harriot applied his method to polynomials of higher degree. By the time Carl Gauss
developed his Fundamental Theorem of Algebra two hundred years later, Harriot's ideas were already wellestablished in the mathematical community. He was also prominent in astronomy, being one of the first to
discover sunspots.
Quadratics → Quadratic Equations → 1. Factoring
Quadratic equations can sometimes be factored into the product of two linear expressions. If we are able to
manage this, then solving a quadratic equation becomes very simple.
First of all, we need to write the quadratic equation in standard form:
ax2 + bx + c
=
0
standard form of quadratic equation
If a quadratic equation in standard form can be factored, we make use of the Zero-Factor Property to find the
solutions to the quadratic equation. Recall that the Zero-Factor Property states that if the product of two factors
equals zero, then either one of the factors is zero or both of the factors is zero. Therefore, we set each of the
linear factors of the quadratic equation equal to zero and solve for x. The values that we find for x are the
solutions to the quadratic equation.
It is easier to see this in practice than it is to explain it.
Solve 2x2 + 9x + 7 = 3 for x.
solution
First of all, we notice that this is not in standard form, so we have to tweak the equation a bit:
2x2 + 9x + 7
=
3
given equation
2x2 + 9x + 4
=
0
subtract 3 from both sides to get it into standard form
Now that it is in standard form, we want to factor the equation:
2x2 + 9x + 4
=
0
standard form
(2x + 1)(x + 4)
=
0
factored form
Once we have this equation in factored form, we note that the only way the left hand side can be equal to the
right hand side is if one or the other of the factors is equal to zero. We set each of the factors equal to zero and
solve for x. Then each solution is one of two possible solutions for this equation.
Quadratics
Factoring – 253
2x + 1
=
0
set 1st factor equal to zero
2x
=
–1
subtract 1 from both sides
x
=
−
x+4
=
0
set 2nd factor equal to zero
x
=
–4
subtract 4 from both sides
the other solution
1
2
divide both sides by two
one solution
Thus we have two possible solutions: x = –1/2 and x = –4.
answer
Check these solutions in the original equation.
254 – Factoring
Quadratics
Advanced
To solve quadratic equations by extracting the square roots.
extracting square roots – the process of solving a quadratic equation by isolating the x2 term on one side,
constant terms on the other side, and then finding the square root of both sides.
Robert Recorde (1510-1558) did more than simply invent the equals sign, "=". He established the English school
of mathematical writers. He was the first English writer on arithmetic, geometry, and astronomy. Many of
Recorde's writings were in poetic form to help students remember the rules of operation. Because he published
his work in English, he had only limited circulation in the rest of Europe, which used Latin and/or Greek as the
standard language for mathematics. Finally, in his work, The Whetstone of Witte, he explains how to extract the
square root.
Quadratics → Quadratic Equations → 2. Extracting Square Roots
Sometimes we find we have a very simple quadratic equation of the form
u2
=
d
where u is an algebraic expression and d > 0. We can use factoring to solve equations of this type.
First get the equation into standard form:
u2
=
d
quadratic equation
u2 – d
=
0
subtract d from both sides
standard form
While it may not be immediately obvious, the expression u2 – d is actually the difference of two perfect squares.
Thus, we can quickly factor this expression as:
u2 – d
=
0
standard form
( u − d)( u + d)
=
0
factor difference of two squares
Then we set each factor equal to zero and solve for u:
u− d
=
u
=
u+ d
=
0
set 2nd factor equal to zero
u
=
− d
solve for u
0
set 1st factor equal to zero
d
solve for u
Since the two solutions only differ in sign, we can compact the solutions into one expression, using a "plus or
minus sign" ( ):
u
Quadratics
=
u is equal to plus or minus the square root of d
Extracting Square Roots – 255
In practice, we can avoid going through all the steps above and skip to the end. Solving an equation of the form
u2 = d without going through the process of factoring is extracting square roots .
Extracting Square Roots
The equation u2 = d, where d > 0, has exactly two solutions:
u
=
and
u
=
The solutions can also be written as:
u
=
Solve the following equations by extracting the square roots:
a.)
5x2
=
15
b.)
(x – 12)2
=
18
solution
a.)
b.)
Note:
While we could go through the business of factoring, it is simpler to try and extract the square roots.
First, we need to isolate x2 by itself. Then we will take the square root of both sides to obtain our
solutions for x:
5x2
=
15
given equation
x2
=
3
divide both sides by 5
x
=
take square root of both sides
answer
Here we have u = (x – 12)2. However, the process is exactly identical, except that we have to go
through one extra step to solve this equation for x.
(x – 12)2
=
x – 12
=
take square root of both sides
x
=
add 12 to both sides
answer
18
given equation
While we can simplify 18 to 3 2, there is no real good reason to do so in the context of this problem.
256 – Extracting Square Roots
Quadratics
Advanced
To solve quadratic equations by completing the square.
No definitions on this page.
The Hindu-Arabic number system was introduced into Europe in the 12th century via Latin translations of Arabic
texts written by al-Khwarizmi. Among the contents of these works are a "new" method of calculation that
requires repeated erasure and shifting of numerals as the calculation is performed. It was called "dust-board"
calculation because the easiest way to do it was with a piece of slate and a lump of chalk - tools still used in
classrooms today in the form we know as the chalkboard.
Quadratics → Quadratic Equations → 3. Completing the Square
Every so often, we find ourselves confronted with a quadratic equation that is neither factorable nor are we able
to extract any square roots from it. Or can we? Consider the equation
x2 – 6x + 2
=
0
We cannot factor this equation at all. Nor can we take the square root of both sides. How can we possibly find
the solutions to such an equation? The answer lies in transforming the equation into an equivalent equation that
we can solve. First, let's subtract 2 from both sides and see what happens:
x2 – 6x + 2
=
0
original equation
x2 – 6x
=
–2
subtract 2 from both sides
What we want to do is add some number to both sides such that the left hand side will become a perfect square
trinomial. In other words, we want the left hand side to look like (x – a)2 where a is some real number.
Remember that we have to add the same number to both sides of the equation to maintain equality. What if we
add the number 9 to both sides?
x2 – 6x + 9
=
–2 + 9
add 9 to both sides
(x – 3)2
=
7
factor left hand side
simplify right hand side
Once we have the equation in the form just given, we can now apply the method of Extracting Square Roots to
find the solutions to this equation:
x
=
solutions to equation
How did we know to add 9 to both sides of the equation? We looked at the coefficient of the x-term. We know
that when we square a linear factor of the form (x – a), the middle term of the resulting trinomial is given by 2a
and the last term becomes a2. In our example above, we took the number 6 (coefficient of x), divided it by 2 to
get 3 and then squared the 3 to get 9.
This process is formally known as Completing the Square.
Quadratics
Completing the Square – 257
Completing the Square
To complete the square for the expression
x2 + bx
add (b/2)2, which is the square of half the coefficient of x. Consequently
=
Note:
completing the square
If you add a quantity to one side of an equation in order to complete the square, remember to add the
same quantity to the other side of the equation.
Solve the quadratic equation by completing the square.
x2 + 8x + 8
=
–6
solution
First, group all the constant terms onto one side of the equation. Then add the square of half the coefficient of
b, which in this case means we divide 8 (coefficient of x-term) by 2 and square it to get 16. Then we convert
the polynomial to factored form and extract the square roots.
Note:
x2 + 8x + 8
=
–6
original equation
x2 + 8x
=
–14
group all constants on right hand side of equation
x2 + 8x + (4)2
=
–14 + (4)2
add (half of 8)2
(x + 4)2
=
–14 + 16
perfect square trinomial on left hand side
(x + 4)2
=
2
simplify
x+4
=
take square root of both sides
x
=
subtract 4 from both sides
answer
In this example, the leading coefficient is 1. In the eventuality that the leading coefficient is not 1,
you must divide both sides of the equation by the leading coefficient before completing the square.
For instance, to complete the square for 5x2 – 2x + 7 = 0, first divide each of term by the leading
coefficient 5:
2
x −
2
5
x+
7
5
=
0
divide each term by leading coefficient 5
and then proceed as in the example above.
258 – Completing the Square
Quadratics
Advanced
To use the Quadratic Formula to solve quadratic equations.
Quadratic Formula - an important formula that can be used to solve any quadratic equation for the independent
variable.
imaginary number - a number that is the square root of a negative real number.
Girolamo Cardano was an Italian physician, mathematician, and astrologer. His book, Ars Magna (published in
1545), is considered a fundamental work in the history of algebra. He was also one of the first mathematicians to
work extensively with imaginary numbers, which he called 'fictitious numbers'.
Quadratics → Quadratic Equations → 4. Quadratic Formula
One of the most important formulas used in algebra is the infamous Quadratic Formula. With this little handy,
dandy formula, we can solve any and all quadratic (second-degree) polynomial equations. This shortcut for
solving quadratics results from another technique used to solve quadratics, namely Completing the Square.
However, instead of using the Completing the Square technique on an individual quadratic equation, we can use
it once in a general setting, thus allowing us to apply the result to any quadratic equation.
Let's see how this works. Consider the following general second-degree polynomial:
ax2 + bx + c
=
0
general second-degree polynomial
Now we wish to solve this equation for x. From the Fundamental Theorem of Algebra and the Linear
Factorization Theorem, we know that we will end up with two roots.
ax2 + bx + c
=
0
given equation
ax2 + bx
=
–c
subtract c from both sides
=
−
=
−
2
x +
2
x +
b
a
x+
b
a
x
⎛ b⎞
⎜ ⎟
⎝ 2a ⎠
2
=
c
+
a
⎛ b⎞
⎜ ⎟
⎝ 2a ⎠
b − 4ac
4a
2
2
complete the square,
remembering to add the same quantity to both sides
convert the left-hand side to its factored form
simplify the right-hand side
2
b
=
2a
b − 4ac
4a
x
Quadratics
divide both sides by a
a
2
2
⎛x + b ⎞
⎜
⎟
2a ⎠
⎝
x+
c
=
2
take the square root of both sides
isolate the x-term
Quadratic Formula – 259
x
=
simplify to find the solutions for x
x
=
since the quantity 2 | a | is the same as 2a,
we can omit the absolute value sign and simplify
Quadratic Formula
The Quadratic Formula
The solutions of a quadratic equation in standard form
ax2 + bx + c
=
0
a≠0
are given by the Quadratic Formula:
x
Note:
=
Quadratic Formula
It is sometimes helpful to remember how to say this formula in plain English: "Negative b, plus or
minus the square root of b squared minus 4ac, all divided by 2a."
Solutions of a Quadratic Equation
The solutions of a quadratic equation ax2 + bx + c = 0, a ≠ 0, can be classified as follows:
Note:
1.
If the discriminant b2 – 4ac is positive, then the quadratic equation has exactly two distinct real
solutions.
2.
If the discriminant b2 – 4ac is zero, then the quadratic equation has exactly one repeated
solution.
3.
If the discriminant b2 – 4ac is negative, then the quadratic equation has no real solution.
If the discriminant is negative (as in classification 3 above), then its square root is imaginary and the
Quadratic Formula yields two complex solutions.
260 – Quadratic Formula
Quadratics
The ancient Greeks first discovered conic sections over 2300 years ago. However, they were mostly interested
in the geometric properties of conics. It wasn't until Newton and his chums rediscovered conics in the
seventeenth century that conic sections began making their appearance in calculus.
There are three main ways of defining a conic section. The first, the geometric interpretation known to the
Greeks, is a plane intersecting a double-napped cone. There are actually seven conic sections, but three of them
are known as degenerate conics because they don't really resemble a conic. The degenerate conics, namely a
point, a line and two intersecting lines, are formed when the intersecting plane passes squarely through the
vertex of the cone. The other four, namely the familiar conics of the parabola, the circle, the ellipse, and the
hyperbola, are all formed by an intersecting plane that does not pass through the vertex.
Geometric Interpretation of Conic Sections
A second way of defining conic sections is the algebraic method of representing a conic section by a seconddegree polynomial of the form:
Ax2 + Bxy + Cy2 + Dx + Ey + F
=
0
general second-degree polynomial equation
Finally, the third and potentially most useful definition is by defining each conic section by a locus of points
satisfying some geometric property. For example, a circle is the locus, or collection, of all points (x, y)
equidistant from some fixed point (h, k) in a plane. Using this locus definition, we obtain the standard equation
for a circle:
(x – h)2 + (y – k)2
=
r2
standard equation of circle
We use similar definitions for the other conic sections. A parabola is defined as the collection of points that are
equidistant from both the focus and a fixed line. We will discuss each definition in much more detail under the
relevant thought.
Conic sections have a wide range of applications, particularly in physics and engineering. For example, a
satellite antenna used to pick up television signals is in the shape of a parabola because a parabolic shape
reflects all incoming (parallel) signals so that they pass through the focus of the parabola. Thus, an amplifier
located at the focus can take all the incoming signals and "boost" the signal so that you can watch television. A
flashlight works on the same principle in reverse. The bulb emits beams which are reflected by the parabolic
reflector to travel parallel outwards, thus the flashlight is able to shine more light on a surface than if it had no
reflector.
Ellipses are the normal shape of one object orbiting another in space (actually all the conic sections are possible,
but only an ellipse will form a true orbit). Ellipses are also used in architecture, such as bridges and other arched
objects.
Hyperbolas are often used in long-range detections systems such as radar.
Quadratics
Conic Sections – 261
262 – Conic Sections
Quadratics
Advanced
To learn the algebraic definition of a circle.
circle – a plane curve consisting of all points at a given distance from a fixed point in the plane.
center – the point in the plane from which all other points on the circle are defined.
radius – the distance between the center and a point on the circle.
While it is relatively simple to show that x2 + y2 = r2, mathematicians have long debated whether or not a solution
can be found if '2' is replaced by any larger natural number. Pierre de Fermat (1601-1665) stated emphatically
that there are no positive whole numbers that solve xn + yn = zn when n is a natural number greater than 2.
However, he left no proof for his statement. With the advent of computers, we have been able to show that
Fermat's Last Theorem holds for at least the first 4 million natural numbers.
Quadratics → Conic Sections → Circles
Standard Equation of a Circle
This is the first sort of conic section usually taught in algebra. Most people are already familiar with the shape
of a circle. It is a perfectly round shape that can be described as a polygon with only one side or infinitely many
sides, depending on how you look at it.
Now we want to look at a circle in algebraic terms. Algebraically speaking, a circle is defined as the set of all
points in a plane that lie an equal distance from a fixed point elsewhere in the plane. The fixed point is called
the center of the circle. The distance from the center to the set of equidistant points is called the radius.
Consider the diagram of a circle shown below.
A point (x, y) is on the circle if and only if its distance from the center (h, k) is some fixed quantity r. If we
move the point (x, y) along the circle, its distance from the center will always be r. The Distance Formula,
which defines the distance between any two points in the plane, gives us the quantity r:
=
Quadratics
r
Distance Formula
Circles – 263
Now, we don't like looking at radical signs, so we will get rid of the radical sign by squaring both sides of the
Distance Formula. The result is the standard form of the equation of a circle:
Standard Form of the Equation of a Circle
The point (x, y) lies on the circle of radius r and center (h, k) if and only if
(x – h)2 + (y – k)2
=
r2
standard form of the equation of a circle
For a circle centered on the origin, the point (h, k) = (0, 0).
The standard form of a circle whose center is at the origin is:
x2 + y2
Note:
=
r2
standard form of the equation of a circle
whose center is the origin
If we have the expression (x + h)2, then this is equivalent to (x – (–h))2, which indicates that the xcoordinate of the center lies to the left of the y-axis. A similar fact holds for the expression (y + k)2 =
(y – (–k))2, thus the y-coordinate of the center is below the x-axis.
As always, an example right about now might clarify what we have explained so far. One of the major
advantages of describing a circle with an equation is that we can easily graph the equation. Similarly, if we
know some basic facts about a particular circle, we can quickly determine the equation describing that circle.
Graph the circle centered at (3, 2) and which has a radius of 2.
solution
In this problem, we are given the center of the circle, along with its radius. From the standard form of the
equation of a circle, we now know three of the five terms. The x- and y-terms are completely arbitrary. Let's
write down what we know:
h
=
3
x-coordinate of center of circle
k
=
2
y-coordinate of center of circle
r
=
2
radius of circle
Thus our equation has the following form:
(x – 3)2 + (y – 2)2
=
4
standard form of an equation of a circle
To graph this on the Cartesian Plane, you need to solve for y. Note that we will actually obtain two possible
solutions for y. This stems from the fact that this equation is quadratic (i.e. the y-term is raised to the second
power), thus we will end up with two solutions from the Fundamental Theorem of Algebra.
(y – 2)2
=
y–2
=
take square root of both sides
y
=
add 2 to both sides to obtain y
=
rearrange terms slightly
264 – Circles
4 – (x – 3)2
subtract (x – 3)2 from both sides
Quadratics
This can be split into two different solutions:
y
2
one possible solution
2
other possible solution
=
2 + 4 − ( x − 3)
=
2 − 4 − ( x − 3)
or
y
When we graph each one of these possible solutions, we end up with only half a circle. This is because a circle
is not a function (it fails the Vertical Line Test), so we have to graph each half of the circle separately. Using an
automatic graphing utility, we discover that our circle looks like:
answer
Recognizing a Circle Equation
Circles can be hidden in seemingly unrelated equations. Suppose we had the following equation.
x2 + 4x + y2 + 6y – 12
=
0
funky equation
At first glance, this seems totally unrelated to the equation of a circle in standard form, right? Wrong! It is
actually the same equation in disguise. Let's examine why.
Suppose we add "12" to both sides in order to get the constant on one side.
x2 + 4x + y2 + 6y
=
12
add 12 to both sides
Now here is where we have to get creative. Let's use one of the methods used for solving quadratic equations,
namely completing the square. However, we use this method twice, once for the terms involving x and once for
the terms involving y.
(x2 + 4x + ?) + (y2 + 6y + ?)
=
12
start completing the square
(x2 + 4x + 4) + (y2 + 6y + 9)
=
12 + 4 + 9
add the same quantities to both sides
(x + 2)2 + (y + 3)2
=
25
complete the square
The last step above is, in fact, the equation of a circle in standard form. Here we have a circle with radius 5, and
center located at (–2, –3).
Quadratics
Circles – 265
266 – Circles
Quadratics
Advanced
To define a parabola using algebraic equations.
parabola – the set of all points in the plane of a line l and a point F not on l whose distance from F equals its
distance from l.
directrix – a line associated with a parabola such that the distance from it to any point on the parabola is equal
to the distance from that point to the focus.
focus (parabola) – the point along with the directrix from which a point on a parabola is equidistant.
vertex – the intersection of a parabola with its axis of symmetry.
axis of symmetry – the line passing through both the focus and the vertex.
The Greek mathematician Archimedes (287-212 B.C.) is among the first to work with parabolas. In his work
Quadrature of the parabola, he finds the area of a segment of a parabola cut off by any chord. There are various
stories related about his death. One rumor has it that he was killed in 212 B.C. when the Romans sacked the city
of Syracuse (his hometown). Apparently he was so absorbed in a math problem that he refused to accompany a
Roman soldier until he had finished the problem. Angered, the Roman soldier ran Archimedes through with his
sword.
Quadratics → Conic Sections → Parabolas
If we graph the function f(x) = ax2 + bx + c on the Cartesian plane, we find that the graph is a distinctive shape
known as a parabola. However, this equation is only good for parabolas that open upwards or downwards. We
can also define "sideways" parabolas, by which we mean a parabola that opens to the left or right.
What we would like to have is a general, all-purpose definition of a parabola that is independent of orientation
or location in the plane. Recall that we defined a circle as the locus of points equidistant from a fixed point.
Now we will define a parabola in a similar manner, using some sort of distance relation as the quantity that truly
defines a parabola. The general definition is as follows:
Geometric Definition of a Parabola
A parabola is the set of all points (x, y) equidistant from a fixed line called the directrix
and a fixed point called the focus not on the line.
Quadratics
Parabolas – 267
The midpoint between the focus and the directrix is called the vertex. We indicate the distance between the
focus and the vertex by the letter p. The line passing through both the focus and the vertex is called the axis of
the parabola. A parabola is symmetric with respect to its axis.
The definition above gives us the general idea of what a parabola looks like, but we are actually more interested
in specific parabolas. To get specific, we need some equations.
Standard Equation of a Parabola
The standard form of the equation of a parabola with vertex (0, 0) and directrix y = –p is
x2
=
4py
p≠0
vertical axis
For directrix x = –p, the equation becomes
y2
=
4px
p≠0
horizontal axis
The focus is on the axis p units (directed distance) from the vertex.
If the vertex is located at the origin, then (h, k) = 0, and the equations above reduce to:
(x – 0)2
=
4p(y – 0)
vertical axis, vertex located at origin (0, 0)
x2
=
4py
simplify
(y – 0)2
=
4p(x – 0)
horizontal axis, vertex located at origin (0, 0)
y2
=
4px
simplify
and
The four basic orientations of a parabola are: up, down, right or left. Each orientation is determined by the
direction it "opens." The parabola above, for example, opens upward. The factor that determines whether it
opens up/down or right/left is the sign of p. If the parabola is on the vertical axis and p > 0, then the parabola
opens up. If p < 0, then it opens downward.
When the parabola is on the horizontal axis and p > 0, then the parabola opens to the right. When p < 0, the
parabola opens to the left.
Note:
A parabola that opens to the right or left is not a function because it fails the Vertical Line Test, while a
parabola that opens up or down is a function.
Proof:
Since the two cases of a parabola are similar, we will only prove the vertical axis case. First, we suppose that
the directrix (y = –p) is parallel to the x-axis. We make the assumption that p > 0. Since p is the directed
distance from the vertex to the focus, the focus must lie above the directrix. Now, according to the Geometric
Definition of a Parabola, the point (x, y) on the parabola must be equidistant from (0, p) and the line y = –p. We
can therefore use the Distance Formula to show that they are the same:
2
2
=
y+p
Distance Formula
x2 + (y – p)2
=
(y + p)2
square both sides
x2 + y2 – 2py + p2
=
y2 + 2py + p2
expand binomials
x2
=
y2 – y2 + 2py + 2py + p2 – p2
move all terms except x2 over
to right hand side
x2
=
4py
simplify
( x − 0) + ( y − p )
268 – Parabolas
proof
Quadratics
Find the vertex and focus of the parabola and sketch its graph.
x2 + 8y
=
0
solution
First, we need to rewrite the equation so that it is in standard form. All we need to do is move the 8y onto the
other side of the equation:
x2 + 8y
=
0
given equation
x2
=
–8y
subtract 8y from both sides
(x – 0)2
=
(4)(–2)(y – 0)
equation of parabola in standard form
Based on the equation of a parabola in standard form, we can see that the vertex (h, k) = (0, 0). Furthermore,
the distance between the vertex and the focus is –2 units. Since this parabola has a vertical axis, this means it
opens downward and the focus is located –2 units below the vertex. In other words, we have:
vertex
=
(0, 0)
and
focus
=
(0, –2)
answer
To graph this parabola, we plot a few points on either side of the axis and then draw a line connecting them:
x
–8
–4
–2
0
2
4
8
y
–8
–2
–0.5
0
–0.5
–2
–8
answer
Applications of Parabolas
Parabolic motion is one of the most common applications of parabolas. What do we mean by "parabolic
motion"? As it turns out, if you throw an object across the room, the path it takes will be a parabola. This is
even true if you simply roll a ball off a table. In that case, the path of the ball will be one half of a parabola that
is vertically oriented and opens downward.
One of the most useful properties of parabolas is the property of reflectivity. In physics, a surface is called
reflective if the tangent line at any point on the surface makes equal angles with an incoming ray (such as light)
and the resulting outgoing ray. The incoming angle is known as the angle of incidence and the outgoing angle
is the angle of reflection. A flat mirror is one common example of a reflective surface. If a parabola is revolved
about its axis, then we have another reflective surface. The key property of parabolic reflectors is that all
incoming rays parallel to the axis are reflected to pass through the focus of the parabola. This is the operating
Quadratics
Parabolas – 269
principle behind radio telescopes and satellite dishes, which gather all incoming radio waves to one place: the
focus. If the light source is located at the focus, as in a flashlight, then all outgoing rays are parallel.
Finally, objects traveling through the solar system sometimes travel in a parabolic orbit. This may seem
contradictory since most people think of an "orbit" as being a circular or elliptical path. An elliptical or circular
orbit means that a celestial object, such as the Moon, is trapped and cannot escape the planet's gravitational
field. A parabolic orbit, such as that traveled by many comets, means that the comet passes very close to some
body (like the Earth), and then flies back out into space, never to be seen again. When the Pioneer spacecraft
was launched, it was given a parabolic orbit so that it could escape Earth's gravitational field and leave the solar
system. As it passed by the outer planets, we made sure that each time it was given a new parabolic orbit so
that it would not become trapped in orbit around one of the four gas giants.
270 – Parabolas
Quadratics
Advanced
To learn the relationship between the center, foci, and intercepts of an ellipse and the equation of an
ellipse.
ellipse – the set of points P in a plane which satisfy PF1 + PF2 = d, where F1 and F2 (its foci) are any two fixed
points and d (its focal constant) is a constant with d > F1F2.
focus (plural foci) – fixed points from which the sum of distances to a point on the conic is a constant.
vertices (ellipse) – endpoints of the major axis of an ellipse.
major axis – the line segment joining the two foci of an ellipse and has endpoints at the vertices of the ellipse.
minor axis – line segment perpendicular to the major axis, passing through the center of the ellipse.
semi-major axis – line segment that is half the length of the major axis. Its endpoints are the center of the
ellipse and one of the vertices.
semi-minor axis – a line segment half the length of the minor axis, perpendicular to the semi-major axis.
eccentricity – a measure of the ovalness of an ellipse.
A branch of mathematics known as elliptic geometry does not, in fact, derive its name from the ellipse. Instead,
the name comes from the Greek word elleiptis, meaning to lack or fall short. Like hyperbolic geometry, the name
elliptic geometry was coined by the geometer Felix Klein, creator of the Klein bottle. In elliptical geometry, there
are no lines that pass through the point P and are parallel to the line L, unlike Euclidean geometry, where only 1
line passes through P and is parallel to L.
Quadratics → Conic Sections → Ellipses
Back around 1500 A.D., Nicholas Copernicus came up with his bizarre (at the time) notion that the Earth
traveled around the Sun. In his day, it was believed that the Earth traveled in a perfectly circular orbit.
Unfortunately, the mathematics at the time was unable to accurately describe the observed motions of the
heavens. All sorts of wild attempts were made to fit the math to the observations, such as the idea of
"epicycles", where the celestial objects traveled in a circular path on a circular path. For theological reasons, it
was incomprehensible that the motions of the planets should be in any shape other than a circle.
However, around 1660 A.D., a bright young astronomer from Germany by the name of Johannes Kepler finally
arrived at the correct assumption: the planets travel in elliptical orbits. The Sun is at one of the foci of the
ellipse. He is also responsible for Kepler's Three Laws of Motion, all dealing with ellipses and every one fitting
the observed motions of the planets.
As with all the other major conic sections, we can define an ellipse in geometric terms. We can also use
algebraic equations to describe a particular ellipse.
Geometric Definition of an Ellipse
An ellipse is the set of all points (x, y) the sum of whose distances from two fixed points (foci) is
constant.
The line through the foci intercepts the ellipse as two points called the vertices (vertex is singular).
The chord joining the vertices is called the major axis. The chord perpendicular to the major axis that
passes through the center of the ellipse is called the minor axis. The length of half of the major axis is
known as the semi-major axis (denoted by a) and the length of half of the minor axis is known as the
semi-minor axis (denoted by b).
Quadratics
Ellipses – 271
Now that we have defined an ellipse geometrically, it is time to apply some algebra. The distance between one
vertex and the center of the ellipse can be represented by the letter a. In which case, the entire length of the
major axis then has length 2a. Similarly, we designate the length of the minor axis as 2b, where b is the
distance between the center of the ellipse and the edge of the ellipse perpendicular to the vertex. Using these
two quantities, we can develop a standard equation of an ellipse.
Standard Equation of an Ellipse (Center at the Origin)
The standard equation of an ellipse, with center at the origin (0, 0) and
major and minor axes of lengths 2a and 2b, respectively, where a > b, is:
=
1
major axis is horizontal
minor axis is vertical
=
1
major axis is vertical
minor axis is horizontal
The foci lie on the major axis, c units from the center, where c2 = a2 – b2.
Find an equation of the ellipse with center at the origin.
Vertices: ( 5, 0)
Foci: ( 2, 0)
solution
We are given that the vertices are at (5, 0) and (–5, 0), which means that the major axis of this ellipse is
horizontal. We are also given that the foci are at (2, 0) and (–2, 0). Using this information we already know
that:
a
=
5
distance from vertex to center
c
=
2
distance from focus to center
272 – Ellipses
Quadratics
In order to find b, we use the relationship c2 = a2 – b2:
c2
=
a2 – b2
22
=
52 – b2
substitute in a = 5 and c = 2
4
=
25 – b2
simplify
b2
=
21
subtract 4 from both sides and add b2 to both sides
b
=
take square root of both sides
Now in our standard form of the equation of an ellipse, we have:
x
2
a
x
2
2
2
(
2
2
2
2
y
2
+
=
1
standard form of equation of ellipse
=
1
substitute in for a and b
=
1
simplify
answer
2
21)
2
4
y
b
y
+
x
+
21
Eccentricity
Now that we have defined ellipses, how can we tell one ellipse from another without long involved
calculations? One way is by looking at just how "flattened" the ellipse is. Consider the orbits of the planets.
The reason why early astronomers had difficulty determining that the shape of the orbits was elliptical was
because the foci of the planets' orbits are relatively close to the center of the solar system, thus the orbits appear
to be nearly circular. To measure the ovalness of an ellipse, we use the concept of eccentricity.
Definition of Eccentricity
The eccentricity e of an ellipse is given by:
e
=
eccentricity
Furthermore, for an ellipse, 1 > e > 0.
Since the foci of an ellipse are located on the major axis between the center and the vertices, it follows that
0
<
c
<
a
If c = 0, then the foci are identical to the center, and we have a circle. Recall that c2 = a2 – b2, so if c = 0, then
we have:
02
=
a2 – b2
let c = 0
b2
=
a2
axes are equal to each other
Quadratics
Ellipses – 273
which gives us, in the standard equation of an ellipse:
x
2
a
2
+
y
a
2
2
x2 + y2
=
1
ellipse where a2 = b2
=
a2
multiply both sides by a2
This is just the standard equation of a circle with r = a.
As it turns out, eccentricity is a property shared by all conic sections, and, in fact, can be used to define a conic
section.
The area of an ellipse is given (exactly) by:
A
Note:
=
πab
If a = b, then we just have the area of a circle: A = r2.
274 – Ellipses
Quadratics
Advanced
To learn the geometric definition and standard equations associated with hyperbolas.
foci (hyperbola) – the two points from which the difference of distances to a point on the hyperbola is constant.
focal radius – a line connecting a point and a focus.
difference of focal radii – the absolute value of the difference of the distances from a point on a hyperbola to
the two foci of the hyperbola.
transverse axis – the line passing through both the vertices and the foci of a hyperbola.
asymptotes (hyperbola) – two lines that are approached by the points on the branches of a hyperbola as the
points get farther from the foci. The hyperbola never touches the asymptotes.
conjugate axis – a line perpendicular to the transverse axis and passing through the center of the hyperbola.
eccentricity (hyperbola) – a measure of the flatness or roundedness of the branches of a hyperbola.
A branch of mathematics known as hyperbolic geometry does not, in fact, derive its name from the hyperbola.
Instead, the name comes from the Greek word hyperbole (note spelling), which means excessive. In hyperbolic
geometry it can be proved that there exists more than one line passing through a point P and parallel to a line L,
whereas in typical Euclidean geometry (the geometry most of us are familiar with), only 1 line can pass through P
and be parallel to a given line L. Felix Klein, creator of the topological oddity of the Klein Bottle, is credited with
coining the term hyperbolic geometry.
Quadratics → Conic Sections → Hyperbolas
Remember that we defined an ellipse as the sum of the distances between the foci and a point on the ellipse. A
hyperbola, on the other hand, is like an inside-out ellipse. We define a hyperbola in terms of the difference of
the distances between the foci and a point on the hyperbola.
Geometric Definition of a Hyperbola
A hyperbola is the set of all points P = (x, y) in the plane such that the difference
of the distances from P to two fixed points F1 and F2 (the foci ) is a given constant.
The line connecting P and a focus is known as a focal radius. Thus
the given constant is sometimes called the difference of focal radii .
Like ellipses, the line through the two foci intersects the hyperbola at two points, which we will again call the
vertices. The line segment connecting the vertices is called the transverse axis.
Quadratics
Hyperbolas – 275
Just like every other conic section, we have a standard form for the equation of a hyperbola
based on its orientation (vertical or horizontal).
Standard Equation of a Hyperbola (Center at the Origin)
The standard form of the equation of a hyperbola with center at (h, k) is:
=
1
transverse axis is horizontal
=
1
transverse axis is vertical
The vertices are a units from the center and the foci are c units from the center.
Also, b2 = c2 – a2.
When attempting to draw a hyperbola, useful aids to sketching are the asymptotes of the hyperbola. Every
hyperbola has two asymptotes that intersect at the center of the hyperbola. These asymptotes form the
diagonals of a rectangle of dimensions 2a by 2b, with center at the origin (0, 0). Remember that 2a is the length
of the line segment connecting the vertices. 2b is the length of the line segment connecting (0, b) and (0, –b)
[horizontal orientation] and is called the conjugate axis. If the hyperbola is oriented vertically, then the
conjugate axis connects (b, 0) and (–b, 0). In either case, the conjugate axis is perpendicular to the transverse
axis and passes through the center of the hyperbola.
Asymptotes of a Hyperbola (Center at the Origin)
For a horizontal transverse axis, the equations of the asymptotes are
y
=
and
y
=
horizontal transverse axis
For a vertical transverse axis, the equations of the asymptotes are
y
276 – Hyperbolas
=
and
y
=
vertical transverse axis
Quadratics
Graphically, we represent the asymptotes as follows:
Note the rectangular box shape formed by the asymptotes. Remember that one of the defining characteristics of
an asymptote is that the graph of the hyperbola never, ever crosses the asymptote. In fact, the hyperbola never
even touches an asymptote. An asymptote is a line that comes arbitrarily close to the hyperbola as the branches
of the hyperbola extend out to infinity. Usually when we sketch a hyperbola, we first plot the vertices, then we
plot the endpoints of the conjugate axis, draw a rectangle the connects all four points, draw the asymptotes, and
last we sketch the branches of the hyperbola outward from the vertices such that they approach the asymptotes
(but never touch or cross the asymptotes!).
a.)
Find an equation of the specified hyperbola with center at the origin.
Vertices: (0, 4)
b.)
Foci: (0, 5)
Use the asymptotes to help sketch the hyperbola.
solution
a.)
The vertices are located at (0, 4) and (0, –4), so the distance between the vertices and the center is
a = 4. The foci are located at (0, 5) and (0, –5) which gives us c = 5. To find b, we use the
relationship:
b2
c2 – a2
=
foci relationship
Substituting in a = 4 and c = 5, we get:
b2
=
52 – 42
substitute in a = 4 and c = 5
=
25 – 16
simplify
=
9
We notice that the vertices are located along the y-axis, so we have a vertical transverse axis and our
equation in standard form for this hyperbola is:
y
2
16
Quadratics
−
x
2
25
=
1
transverse axis vertical
answer
Hyperbolas – 277
b.)
Now we need to find the asymptotes in order to graph this hyperbola. Since we have a vertical
transverse axis, we use the relation:
y
=
a
b
x
and
y
=
a
− x
b
In this case, a = 4 and b = 5, so we have:
y
=
4
5
x
and
y
=
4
− x
5
Using these asymptotes, we can sketch the hyperbola as shown below. The hyperbola is represented
by the red lines.
answer
Eccentricity of a Hyperbola
Like ellipses, there is a special relationship between the distance from the center to a focus and the distance
from the center to a vertex. This relationship is known as eccentricity and is defined for hyperbolas in exactly
the same way we define eccentricity for ellipses. The difference is that whereas eccentricity for an ellipse is
defined as 1 > e > 0, eccentricity for a hyperbola is defined such that e > 1.
Definition of Eccentricity of a Hyperbola
The eccentricity of a hyperbola is given by the ratio
e
Note:
=
This is the same definition for the eccentricity of an ellipse. However, in the case of a hyperbola,
c > a, so it follows that e > 1. If the eccentricity is large, then the branches of the hyperbola are
nearly flat. As e gets closer and closer to 1, the branches of the hyperbola are more pointed.
278 – Hyperbolas
Quadratics
Applications of Hyperbolas
There are a variety of applications involving hyperbolas. As with any conic section, they can be used to trace
the path of celestial objects. Many comets have hyperbolic orbits. Comets with hyperbolic (or parabolic) orbits
can only be seen once when they pass by the Earth, but comets with elliptical (or circular) orbits, such as the
famous Halley's comet can be seen periodically. In the case of Halley's comet, we see it every 76 years or so
(its next visit is in 2061). The sun serves as one focus of the orbit. The type of orbit for a comet or other
celestial body can be determined as follows:
1.
Ellipse
v
<
2.
Parabola
v
=
3.
Hyperbola
v
>
2GM
p
2GM
p
2GM
p
In all three cases, p is the distance between the vertex and the focus of the comet's orbit (in meters), v is the
velocity of the comet at the vertex (in meters per second), M is the mass of the sun (M 1.991 x 1030 kg), and G
is Newton's Gravitational Constant (G ≈ 6.67 x 10−11 m3/(kg s2)).
Note that the velocity is actually the escape speed necessary to escape the pull of the Sun's gravity. If the speed
is too little, then the comet is trapped in an elliptical orbit, never to leave. If the speed is greater than or equal to
the escape velocity, then the comet will eventually leave the solar system, never to return.
One very useful application of using escape velocities is in planning missions to other planets. In order to
escape Earth's gravity, a satellite or probe circles the Earth a few times to gain speed, when its speed is great
enough, it eventually leaves on its journey to another planet (such as Mars). When it reaches Mars, it uses the
gravitational pull of Mars in order to slow itself down to a velocity less than the escape velocity so it can remain
in orbit. Using the gravity of a planet to gain speed is often called a "slingshot".
Note:
Due to conservation of momentum, every time a satellite travels around the Earth in order to gain
speed, the rotation of the Earth slows down just a tiny bit. Similarly, when the satellite slows down
around Mars, the rotation of Mars speeds up just a fraction. Mother Nature never gives away energy
for free, although modern science is trying to find a way around that.
Quadratics
Hyperbolas – 279
280 – Hyperbolas
Quadratics
Advanced
To summarize the equations we use to translate conic sections around in the Cartesian plane.
No definitions on this page.
Archimedes of Syracuse (287-212 B.C.) was one of the greatest thinkers of the ancient world. He established
principles for plane and solid geometry, discovered the concept of specific gravity, and incidentally discovered
the area of a circle is equal to π times its radius squared, that is, A = π r2. He also used a sophisticated
technique to find the area under a curve, which anticipated the development of integral calculus by about 2000
years.
Quadratics → Conic Sections → Translation of Conics
In all the other thoughts under the thought Conic Sections, we look at various shapes such as circles, parabolas,
ellipses and hyperbolas. In order to introduce you to the fundamental concepts, we leave each shape centered at
the origin. Now it is time to start monkeying around with the equations and moving the conic sections around
in the Cartesian plane.
The basic form of the equations remains virtually unchanged, except that everywhere we see an x, we plug
in (x – h) and everywhere we see a y, we plug in (y – k). In general, the coordinate point (h, k) represents the
"new" center of the conic section. When (h, k) = (0, 0), the conic section is centered on the origin. If h = 0
and k 0, then the conic section is shifted along the vertical axis. If h 0 and k = 0, then the conic section is
shifted along the horizontal axis.
Anytime you are confronted with a conic section that is not centered at the origin, this is the page you want to
come back to for reference.
Quadratics
Translation of Conics – 281
Standard Forms of Equations of Conics
Circle: Center = (h, k)
Radius = r
(x – h)2 + (y – k)2
=
r2
standard equation of a circle
Parabola: Vertex = (h, k)
Directed distance from vertex to focus = p
(x – h)2
=
4p(y – k)
parabola opens up or down
(y – k)2
=
4p(x – h)
parabola opens left or right
Ellipse: Center = (h, k)
Major axis length = 2a
Minor axis length = 2b
( x − h)
b
2
2
+
( y − k)
a
2
=
1
major axis
horizontal
=
1
major axis vertical
2
Hyperbola: Center = (h, k)
Transverse axis length = 2a
Conjugate axis length = 2b
Note:
=
1
transverse axis
horizontal
=
1
transverse axis vertical
If h < 0, then the conic is shifted to the left. If h > 0, then the conic is shifted to the right.
If k < 0, the conic is shifted downward. If k > 0, then the conic is shifted upward.
Rather than give an extended example of how each equation can be used on this page, we have subdivided
examples of each translation into separate thoughts in the Algebra Brain.
Translations of conics have important applications in the real world. When we trace the trajectory of a
projectile, it travels in a parabolic path. The vertex of the parabola is often far from the origin. Usually, we
assign the origin to be the starting point of the projectile and then figure out the rest of the problem from there.
Another real world example involves celestial mechanics. Orbits of planets are elliptical with the sun at one
focus. If we place the origin so that it coincides with the sun, then problems involving orbital mechanics
become slightly easier to solve.
For more in-depth examples of how to move conic sections around the Cartesian plane, check out the following
thoughts:
Circle Translation
Parabola Translation
Ellipse Translation
Hyperbola Translation
282 – Translations of Conics
Quadratics
Advanced
To move circles around in the Cartesian plane.
No definitions on this page.
Apollonius of Perga (c.262-170 B.C.) is perhaps the greatest Greek geometer since Euclid. His work, Conics, is
considered to be the greatest achievement of Greek geometry. In Conics, Apollonius set forth new ideas on how
to subdivide the cone to produce circles. He even improved on Archimedes calculation of π.
Quadratics → Conic Sections → Translation of Conics → Circle Translation
In the previous thought, we introduced the notion of moving a conic section around in the plane. Now we are
going to focus on a specific example, namely the circle.
For your convenience, we will once again provide you with the standard equation of a circle:
Standard Equation of a Circle
The point (x, y) lies on the circle of radius r and center (h, k) if and only if
(x – h)2 + (y – k)2
=
r2
standard form of the equation of a circle
For a circle centered on the origin, the point (h, k) = (0, 0).
Find the standard form of the equation of the specified circle:
Center: (3, –2)
Solution Point: (–1, 1)
Graph the circle.
solution
We are given two pieces of data about this circle. We know its center and we know a given point located on
that circle. All we need to do is find the radius of that circle. To do that we apply the distance formula, since
the distance between the center and the solution point will be equal to the radius. We let C = (h, k) = (3, –2) and
P = (x1, y1) = (–1, 1).
Quadratics
Circle Translation – 283
( x1 − h ) 2 + ( y1 − k ) 2
=
r
Distance Formula
2
=
r
substitute in (h, k) = (3, –2) and (x1, y1) = (–1, 1)
2
=
r
simplify
16 + 9
=
r
simplify
25
=
r
simplify
5
=
r
take positive square root
2
( −1 − 3) + ( 1 + 2)
2
( −4) + ( 3)
Thus we now know that the radius of the circle is 5. Using our standard equation of a circle, we can thus define
this circle as:
(x – h)2 + (y – k)2
=
r2
standard equation of circle
(x – 3)2 + (y + 2)2
=
25
substitute in (h, k) = (3, –2) and r = 5
answer
The graph of this circle looks like:
answer
284 – Circle Translation
Quadratics
Advanced
To move parabolas around in the Cartesian plane.
No definitions on this page.
The eighth volume of Apollonius's work, Conics has long since vanished from the world, along with a number of
other writings. Scholars are now trying to piece them back together. Among his writings, he mentions "burning
mirrors", wherein he disproved the notion that parallel rays of light could be focused by a spherical mirror.
Quadratics → Conic Sections → Translation of Conics → Parabola Translation
Moving a parabola around the Cartesian plane is very similar to moving a circle around. The defining point of a
circle is its center, whereas for a parabola, we define it in terms of its vertex. The basic equation for a parabola
translated about the Cartesian plane is very similar to the equation of a parabola with its vertex at the origin.
The only difference is that we replace x with (x – h) and y with (y – k). Now we will show an example or two
on how to work with parabolas moved around the plane. For convenience, we once again provide you with the
standard equation of a parabola.
Standard Equation of a Parabola
The standard form of the equation of a parabola with vertex (h, k) and directrix y = –p is
(x – h)2
=
4p(y – k)
p≠0
vertical axis
For directrix x = –p, the equation becomes
(y – k)2
=
4p(x – h)
p≠0
horizontal axis
The focus is on the axis p units (directed distance) from the vertex.
Find the vertex, focus, and directrix of the parabola and sketch its graph.
⎛x +
⎜
⎝
1⎞
⎟
2⎠
2
=
4(y – 3)
given parabola
solution
Since our parabola equation is in standard form, we can immediately find the vertex, which is located at:
⎛ − 1 , 3⎞
⎜ 2 ⎟
⎝
⎠
coordinates of the vertex
answer
Our equation also tells us that p = 1. Since x is the variable that is squared, we know that this has a vertical axis.
Therefore, the focus is located p units above the vertex (because p is positive). The focus lies at:
⎛ − 1 , 4⎞
⎜ 2 ⎟
⎝
⎠
Quadratics
coordinates of the focus
answer
Parabola Translation – 285
Finally, by definition the directrix lies p units below the vertex, so we know that the directrix is given by the
line:
y
=
equation of directrix
answer
2
Using these three quantities, we can now sketch the graph of the parabola:
answer
Find an equation of the parabola with the given parameters:
Vertex: (3, 2)
Focus: (–1, 2)
Graph the parabola.
solution
Note that the y-coordinates of both the vertex and the focus are the same. That tells us that this parabola has a
horizontal axis of symmetry. We are given that the vertex (h, k) is (3, 2). Now we need to find p in order to
find an equation. We can find p by taking the difference of the x-coordinates of the vertex and the focus. In
this case, the difference is 4. Since the focus is located to the left of the vertex on the Cartesian plane, we know
that this parabola opens up to the left. Therefore, p = –4.
Using the vertex and p, we can now find our equation:
(y – k)2
=
4p(x – h)
standard equation of parabola
opens to the right or left
(y – 2)2
=
4(–4)(x – 3)
substitute in (h, k) = (3, 2) and p = –4
(y – 2)2
=
–16(x – 3)
simplify
answer
Now we use this equation to help us graph the parabola:
286 – Parabola Translation
Quadratics
answer
Quadratics
Parabola Translation – 287
288 – Parabola Translation
Quadratics
Advanced
To move ellipses around in the Cartesian plane.
No definitions on this page.
Apollonius of Perga's (c.262-170 B.C.) most important contribution to the study of conics was the fact that he
used a single cone to produce all conic sections. He split the cone along two fixed lines called the latus
transversum and the latus erectum. These eventually became a coordinate system and frame of reference.
Thus geometry could do the work that algebra does now.
Quadratics → Conic Sections → Translation of Conics → Ellipse Translation
Translation of ellipses is very similar to translations of other conic sections, particularly circles, since ellipses
and circles are directly related. Once again, we replace all instances of x with (x – h) and all instances of y with
(y – k), where (h, k) is the location of the center of the ellipse.
For convenience, we give you the standard equations of an ellipse again, this time not centered at the origin.
Standard Equation of an Ellipse
The standard equation of an ellipse, with center (h, k) and major
and minor axes of lengths 2a and 2b, respectively, where a > b, is:
=
1
major axis is horizontal
minor axis is vertical
=
1
major axis is vertical
minor axis is horizontal
The foci lie on the major axis, c units from the center, where c2 = a2 – b2.
Find the center, foci, and vertices of the ellipse, and sketch its graph.
( x − 1)
9
2
+
( y − 5)
25
2
=
1
solution
First, we notice that the number under the y-term is larger than the number under the x-term. Recall that for an
ellipse, a > b, therefore if we compare our given equation to the standard form of the equation, we realize that
this ellipse has a vertical major axis and a horizontal minor axis.
The second thing we can find just by looking at our equation is the center. Again, we compare our given
equation to the standard equation. The center is located at (h, k). In our given equation, h = 1 and k = 5, so the
center is located at:
(1, 5)
Quadratics
center
answer
Ellipse Translation – 289
To find the vertices, we need to know a. We already stated that a2 is the larger of the denominators, by
definition, so we can conclude that:
a
=
=
25
5
The vertices of the ellipse are located a units away from the center. Our ellipse has a vertical major axis, so the
vertices are located 5 units above the center and 5 units below the center, at:
(1, 5 – 5)
=
(1, 0)
one vertex
(1, 5 + 5)
=
(1, 10)
other vertex
answer
Finally, we need to find the foci of the ellipse. According to the algebraic definition of an ellipse given above,
the foci lie c units from the center (in this case, above and below the center because our major axis is vertical),
where
c2
=
a2 – b2
foci equation
c2
=
25 – 9
substitute in a2 = 25 and b2 = 9
c2
=
16
simplify
c
=
4
take square root of 16
Thus the foci are located at
(1, 5 + 4)
=
(1, 9)
one focus
(1, 5 – 4)
=
(1, 1)
other focus
answer
Once we have determined the center, vertices, and foci of this ellipse, we can now graph it:
answer
290 – Ellipse Translation
Quadratics
Find an equation for the ellipse with the given features:
Vertices: (0, 2) and (4, 2)
Minor axis length of 2
Graph the ellipse.
solution
We are given very little information about this ellipse, but believe it or not, we can find everything out about
this ellipse we need just from these two pieces of information.
First, let's deal with the vertices. By definition, the center of the ellipse is located at the midpoint of the line
segment connecting the two vertices (the major axis). Therefore, if we use the Midpoint Formula, we can easily
find the center of the ellipse:
⎛ x1 + x2 y1 + y2 ⎞
,
⎜
⎟
2
⎝ 2
⎠
=
⎛ 0 + 4, 2 + 2⎞
⎜ 2
⎟
2 ⎠
⎝
Midpoint Formula
=
⎛ 4, 4⎞
⎜ 2 2⎟
⎝
⎠
simplify
=
(2, 2)
simplify
center of the ellipse
We are given that the length of the minor axis is 2, which is equal to 2b, where b is the length of the line
segment connecting the center and the endpoint of the minor axis. This implies that b = 1.
The length of the major axis is defined as 2a, so a is the distance between the center and the vertices. In this
case, the length of the major axis is 4, so then a = 2.
One last thing we have to determine is which way the major axis is oriented (vertical or horizontal). Since the
vertices both have the same y-coordinate, and by definition are located on the major axis, we now know that the
major axis is aligned horizontally. Thus, we use the first standard equation given in the definition above:
( x − h)
a
2
+
2
( x − 2)
b
2
+
2
2
( x − 2)
4
( y − k)
2
2
( y − 2)
=
1
standard equation of ellipse
major axis vertical
=
1
substitute in (h, k) = (2, 2), a = 2, b = 1
=
1
simplify
answer
2
2
1
2
+
( y − 2)
1
2
The graph of this ellipse looks is shown below:
Quadratics
Ellipse Translation – 291
answer
292 – Ellipse Translation
Quadratics
Advanced
To move hyperbolas around in the Cartesian plane.
No definitions on this page.
Among the 400 theorems in Apollonius's Conics are ideas for cataloging "new" kinds of closed conics called
parabolas, ellipses, and hyperbolas. He also sets aside the Pythagorean distaste for infinities, infinitesimals, and
infinite sets. This new way of thinking would influence later mathematicians and pave the way for infinitesimal
calculus. Finally, his notions of epicircles, epicycles and eccentrics would influence Ptolemaic cosmology until
Johannes Kepler (1571-1630) discovered the correct motions of the planets: elliptical orbits.
Quadratics → Conic Sections → Translation of Conics → Hyperbola Translation
When we first introduced the notion of a hyperbola, we confined our discussion to a hyperbola whose center is
located at the origin. However, not all hyperbolas can be written so that their center is at the origin and
oftentimes is serves our purposes better to locate the center of the hyperbola somewhere else.
Like translations of circles, parabola, and ellipses, translation of hyperbolas in the Cartesian plane does not
involve a really radical (not in the mathematical sense) change to our original hyperbola equation. Again, we
simply replace all instances of x with (x – h) and we replace all instances of y with (y – k). Note that this applies
to both the standard equations of the hyperbola and the equations of the asymptotes. Since we shift the center
of the hyperbola to the point (h, k), we also shift the intersection of the asymptotes, thus the asymptotes
themselves are shifted by exactly the same amount. This will hopefully become clearer when we work an
example.
We now provide the equations for a hyperbola whose center is located at (h, k). If (h, k) = (0, 0), then the center
is located at the origin, and we obtain the same hyperbola equations as we did before.
Standard Equation of a Hyperbola
The standard form of the equation of a hyperbola with center at (h, k) is:
=
1
=
1
transverse axis is horizontal
transverse axis is vertical
The vertices are a units from the center and the foci are c units from the center. Also, b2 = c2 – a2.
The equations for the asymptotes are then as follows:
Quadratics
Hyperbola Translation – 293
Asymptotes of a Hyperbola
For a horizontal transverse axis, the equations of the asymptotes are
(y – k)
=
and
(y – k)
=
For a vertical transverse axis, the equations of the asymptotes are
(y – k)
=
and
(y – k)
=
Find the center, vertices, and foci of the hyperbola and sketch its graph. Sketch the asymptotes as an aid in
obtaining the graph of the hyperbola.
( x + 1)
4
2
−
( y − 2)
2
=
1
1
solution
We can find the center of the hyperbola just by looking at the given equation. If we compare the given
hyperbola equation to the standard equation for a hyperbola, we can see that the center is located at (–1, 2).
(Remember that x + h = x – (–h)):
Center:
(–1, 2)
center of the hyperbola
answer
Since the x-term is first in the hyperbola equation, we know that this hyperbola has a horizontal transverse axis.
We need to know this in order to find the vertices and foci. To find the vertices, we need a, which is the square
root of the denominator under the x-term:
a2
=
a
=
4
2
take square root of both sides
The number a tells us how many units from the center the vertices are located. In this case, the vertices are
located 2 units to the left and right of the center (because it is aligned horizontally). Thus, the vertices are
located at:
(–1 – 2, 2)
=
(–3, 2)
one vertex
(–1 + 2, 2)
=
(1, 2)
the other vertex
answer
Once we have found the vertices, we can concentrate on the foci, which are each located c units away from the
center of the hyperbola. The number c is found from the formula:
c2
=
a2 + b2
foci equation
=
4+1
substitute in a2 = 4 and b2 = 1
c2
=
5
simplify
c
=
5
294 – Hyperbola Translation
take square roots of both sides
Quadratics
Note that the square root of 5 is larger than 2, or a, which is as it should be. For a hyperbola, c > a. Thus, the
foci are located at:
( −1 +
5 , 2)
≈
(1.236, 2)
one focus
( −1 −
5 , 2)
≈
(–3.326, 2)
other focus
answer
We are almost ready to sketch our hyperbola. All we need now are the equations for our asymptotes to help us
sketch the branches of the hyperbola. According to our asymptote equations from the top of the page, they
should take the form:
(y – k)
b
=
a
( x − h)
and
(y – k)
=
b
− ( x − h)
a
because the hyperbola has a horizontal transverse axis. If we substitute in (h, k) = (–1, 2), a = 2 and b = 1, we
obtain:
(y – 2)
=
y
=
1
2
1
2
( x + 1)
x+
5
2
substitute in (h, k) = (–1, 2), a = 2 and b = 1
rewrite equation in Slope-Intercept Form
one asymptote
and
(y – 2)
=
1
− ( x + 1)
2
substitute in (h, k) = (–1, 2), a = 2 and b = 1
y
=
1
3
− x+
2
2
rewrite equation in Slope-Intercept Form
the other asymptote
When we finally sketch the graph of this hyperbola, we end up with:
answer
Although we have cluttered up our hyperbola considerably, labeling all points of interest, it is not always
necessary to do so. If we remove all the extraneous information, our hyperbola looks like:
Quadratics
Hyperbola Translation – 295
where the red lines represent the branches of the hyperbola and the blue lines represent the asymptotes.
296 – Hyperbola Translation
Quadratics
Unit 5
Matrices and Determinants
Matrices and Determinants – 297
298 – Matrices and Determinants
Advanced
To learn the basic definition of a matrix.
matrix – a rectangular array of elements used to facilitate the study of problems in which the relation between
these elements is fundamental.
entry (matrix) – the object in a particular row and column of a matrix.
rows – horizontal lines of numbers and variables in a matrix.
columns – vertical lines of numbers or variables in a matrix.
main diagonal entries – matrix entries that form a diagonal line from the top left corner to the lower right.
Arthur Cayley (1821-1895) was one of the most prolific mathematicians of the 19th century. Besides being a
mathematician, he also practiced law for 14 years, while at the same time writing over 300 mathematical papers.
He is most famous for his invention of the algebra of matrices. This is partly a result of attempting to solve
problems involving "n-dimensional" geometry, a term he coined. He saw that increasing dimensionality is the
same as adding another variable to a linear equation. Solving such equations is child's play with matrix algebra,
but considerably more difficult with traditional techniques such as the Method of Substitution.
Matrices
In everyday life, we tend to arrange objects in an orderly fashion. In mathematics, we do the same thing. An
orderly arrangement of numbers and/or expressions is known as a matrix. In this section of the Algebra Brain,
you will learn about the properties of matrices (plural of "matrix"). You will also learn how to manipulate
matrices using properties of addition and multiplication. Finally, you will learn how to use matrices to solve
actual problems that occur in real life.
Before we describe matrices in detail, we need a working definition of what a matrix actually is:
Definition of a Matrix
If m and n are positive integers, an m x n matrix (read “m by n”) is a rectangular array
in which each entry, aij, of the matrix is a number.
An m x n matrix has m rows (horizontal lines) and n columns (vertical lines).
A matrix having m rows and n columns is said to be of order m x n. If m = n,
the matrix is square and of order n. For a square matrix, the entries a11, a22, . . ., ann
are the main diagonal entries.
Matrices and Determinants
Matrices – 299
a.
c.
Order: 2 x 1
b.
Order: 2 x 3
2
3
3
0 2 2
Order: 3 x 2
4
5
4
d.
5 0
Order: 4 x 4 (square matrix)
3
1 1 2 2
1
3 3 4 4
2
5 5 6 6
1
7 7 8 8
Matrices have a wide range of applications. Among the most common use of matrices is to store data. This is
how computers store and process information. All information is stored in an array. Then if the computer
needs a piece of information for some process, it just needs to look at a particular section of the hard drive to
find it. This speeds up processing time considerably.
Another extremely common and important use is to solve many equations with many unknowns simultaneously.
Matrices can greatly simplify the process, although we usually turn to computers to give us a hand with this as
well, as they can do it many times faster than a human.
300 – Matrices
Matrices and Determinants
Advanced
To introduce operations involving matrices and to see how two matrices are equal to one another.
No definitions on this page.
William Rowan Hamilton (1805-1865) was an Irish mathematician and physicist. He showed promise of genius
at an early age, mastering the languages of Greek, Latin, Hebrew, Persian, Arabic, and Sanskrit by the time he
was ten. In mathematics, Hamilton worked to justify the use of negative and imaginary quantities. He used his
ideas about negative numbers to construct the set of rational numbers. His work with imaginary numbers led to
the discovery of the quaternion, or fourth-dimensional equation, in 1843.
Matrices → Matrix Operations
Matrices have a wide range of applications besides solving systems of linear equations. There is a very rich and
diverse branch of mathematics that deals exclusively with matrices. The British mathematician Arthur Cayley
first invented matrices in 1858. He was a Cambridge University graduate and a lawyer by profession. Cayley
and two American mathematicians, Benjamin Peirce (1809 – 1880) and his son Charles S. Peirce (1839 – 1914)
are credited with developing "matrix algebra".
In this section of the Algebra Brain, we will look at some of the fundamental properties and operations
performed with matrices. First, we need to understand how two matrices relate to each other. In other words,
we want to know how to represent matrices and when matrices are equal to each other.
Representation of a Matrix
Matrices can be represented in three standard ways:
1.)
A matrix can be denoted by an uppercase letter such as A, B, C, etc.
2.)
A matrix can be denoted by a representative element enclosed in brackets,
such as [aij], [bij], etc.
3.)
A matrix can be denoted by a rectangular array of numbers, such as:
A
Note:
=
[aij]
=
Two matrices are equal if and only if corresponding entries are equal. That is, they must have the
same order (m x n) and aij = bij for 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Matrices and Determinants
Matrix Operations – 301
Solve for a11, a12, a21, and a22 in the following matrix equation:
⎛ a11 a12 ⎞
⎜
⎟
⎝ a21 a22 ⎠
=
⎛ 4 −7 ⎞
⎜
⎟
⎝3 0 ⎠
solution
Since two matrices are only equal if their corresponding entries are equal, we can conclude
that:
a11
=
4
and
a12
=
–7
and
a21
=
3
and
a22
=
0
answer
302 – Matrix Operations
Matrices and Determinants
Advanced
To learn the definition of the identity matrix.
identity matrix – a matrix that consists of 1’s on its main diagonal and 0’s elsewhere.
James Joseph Sylvester (1814-1897) used the term matrix to denote "'an oblong arrangement of terms
consisting, suppose, of m lines and n columns" because "out of that arrangement, we may form various systems
of determinants." Although Sylvester coined the term, it was his friend Arthur Cayley who put the terminology to
use in his papers of 1855 and 1858.
Matrices → Matrix Operations → Identity Matrix
One of the most useful types of matrices is the identity matrix, defined as follows:
Definition of Identity Matrix
The n x n matrix that consists of 1’s on its main diagonal and 0’s elsewhere
is called the identity matrix of order n and is denoted by
In
=
1 0 0
0
0 1 0
0
0 0 1
0
0 0 0
1
identity matrix
Note than an identity matrix must be square. That is, it has the same number of rows and columns. When the
order is understood to be n, we can simply denote the identity matrix In as I.
Also, if A is an n x n matrix, then the identity matrix has the property that AIn = A and InA = A.
For example:
⎛4 3 0 ⎞ ⎛1 0 0⎞
⎜ 2 −1 −3 ⎟ ⋅ ⎜ 0 1 0 ⎟
⎜
⎟⎜
⎟
⎝7 0 4 ⎠ ⎝0 0 1⎠
=
⎛4 3 0 ⎞
⎜ 2 −1 −3 ⎟
⎜
⎟
⎝7 0 4 ⎠
AI3 = A
=
⎛4 3 0 ⎞
⎜ 2 −1 −3 ⎟
⎜
⎟
⎝7 0 4 ⎠
I3A = A
and
⎛1 0 0⎞ ⎛4 3 0 ⎞
⎜ 0 1 0 ⎟ ⋅ ⎜ 2 −1 −3 ⎟
⎜
⎟⎜
⎟
⎝0 0 1⎠ ⎝7 0 4 ⎠
Matrices and Determinants
Identity Matrix – 303
304 – Identity Matrix
Matrices and Determinants
Advanced
To add/subtract matrices together.
zero matrix – a matrix in which all the elements are zero.
Johannes Kepler (1571-1630) is credited with discovering the correct shape of the orbits of the planets
(elliptical), but he was actually trying to discover the mathematical rules God used to create the world. In his
work Mysterium Cosmographicum, he attempts to explain why there are only six planets. According to Kepler,
this is because "God is always a geometer." Kepler's theory was that each planet was separated from its
neighbors by a regular Euclidean solid. For instance, inside the orbit of Saturn, one could inscribe a cube, which
in turn circumscribed the orbit of Jupiter.
Matrices → Matrix Operations → Matrix Addition
One of the most basic operations of real numbers is to add two real numbers together. Similarly, one of the
most basic operations of matrices is to add two matrices together. However, we have to note a few ground
rules.
First, we can only add matrices of the same order. For example, if we have a 2 x 3 matrix, we can only add it to
another 2 x 3 matrix.
Second, to add two matrices of the same order, we simply add their corresponding entries.
That is all there is to it. The process is summarized below:
Definition of Matrix Addition
If A = [aij] and B = [bij] are matrices of order m x n, their sum is the m x n matrix given by
A+B
=
[aij + bij]
The sum of two matrices of different orders is undefined.
Let's look at a couple of examples:
Add the following matrices:
Note:
a.)
⎛1 1 ⎞ ⎛2 3⎞
⎜
⎟+⎜
⎟
⎝ 3 −4 ⎠ ⎝ 3 5 ⎠
⎛1+ 2 1+ 3 ⎞
⎜
⎟
⎝ 3 + 3 −4 + 5 ⎠
b.)
⎛ 3 2⎞ ⎛ 0 3⎞
⎜ −5 7 ⎟ + ⎜ −2 9 ⎟
⎜
⎟ ⎜
⎟
⎝ −1 0 ⎠ ⎝ −1 4 ⎠
c.)
( −3 2 −1 ) + ( 3 −2 1 )
=
=
=
⎡ 3 + 0 2 + 3⎤
⎢ −5 + ( −2) 7 + 9⎥
⎢
⎥
⎣ −1 + ( −1) 0 + 4⎦
=
⎛3 4⎞
⎜
⎟
⎝6 1⎠
=
answer
⎛3 5⎞
⎜ −7 16 ⎟
⎜
⎟
⎝ −2 4 ⎠
[ −3 + 3 2 + ( −2) −1 + 1 ]
=
answer
(0 0 0)
answer
The solution to (c) is called a zero matrix because all its elements are zero.
Matrices and Determinants
Matrix Addition – 305
Matrix subtraction is defined in similar manner to matrix addition except that we take the difference between
corresponding entries. Again, both matrices must have the same exact dimensions in order for matrix
subtraction to work.
306 – Matrix Addition
Matrices and Determinants
Advanced
To multiply a matrix by a real number.
scalar – a real number.
scalar multiplication – the product of a real number and a matrix.
The English word "sine" comes from a series of mistranslations of the Sanskrit word jya-ardha, often abbreviated
as simply jya or jiva. The Arabs later translated this into an otherwise meaningless Arabic word jiba. Since
Arabic is written without vowels, later writers interpreted jb as jaib, which means "bosom" or "breast". When
Arabic trigonometry was translated into Latin in the 12th century, the translator use the Latin equivalent sinus,
which also means "bosom", and by extension, "fold" (as in a toga over a breast) or a bay or gulf. This Latin word
has now become our English "sine".
Matrices → Matrix Operations → Scalar Multiplication
When we deal with variables and real numbers, we often multiply the two together. The number in front of the
variable is known as a coefficient. However, we can also call constant multiples of real numbers scalars
because they "scale" the number. For instance, if we have the expression 2x, the number 2 "scales" the variable
x. No matter what x is, the result of evaluating the expression is a number that is twice as large as the number
we started with.
Now we are going to do the same thing to matrices. Given a matrix A, if we multiply the entirety of the matrix
by some real number constant, we end up "scaling" all the elements within the matrix. Thus if we have the
expression 3A, what we actually mean is that we multiply each and every element within the matrix A by 3. We
call this process scalar multiplication and define it as follows:
Definition of Scalar Multiplication
If A = [aij] is an m x n matrix and c is a scalar (any real number),
the scalar multiple of A by c is the m x n matrix given by
cA
=
[caij]
scalar multiplication of a matrix
Scalar multiplication is often combined with matrix addition to create matrix expression similar to algebraic
expressions with real numbers. We can take advantage of these expressions to solve matrix equations in much
the same way we solve normal algebraic equations. Let's look at a few examples of simplifying matrix
expressions using scalar multiplication and matrix addition.
For the given matrices find:
a.)
4A + B
b.)
–2B
c.)
A – 3B
A
=
⎛ 2 −2 1 ⎞
⎜4 3 7⎟
⎜
⎟
⎝ 0 −3 8 ⎠
Matrices and Determinants
and
B
=
⎛0 0 3⎞
⎜ 0 −2 1 ⎟
⎜
⎟
⎝ 7 −6 5 ⎠
Scalar Multiplication – 307
solution
Matrix operations tend to follow the same order of operations as normal real numbers. First we perform any
scalar multiplications. Then we perform the addition and subtraction operations in order from left to right.
a.)
b.)
c.)
4A + B
–2B
A – 3B
=
⎛ 2 −2 1 ⎞ ⎛ 0 0 3 ⎞
4⎜ 4 3 7 ⎟ + ⎜ 0 −2 1 ⎟
⎜
⎟ ⎜
⎟
⎝ 0 −3 8 ⎠ ⎝ 7 −6 5 ⎠
multiply each element in A by the scalar 4
=
⎛ 8 −8 4 ⎞ ⎛ 0 0 3 ⎞
⎜ 16 12 28 ⎟ + ⎜ 0 −2 1 ⎟
⎜
⎟ ⎜
⎟
⎝ 0 −12 32 ⎠ ⎝ 7 −6 5 ⎠
add 4A to B
=
⎛ 8 8 −7 ⎞
⎜ 16 10 29 ⎟
⎜
⎟
⎝ 7 −18 37 ⎠
answer
=
⎛0 0 3⎞
−2⋅ ⎜ 0 −2 1 ⎟
⎜
⎟
⎝ 7 −6 5 ⎠
multiply each element in B by the scalar –2
=
⎛ 0 0 −6 ⎞
⎜ 0 4 −2 ⎟
⎜
⎟
⎝ −14 12 −10 ⎠
answer
=
⎛ 2 −2 1 ⎞
⎛0 0 3⎞
⎜ 4 3 7 ⎟ − 3⋅ ⎜ 0 −2 1 ⎟
⎜
⎟
⎜
⎟
⎝ 0 −3 8 ⎠
⎝ 7 −6 5 ⎠
multiply each element in B by the scalar 3
=
⎛ 2 −2 1 ⎞ ⎛ 0 0 9 ⎞
⎜ 4 3 7 ⎟ − ⎜ 0 −6 3 ⎟
⎜
⎟ ⎜
⎟
⎝ 0 −3 8 ⎠ ⎝ 21 −18 15 ⎠
subtract 3B from A
=
⎛ 2 −2 −8 ⎞
⎜ 4 9 4⎟
⎜
⎟
⎝ −21 15 −7 ⎠
answer
308 – Scalar Multiplication
Matrices and Determinants
Advanced
To multiply two matrices together.
No definitions on this page.
In ancient China, the civil service required its employees to undertake examinations for advancement. These
examinations were largely based on Chinese literature, but some administrative services - surveying, taxation,
calendar making - required knowledge of mathematics. Instead of attempting to find creative solutions to
problems, applicants for offices were encouraged to study "classical" mathematics, tried and true methods for
finding solutions. The examination system often required recitation of passages from mathematical texts as well
as the solving of problems in the same manner as described in those texts.
Matrices → Matrix Operations → Matrix Multiplication
Matrix multiplication is by far the most convoluted of the basic matrix operations. Unlike matrix addition,
where we simply add corresponding entries, matrix multiplication is a combination of multiplication and
addition, and not of corresponding entries, either.
First let's define matrix multiplication. Then we will explain the definition and give a few examples of how it
works.
Definition of Matrix Multiplication
If A = [aij] is an m x n matrix and B = [bij] is an n x p matrix, the product AB is an m x p matrix
AB
=
[cij]
matrix multiplication
where cij = ai1b1j + ai2b2j + ai3b3j + . . . + ainbnj.
In order for AB to be defined, the number of columns of the first matrix must equal the number of rows of the
second matrix. The resulting matrix has the order given by the number of rows of the first matrix and the
number of columns of the second matrix. That is, if A is an m x n matrix, B must be an n x p matrix in order for
matrix multiplication to work. The result will be a matrix of order m x p.
Note:
The definition of matrix multiplication indicates a row-by-column multiplication. That is, we multiply
the entries in the first row of A by the corresponding entries in the first column of B and add up the
results. The final tally goes into the first entry of the resultant matrix. We do the same thing with the
first row of A and the second column of B. That result goes into the first row, second column of
resultant matrix C. And so on.
Although matrix multiplication can quickly become tedious, there are good reasons for using it. In today's
modern world, we often use computers to perform matrix operations. However, we still need to know how to
do it by hand just in case we do not have a computer handy.
Let's find the product of two matrices to see how this process works.
Find the product AB, where
A
=
⎛1 2⎞
⎜
⎟
⎝4 2⎠
Matrices and Determinants
and
B
=
⎛ 2 −1 ⎞
⎜
⎟
⎝ −1 8 ⎠
Matrix Multiplication – 309
solution
Before we do anything else, we need to double check the order of the matrices to make sure that we can
multiply them together. Matrix A is 2 x 2 and matrix B is 2 x 2. Since the number of columns of A equals the
number of rows of B, we conclude that we will be able to multiply the matrices together. Furthermore, we will
obtain a 2 x 2 matrix as the result, in the form:
⎛ 1 2 ⎞ ⋅ ⎛ 2 −1 ⎞
⎜
⎟⎜
⎟
⎝ 4 2 ⎠ ⎝ −1 8 ⎠
=
⎛ c11 c12 ⎞
⎜
⎟
⎝ c21 c22 ⎠
form of resultant matrix
To find the entries of the product matrix C, multiply each row of A by each column of B as follows:
AB
=
⎛ 1 2 ⎞ ⎛ 2 −1 ⎞
⎜
⎟ ⋅⎜
⎟
⎝ 4 2 ⎠ ⎝ −1 8 ⎠
multiply two matrices
=
⎡ ( 1) ( 2) + ( 2) ( −1) ( 1) ( −1) + ( 2) ( 8) ⎤
⎢
⎥
⎣ ( 4) ( 2) + ( 2) ( −1) ( 4) ( −1) + ( 2) ( 8) ⎦
add rows of A to columns of B
=
⎛ 2 − 2 −1 + 16 ⎞
⎜
⎟
⎝ 8 − 2 −4 + 16 ⎠
simplify
=
⎛ 0 15 ⎞
⎜
⎟
⎝ 6 12 ⎠
simplify
answer
Be sure that you understand that the number of columns of A must equal the number of rows of B in order for
this to work. If they are not equal, then we end up multiplying a number by something that does not exist, and
the process fails. In other words, the middle two indices must be the same. The order of the resultant matrix is
given by the outer indices, as indicated below:
Application of Matrix Multiplication
One of the most important applications of this process is solving systems of linear equations. If we have a
system of linear equations that looks like:
a11x1 + a12x2 + a13x3
=
b1
a21x1 + a22x2 + a23x3
=
b2
a31x1 + a32x2 + a33x3
=
b3
we can represent this system as:
AX
=
B
linear system of equations written as matrix multiplication
where the matrix A represents all the coefficients of the unknowns, matrix X represents all the unknowns in the
system, and B represents the constant terms. In this case, A is called the coefficient matrix of the system, and X
and B are column matrices:
310 – Matrix Multiplication
Matrices and Determinants
⎛⎜ a11 a12 a13 ⎞⎟ ⎛⎜ x1 ⎞⎟
⎜ a21 a22 a23 ⎟ ⋅ ⎜ x2 ⎟
⎜ a31 a32 a33 ⎟ ⎜ x3 ⎟
⎝
⎠⎝ ⎠
=
x X
A
⎛⎜ b1 ⎞⎟
⎜ b2 ⎟
⎜ b3 ⎟
⎝ ⎠
=
B
Find matrices A, X, and B such that the following system of linear equations can be represented by the matrix
equation AX = B. Solve the system of equations.
x – 2y + 3z =
–x + 3y –
9
z = –6
2x – 5y + 5z = 17
solution
We represent the matrix A with the coefficients of all the unknowns in the system. We represent matrix X with
the unknowns in the system and matrix B is the constant terms:
⎛ 1 −2 3 ⎞ ⎛ x ⎞
⎜ −1 3 −1 ⎟ ⋅ ⎜ y ⎟
⎜
⎟⎜ ⎟
⎝ 2 −5 5 ⎠ ⎝ z ⎠
=
⎛9⎞
⎜ −6 ⎟
⎜ ⎟
⎝ 17 ⎠
representation of linear system
with matrix equation
Now in order to solve this system of equations, we form the augmented matrix as follows:
⎛ 1 −2 3
⎜ −1 3 −1
⎜
⎝ 2 −5 5
9
⎞
−6 ⎟
⎟
17 ⎠
augmented matrix formed from coefficient matrix and constant matrix
We can now apply either Gaussian elimination with back-substitution or Gauss-Jordan elimination in order to
solve this system. Let's start doing some elementary row operations to convert the matrix to row-echelon form.
We will start by adding Row 2 to Row 1 and putting the result in Row 2:
R1 + R2
⎛ 1 −2 3
⎜0 1 2
⎜
⎝ 2 −5 5
R3 – 2R1
⎛ 1 −2 3
⎜0 1 2
⎜
⎝ 0 −1 −1
R3 + R2
⎛ 1 −2 3
⎜0 1 2
⎜
⎝0 0 1
⎞
⎟
⎟
17 ⎠
9
3
⎞
⎟
⎟
−1 ⎠
9
3
9⎞
3⎟
⎟
2⎠
add first row to second row
and put result in second row
subtract 2 times first row from third row
and put result in third row
add third row to second row
and put result in third row
Now we have transformed the augmented matrix into row-echelon form. At this point, we can either proceed
with back-substitution or continue performing row operations until we obtain a matrix in reduced row-echelon
form. Let's continue with the latter:
Matrices and Determinants
Matrix Multiplication – 311
R2 – 2R3
R1 + 2R2
R1 – 3R3
⎛ 1 −2 3
⎜0 1 0
⎜
⎝0 0 1
9
⎞
−1 ⎟
⎟
2 ⎠
subtract 2 times the third row from the second row
and put the result in the third row
⎛1 0 3
⎜0 1 0
⎜
⎝0 0 1
⎞
−1 ⎟
⎟
2 ⎠
add 2 times the second row to the first row
and put the result in the first row
⎛1 0 0
⎜0 1 0
⎜
⎝0 0 1
⎞
−1 ⎟
⎟
2 ⎠
subtract 3 times the third row from the first row
and put the result in the first row
7
1
Now we can simply read our solution from the reduced row-echelon form of the augmented matrix:
x
=
1
and
y
=
–1
and
z
=
2
answer
We can double check our solutions using matrix multiplication:
⎛ 1 −2 3 ⎞ ⎛ x ⎞
⎜ −1 3 −1 ⎟ ⋅ ⎜ y ⎟
⎜
⎟⎜ ⎟
⎝ 2 −5 5 ⎠ ⎝ z ⎠
=
⎛ 1 −2 3 ⎞ ⎛ 1 ⎞
⎜ −1 3 −1 ⎟ ⋅ ⎜ −1 ⎟
⎜
⎟⎜ ⎟
⎝ 2 −5 5 ⎠ ⎝ 2 ⎠
substitute in x = 1, y = –1, and z = 2 into
our matrix equation
=
⎡ 1( 1) + ( −2) ( −1) + ( 3) ( 2) ⎤
⎢ ( −1) ( 1) + ( 3) ( −1) + ( −1) ( 2) ⎥
⎢
⎥
⎣ ( 2) ( 1) + ( −5) ( −1) + ( 5) ( 2) ⎦
matrix multiplication
=
⎛ 1+ 2+ 6 ⎞
⎜ −1 − 3 − 2 ⎟
⎜
⎟
⎝ 2 + 5 + 10 ⎠
simplify
=
⎛9⎞
⎜ −6 ⎟
⎜ ⎟
⎝ 17 ⎠
simplify
check
The product of A and X is indeed B, as we were given at the beginning of this problem, so we can conclude that
we have found the correct solution to the system of linear equations.
312 – Matrix Multiplication
Matrices and Determinants
Advanced
To identify properties of matrix operations.
zero matrix – a matrix consisting entirely of zeros.
additive identity – a matrix O such that A + O = A, provided the matrices O and A have the same dimensions.
Indian mathematics originated in northern India around 600 B.C. The brahmin, or priestly caste, were the
bearers of religious traditions that grew into Hinduism. Lower castes were considered unworthy of learning, so
the brahmin passed on their traditions to each other in oral form, as opposed to written form. To aid in
memorization of important ideas, many works were put into poetic stanzas. Such works included the basics of
mathematics and astronomy, along with literature and religion.
Matrices → Matrix Operations → Properties of Matrix Operations
Properties of Matrix Addition and Scalar Multiplication
Matrix addition obeys many of the same properties obeyed by addition of real numbers. Similarly, scalar
multiplication of a matrix also obeys properties obeyed by scalar multiplication of real numbers.
Properties of Matrix Addition and Scalar Multiplication
Let A, B, and C be m x n matrices and let c and d be scalars.
1.)
A+B=B+A
Commutative Property of Matrix Addition
2.)
A + (B + C) = (A + B) + C
Associative Property of Matrix Addition
3.)
(cd)A = c(dA)
Associative Property of Scalar Multiplication
4.)
1A = A
Scalar Identity
5.)
c(A + B) = cA + cB
Distributive Property
6.)
(c + d)A = cA + dA
Distributive Property
Remember that we can only add two or more matrices together if they are of the same order. Note also that if A
and B are of the same order, then A – B represents the sum of A and (–1)B. That is,
A–B
=
A + (–1)B
subtraction of two matrices
One very important property of real numbers is that we can add the number 0 to any real number and we obtain
the number we started with as a result. In other words, zero is the additive identity, where c + 0 = c if c is any
real number. A similar property holds true for matrices. If we have the m x n matrix A and we add another m x
n matrix O that consists entirely of zeros (and is therefore called a zero matrix), then A + O = A. The matrix O
is the additive identity for the set of all m x n matrices. For example, the following matrix is the additive
identity for the set of all 2 x 3 matrices:
O
=
⎛0 0 0⎞
⎜
⎟
⎝0 0 0⎠
Matrices and Determinants
zero 2 x 3 matrix
Properties of Matrix Operations – 313
We can use the properties of matrix addition and scalar multiplication in order to solve matrix equations. The
process is fairly similar to that used to solve equations with real numbers. Compare the following methods of
solution:
Real Numbers
(solve for x)
x+a = b
m x n Matrices
(solve for X)
X+A = B
given equation
X + A + (–A)
=
B + (–A)
subtract same quantity
from both sides
b–a
X+O
=
B–A
simplify
b–a
X
=
B–A
simplify
x + a + (–a)
=
b + (–a)
x+0
=
x
=
Let's look at an example of how these properties can be used to help us solve a matrix equation.
Solve the following equation for X given:
A
⎛ −2 −1 ⎞
⎜1 0⎟
⎜
⎟
⎝ 3 −4 ⎠
=
2X + 3A
=
B
and
B
=
⎛0 3⎞
⎜2 0⎟
⎜
⎟
⎝ −4 −1 ⎠
given equation
solution
Before we start substituting in matrices, we want to use properties of matrix addition and scalar multiplication
to isolate the matrix X by itself, just as we normally isolate the variable x when working with real numbers.
We begin by subtracting 3A from both sides of the equation. Then we multiply both sides of the equation by 1/2
(the same as dividing both sides by 2) to get the matrix X by itself. We know that the matrix X will be a 3 x 2
matrix because both A and B are also 3 x 2 matrices.
2X + 3A
=
B
given equation
2X
=
B – 3A
subtract 3A from both sides
X
=
1
2
( B − 3 A)
multiply both sides by 1/2
At this point, we can now replace A and B with their rectangular array forms:
314 – Properties of Matrix Opeations
Matrices and Determinants
X
⎡⎛ 0 3 ⎞
⎛ −2 −1 ⎞ ⎤
⎟
2 0 − 3⋅ ⎜ 1 0 ⎟ ⎥
⎟
⎜
⎟⎥
2⎢⎜
⎣ ⎝ −4 −1 ⎠
⎝ 3 −4 ⎠ ⎦
replace A and B with rectangular array forms
⎡⎛ 0 3 ⎞ ⎛ −6 −3 ⎞⎤
2 0 ⎟ − ⎜ 3 0 ⎟⎥
⎟ ⎜
⎟⎥
2 ⎢⎜
⎣⎝ −4 −1 ⎠ ⎝ 9 −12 ⎠⎦
multiply second matrix by scalar multiple 3
=
1⎢⎜
=
1 ⎢⎜
=
1⎜
⎛ 6
X
=
⎞
−1 0 ⎟
⎜
⎟
2
⎝ −13 11 ⎠
subtract matrices
⎛ 3
⎜
⎜ −1
⎜ 2
⎜ 13
⎜−
⎝ 2
multiply by scalar multiple 1/2
answer
6
⎞
⎟
0 ⎟
⎟
11 ⎟
⎟
2 ⎠
3
Properties of Matrix Multiplication
Matrix multiplication obeys rules of its own, although these properties are similar in nature to those obeyed by
multiplication of real numbers. However, we must always be aware that matrix multiplication only works if our
matrices have the right orders. That is, if we have two matrices A and B, and we want the product AB, then the
number of columns of A must be equal to the number of rows of B, otherwise we cannot multiply them together.
For more details on how matrix multiplication actually works, we refer you to the Matrix Multiplication section
of the Algebra Brain.
Properties of Matrix Multiplication
Let A, B, and C be matrices and let c be a scalar.
Note:
1.)
A(BC) = (AB)C
Associative Property of Matrix Multiplication
2.)
A(B + C) = AB + AC
Distributive Property
3.)
(A + B)C = AC + BC
Distributive Property
4.)
c(AB) = (cA)B = A(cB)
There is no Commutative Property of Matrix Multiplication. That is, AB ≠ BA, except in certain rare
instances.
Matrices and Determinants
Properties of Matrix Operations – 315
316 – Properties of Matrix Opeations
Matrices and Determinants
Advanced
To represent a system of linear equations with a matrix.
augmented matrix – a matrix derived from a system of linear equations, where the last column of the matrix
represents the constant terms of the system.
coefficient matrix – the matrix derived from the coefficients of the variables in a linear system. It does not
include any of the constant terms of the system.
The Chinese mathematical work Jiu zhang suan shu (Nine Chapters on the Mathematical Art), compiled around
200 B.C., contains two methods of solving systems of linear equations. The first method is known as the method
of surplus and deficiency. It is effectively a method for solving two equations with two unknowns. The second
method is virtually identical to Gaussian elimination and is even presented in matrix form.
Matrices → Linear Systems in a Matrix
A matrix can be derived from a system of linear equations (each written in standard form with the constant term
on the right of the equals sign). This matrix is called the augmented matrix of the linear system. The matrix
derived from just the coefficients of the linear system is called the coefficient matrix.
System of Linear Equations
x – 4y +
3z =
5
–x + 3y
–
z =
–3
2x
–
4z =
6
Augmented Matrix
5 ⎞
⎛ 1 −4 3
⎜ −1 3 −1
⎜
⎝ 2 0 −4
Note:
−3 ⎟
6
⎟
⎠
The last column of the matrix corresponds to the right hand side of the linear system.
Coefficient Matrix
⎛ 1 −4 3 ⎞
⎜ −1 3 −1 ⎟
⎜
⎟
⎝ 2 0 −4 ⎠
This matrix just contains the coefficients of the linear system. Note that a 0 is placed for the missing y-variable
in the third equation.
Whenever forming the coefficient or the augmented matrix, begin by vertically aligning the variables in the
equation and then using zeros for the missing variables:
Matrices and Determinants
Linear Systems in a Matrix – 317
System
Line up variables
x + 3y = 9
–y + 4z = –2
x – 5z =
0
318 – Linear Systems in a Matrix
Form Augmented Matrix
x + 3y
= 9
– y + 4z = –2
x
– 5z =
0
⎛1 3 0
⎜ 0 −1 4
⎜
⎝ 1 0 −5
9
⎞
0
⎟
⎠
−2 ⎟
Matrices and Determinants
Advanced
To solve linear systems of equations by applying elementary row operations to the matrix
representation of a given linear system.
elementary row operations – interchanging, multiplying, and adding rows of a matrix to find solutions for each
of the variables in the linear system represented by the matrix.
row-equivalent – two matrices are said to be row-equivalent when one matrix can be obtained from the other
matrix by applying elementary row operations.
row-echelon form – a matrix that has a stair-step pattern with leading coefficients of 1 along the main diagonal
and zeros below the main diagonal.
leading 1 – the first non-zero entry of a row that does not consist entirely of zeros.
reduced row-echelon form – an augmented matrix that has 1's along the main diagonal of the coefficient matrix
and zeros in every entry above and below the main diagonal of the coefficient matrix.
Although Arthur Cayley is credited with developing the algebra of matrices, the word "matrix" was actually coined
in 1850 by James Joseph Sylvester, a friend of Cayley's and a fellow mathematician.
Matrices → Linear Systems in a Matrix → Elementary Row Operations
When you attempt to solve a system of linear equations, you often have to transform one or more equations into
an equivalent equation in order to eliminate one or more variables. This results in an equivalent system of
linear equations. One way to transform a linear equation is to multiply the entire equation by some real number.
Another way is to interchange two equations. Finally, we can add a multiple of one equation to another
equation.
We can adopt these three procedures to work with matrices. In particular, we can represent a system of linear
equations in a matrix. Then we apply the three operations given above to find solutions for each of the
variables in the linear system.
We call these basic procedures elementary row operations.
Elementary Row Operations
There are three basic row operations that may be
performed on systems of linear equations:
1.
Interchange two rows.
2.
Multiply a row by a nonzero constant.
3.
Add a multiple of one row to another row.
These same operations may be performed on linear systems that are represented as an augmented matrix.
Two matrices are row-equivalent if one can be obtained from the other by a series of elementary row
operations.
Elementary row operations are simple to perform, but they can require considerable amounts of pure arithmetic.
It is important to understand what is transpiring in each step. Therefore, it is helpful to indicate what is
happening to each row of the matrix. In the examples of the elementary row operations given below, we use R
to designate a Row and a subscript (1, 2, etc.) to indicate one particular Row. The operation "3R1 →", for
Matrices and Determinants
Elementary Row Operations – 319
instance, tells us that we multiply each element in the first row by 3. The arrow is pointing to the first row, so
we know that we are supposed to place the results of our multiplication in the first row.
a.)
Interchange two rows:
Original Matrix
Operation
New Row Equivalent Matrix
⎛ 4 3 −1 5 ⎞
⎛0 3 1 0⎞
⎜ 4 3 −1 5 ⎟
⎜
⎟
⎝1 8 3 2⎠
⎜0 3 1 0⎟
⎜
⎟
⎝1 8 3 2⎠
b.)
Multiply the first row by 3:
Original Matrix
⎛ 4 3 −1 5 ⎞
⎜0 3 1 0⎟
⎜
⎟
⎝1 8 3 2⎠
c.)
Operation
3R1 →
New Row Equivalent Matrix
⎛ 12 9 −3 15 ⎞
⎜0 3 1 0⎟
⎜
⎟
⎝1 8 3 2⎠
Add –2 times the third row to the second row:
Original Matrix
Operation
⎛ 4 3 −1 5 ⎞
⎜0 3 1 0⎟
⎜
⎟
⎝1 8 3 2⎠
–2R3 + R2 →
New Row Equivalent Matrix
⎛ 4 3 −1 5 ⎞
⎜ −2 −13 −5 −4 ⎟
⎜
⎟
⎝1 8 3 2 ⎠
Here we are multiplying each element in the second row by –2 and adding to the result the corresponding
element in Row 3. The final result is then placed into its corresponding position in Row 2.
Note:
The operation is written next to the row that is being changed.
Although we have given very general examples of elementary row operations, when you begin to study
Gaussian elimination, you will see that this is an extremely powerful tool for quickly solving systems of linear
equations by hand. Consider the following example of solving a system of linear equations using elementary
row operations:
Solve the system of linear equations using elementary row operations
Linear System
x + 2y = 7
2x +
y = 8
solution
To demonstrate the parallel between elementary row operations and normal procedures on how to solve a linear
system without a matrix, we are going to simultaneously solve the system both without a matrix and using an
associated augmented matrix.
320 – Elementary Row Operations
Matrices and Determinants
Linear System
x + 2y = 7
2x +
y = 8
Associated Augmented Matrix
⎛1 2 7⎞
⎜
⎝2 1
⎟
8⎠
Linear System
multiply the first equation by 2
2x + 4y = 14
2x +
Associated Augmented Matrix
multiply the first row by 2
2R1
2 4
14
⎛
⎜
⎝2 1
y = 8
⎞
⎟
8 ⎠
Linear System
subtract the second equation
from the first equation
2x + 4y = 14
Associated Augmented Matrix
subtract Row 2 from Row 1 and
put the result in Row 2
0x + 3y = 6
R2 – R1→
⎛2 4
⎜
⎝0 3
14 ⎞
6
⎟
⎠
Linear System
divide the second equation
by 3 to solve for y
2x + 4y = 14
Associated Augmented Matrix
divide Row 2 by 3 and
put the result in Row 2
0x +
1
y = 2
Linear System
divide the first equation
by 2
x + 2y = 7
0x +
y = 2
/3R2 →
⎛2 4
⎜
⎝0 1
14 ⎞
2
⎟
⎠
Associated Augmented Matrix
divide Row 1 by 2 and
put the result in Row 1
1
/2R1→
1 2
7
⎛
⎜
⎝0 1
⎞
⎟
2⎠
At this point, we can back-substitute in y = 2 into the first equation and then solve the first equation for x to find
out that x = 3. Therefore, the solution to this system of linear equations is:
x
=
3
and
y
=
2
answer
The last step in the example above illustrates one reason why we use elementary row operations. Notice that in
order to find the solution for y, all we had to do was read across the bottom row because there is only a 0 before
the y. This last matrix is said to be in row-echelon form. The term echelon refers to the stair-step pattern
formed by non-zero elements in the matrix. In this form, a matrix must have the following properties:
Row-Echelon Form and Reduced Row-Echelon Form
A matrix in row-echelon form has the following properties:
1.
All rows consisting entirely of zeros occur at the bottom of the matrix.
2.
For each row that does not consist entirely of zeros,
the first non-zero entry is a 1 (called a leading 1).
3.
For two consecutive (nonzero) rows, the leading 1 in the higher row
is farther to the left than the leading 1 in the lower row.
A matrix in row-echelon form is in reduced row-echelon form if every column
that has a leading 1 has zeros in every position above and below its leading 1.
Matrices and Determinants
Elementary Row Operations – 321
Words fail to adequately describe just what row-echelon form is all about, so let's take a look at a few examples:
The following matrices are in row-echelon form:
a.)
⎛1 0 0 0⎞
⎜0 1 1 5⎟
⎜
⎟
⎝0 0 1 7⎠
b.)
c.)
⎛1 0 2 1 1 ⎞
⎜0 0 1 4 7 ⎟
⎜
⎟
⎝ 0 0 0 1 −9 ⎠
d.)
⎛⎜ 1
⎜0
⎜0
⎜0
⎝
0 0 9⎞
⎟
1 0 1⎟
0 1 5⎟
⎟
0 0 0⎠
The matrix in (d) also happens to be in reduced row-echelon form. The following matrices are not in rowechelon form (reduced or otherwise).
e.)
⎛ 1 4 3 12 ⎞
⎜0 2 8 6 ⎟
⎜
⎟
⎝ 0 0 1 −5 ⎠
f.)
⎛ 1 2 −1 2 ⎞
⎜0 0 0 0 ⎟
⎜
⎟
⎝ 0 1 2 −4 ⎠
We can easily transform (e) into row-echelon form by multiplying R2 by 1/2. We can transform (f) into rowechelon form by interchanging R2 and R3.
322 – Elementary Row Operations
Matrices and Determinants
Advanced
To solve systems of linear equations using Gaussian elimination.
row-echelon form (matrices) – a matrix that has a stair-step pattern with 1's along the main diagonal, and all
zeros below the main diagonal.
inconsistent – a system of linear equations that has no solution.
In 1801, Karl Friedrich Gauss (1777-1855) rediscovered the "lost" asteroid Ceres. Ceres was originally
discovered by Giuseppe Piazzi of Palermo, Italy earlier in the year. Unfortunately, he was unable to keep track
of the strange object in the sky and so Ceres eventually disappeared. He believed that he had found a comet,
but Gauss examined Piazzi's calculations and this examination allowed another astronomer, Baron Francis
Xavier von Zach, to rediscover Ceres at the end of the year. Shortly thereafter, a fair number of asteroids were
subsequently discovered. Now we know of thousands, if not millions, of asteroids in orbit between the planets
Jupiter and Mars.
Matrices → Linear Systems in a Matrix → Elementary Row Operations →
Gaussian Elimination - Matrices
In the previous thought, we looked at transforming matrices by manipulating the rows of matrices. Now we are
going to examine in detail just why we do this. Recall that we can write a system of linear equations in an
augmented matrix. We use a process called Gaussian Elimination with Back-Substitution to transform the
matrix into a form where we can begin solving the system of linear equations. In effect, we reduce the matrix to
row-echelon form. If the system of linear equations does have a solution set, then the last row of the matrix will
contain only two nonzero elements. The next to last element in the row represents the final variable in the
linear system. The last element in the row represents the value of that variable.
It is much easier to see the process than it is to explain it. First we will give you the general rules for Gaussian
Elimination with Back-Substitution. Then we will demonstrate this process on a system of linear equations.
Gaussian Elimination with Back-Substitution
1.)
Write the augmented matrix of the system of linear equations
2.)
Use elementary row operations to rewrite the augmented matrix in row-echelon form.
3.)
Write the system of linear equations corresponding to the matrix in row-echelon form
and use back-substitution to find the solution set.
Now that we have given the rules for the procedure, it is time to get our hands dirty and work an example.
When applying this procedure, it is important to keep track of the order in which the operations are being
performed. In general, it is best to start with the first column, using elementary row operations to obtain zeros
below the leading 1. Then we move on to the second column, again trying to obtain zeros below the leading 1,
and so on.
Matrices and Determinants
Gaussian Elimination – 323
Use Gaussian elimination with back-substitution to solve the following system of linear
equations.
x
3x +
– 3z = –2
y – 2z =
2x + 2y +
z =
5
4
solution
According to the rules for Gaussian elimination, we first need to represent this with a matrix. We use the
coefficients of the variables as the elements in our matrix. The last column of the matrix represents the constant
terms on the right hand side of the equations. We end up with the following matrix representation of our linear
system:
⎛ 1 0 −3
⎜ 3 1 −2
⎜
⎝2 2 1
−2 ⎞
⎟
⎟
4 ⎠
5
matrix representation of our linear system
First of all, notice that the top row has a "1" as our first element. That suits us just fine, so we are not going to
mess with this row. We will start by transforming the "3" in the second row into a "0". To do that, we subtract
R2 from 3R1 and put the result in the second row:
3R1 – R2
⎛ 1 0 −3
⎜ 0 −1 −7
⎜
⎝2 2 1
−2
⎞
−11 ⎟
⎟
4 ⎠
subtract the second row from 3 times the first row
Before we go any further, we should remind you TO BE EXTREMELY CAREFUL ABOUT YOUR SIGNS! Sign
errors are all too frequent when working with elementary row operations, so it is a very, very good idea to
double check each step. This is one reason why we like to use computers to solve this sort of problem. Okay,
back to the problem. Our next step is to get a "0" as the first element in R3. We can do that if we subtract R3
from 2R1 and put the result in the third row:
2R1 – R3
⎛ 1 0 −3
⎜ 0 −1 −7
⎜
⎝ 0 −2 −7
−2
⎞
−11 ⎟
⎟
−8 ⎠
subtract the third row from 2 times the first row
Our next step is to somehow transform the "–2" in the last row into another "0". We do that by subtracting R3
from 2R2 and putting the result in the last row. Again, we need to be VERY careful about our signs.
2R2 – R3
⎛ 1 0 −3
⎜ 0 −1 −7
⎜
⎝ 0 0 −7
−2
⎞
−11 ⎟
⎟
−14 ⎠
subtract the third row from 2 times the second row
In order to get the third element in row 3 to be a "1", we multiply R3 by –1/7:
⎛ 1 0 −3
⎜ 0 −1 −7
⎜
⎝0 0 1
324 – Gaussian Elimination
−2
⎞
−11 ⎟
⎟
2 ⎠
multiply the third row by –1/7
Matrices and Determinants
Although at first glance, it may seem like we have reduced the matrix to row-echelon form, we are not quite
done yet. Remember that the diagonal of a matrix in row-echelon form consists only of "1's". We have a "–1"
in the second row. Getting rid of it is simply a matter of multiplying R2 by –1:
⎛ 1 0 −3
⎜0 1 7
⎜
⎝0 0 1
–1R2
−2 ⎞
11 ⎟
⎟
2 ⎠
multiply the second row by –1
Now that we have converted our matrix into row-echelon form, all we need to do is look at the third row to
determine that:
z
=
2
third linear equation
If we rewrite our matrix in row echelon form as a system of linear equations, we have:
x
– 3z = –2
y + 7z = 11
z =
2
Using back-substitution, we can easily determine that:
x – 3z
=
–2
first linear equation
x – 3(2)
=
–2
substitute in z = 2
x–6
=
–2
simplify
x
=
4
solve for x
y + 7z
=
11
second linear equation
y + 7(2)
=
11
substitute in z = 2
y + 14
=
11
simplify
y
=
–3
solve for y
and that
Therefore, our solution set is:
x
=
4
and
y
=
–3
and
z
=
2
answer
If we had obtained a row that was zeros except for the last entry, then that would mean our system of equations
has no solution. At that point, we do not need to continue any further. Such a system is called inconsistent.
Matrices and Determinants
Gaussian Elimination – 325
326 – Gaussian Elimination
Matrices and Determinants
Advanced
To solve systems of linear equations using Gauss-Jordan elimination.
reduced row-echelon form – a matrix which has 1's (or zeros) along the main diagonal and zeros above and
below the main diagonal.
Karl Friedrich Gauss (1777-1855) was extremely fond of newspapers and magazines, still a novelty item in the
early 19th century. Students at the Unversity of Göttingen in Germany referred to Gauss as the "newspaper
tiger" for his habit of staring down anyone who dared to take a newspaper he wanted first.
Matrices → Linear Systems in a Matrix → Elementary Row Operations →
Gaussian Elimination – Matrices → Gauss-Jordan Elimination
In the previous thought, we used Gaussian elimination with back-substitution to transform a matrix representing
a system of linear equations into a matrix in row-echelon form. Now we will discuss a second method of
solving systems of linear equations that is virtually identical to Gaussian elimination. The difference is that we
are now going to transform a matrix into reduced row-echelon form. Recall that a matrix in reduced rowechelon form has every column with a leading one also having a zero above and below the leading one. This is
really just an extension of Gaussian elimination. We simply keep applying elementary row operations until we
arrive at a matrix in reduced row-echelon form. Then the solutions of the system of linear equations are given
explicitly without resorting to back-substitution.
Let's use the example from the previous thought to illustrate this concept.
In the previous thought, we started with the following system of linear equations:
x
3x +
– 3z = –2
y – 2z =
2x + 2y +
z =
5
4
After apply elementary row operations, we transformed the matrix representing this system into:
⎛ 1 0 −3
⎜0 1 7
⎜
⎝0 0 1
−2 ⎞
11 ⎟
⎟
2 ⎠
row-echelon matrix representing
system of linear equations
In that example, we used back-substitution to find the solution to the system. Now use Gauss-Jordan
elimination to find the solution.
solution
Remember that we want the coefficient part of the augmented matrix to consist of only "1's" along the diagonal
and zeros everywhere else. In order to do that, we are going to continue performing elementary row operations.
We will start be getting rid of the "–3" in the top row. We can do that by adding 3R3 to R1 and placing the
result in R1:
Matrices and Determinants
Gauss-Jordan Elimination – 327
R1 + 3R3
⎛1 0 0
⎜0 1 7
⎜
⎝0 0 1
4
⎞
11 ⎟
⎟
2 ⎠
add 3 times the third row to the first row
That takes care of the first row. Now we need to turn that pesky "7" in the second row into a "0". Again, we
will make use of the third row, subtracting 7R3 from R2 and placing the result in the second row:
⎛1 0 0
⎜0 1 0
⎜
⎝0 0 1
R2 – 7R3
⎞
−3 ⎟
⎟
2 ⎠
4
subtract 7 times the third row from the second row
At this point, in order to find the solution to the system of linear equations, all we have to do is read it directly
from the matrix. If we convert the matrix back to a system of linear equations, we get:
x
=
4
and
y
=
–3
and
z
=
2
answer
Compare the solutions found by this method to normal Gaussian elimination with back-substitution. Although
we tend to use integers in our elementary row operations, there is no reason why we cannot use fractions if we
need to. Fractions tend to be rather messy, so we usually try to find a way to avoid them by judiciously
choosing elementary row operations.
328 – Gauss-Jordan Elimination
Matrices and Determinants
Advanced
To define the inverse of a square matrix.
invertible – a matrix with an inverse.
nonsingular – a matrix with an inverse. This is synonymous with invertible.
singular – a matrix without an inverse.
Edmond Halley (1656-1742), after whom Halley's Comet is named, was a noted English astronomer and
geophysicist. He theorized that comets travel in elliptical paths and are therefore subject to Kepler's Three Laws
of Planetary Motion. He was absolutely correct in this and proved it by predicting that the comet of 1682 would
once again be visible in 1758. It was this comet, which did indeed return close to Halley's prediction, that is now
known as Halley's Comet.
Matrices → Inverse of a Square Matrix
Before we talk about the inverse of a matrix, let's first review the inverse of a real number. Consider the
equation ax = b. In order to solve this equation for x, we divide both sides by a, provided that a ≠ 0:
ax
=
x
=
b
given equation
b
divide both sides by a, if a ≠ 0
a
However, we can rewrite the division by a as multiplying b by the quantity a–1. In other words, we actually
multiply both sides by the quantity a–1:
ax
=
b
given equation
(a–1a)x
=
a–1b
multiply both sides by "a–1" (read "a inverse")
(1)x
=
a–1b
recall that a(a–1) = a/a = 1
x
=
a–1b
The number a–1 is called the multiplicative inverse of a because a–1a = 1. The definition of the multiplicative
inverse of a matrix is similar:
Definition of the Inverse of a Square Matrix
Let A be an n x n matrix. If there exists matrix A–1 such that
AA–1
=
In
=
A–1A
then A–1 is called the inverse of A.
Note:
Recall that it is not always true that AB = BA, even if both products are defined. However, if A and B
are both square matrices and AB = In, then it can be shown that BA = In and hence B would be the
inverse of A.
Matrices and Determinants
Inverse of a Square Matrix – 329
Show that B is the inverse of A.
A
⎛2 1⎞
⎜
⎟
⎝5 3⎠
=
and
B
=
⎛ 3 −1 ⎞
⎜
⎟
⎝ −5 2 ⎠
solution
In order to demonstrate that these two matrices are inverses of each other, we multiply them together. First we
take the product AB. Then we take the product BA. If we obtain I2 both times, then we know that B is indeed
the inverse of A.
AB
=
⎛ 2 1 ⎞ ⎛ 3 −1 ⎞
⎜
⎟ ⋅⎜
⎟
⎝ 5 3 ⎠ ⎝ −5 2 ⎠
multiply matrices
=
⎡ 2( 3) + 1( −5) 2( −1) + 1( 2) ⎤
⎢
⎥
⎣ 5( 3) + 3( −5) 5( −1) + 3( 2) ⎦
multiply rows of A by columns of B
=
⎛ 6 − 5 −2 +
⎜
⎝ 15 − 15 −5 +
simplify
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
2⎞
⎟
6⎠
simplify
Now we find the product BA:
BA
=
⎛ 3 −1 ⎞ ⎛ 2 1 ⎞
⎜
⎟ ⋅⎜
⎟
⎝ −5 2 ⎠ ⎝ 5 3 ⎠
multiply matrices
=
⎡ 3( 2) + ( −1) ( 5) 3( 1) + ( −1) ( 3) ⎤
⎢
⎥
⎣ −5( 2) + 2( 5) −5( 1) + 2( 3) ⎦
multiply rows of B by columns of A
=
⎛ 6−5 3−3 ⎞
⎜
⎟
⎝ −10 + 10 −5 + 6 ⎠
simplify
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
simplify
Since AB = In = BA, we can conclude that B is indeed the matrix inverse of A.
answer
If a matrix A does have an inverse, then it is called invertible (or nonsingular). If it does not have an inverse,
then it is called singular. A nonsquare matrix cannot have an inverse. To see why, consider the situation
where matrix A is of order m x n and where matrix B is of order n x m, where m ≠ n. The product AB will then
be of order m x m and the product BA is of order n x n. Since they are of different orders, there is no way they
can produce the same product Im (or n).
Most of the time we are not given the inverse of a matrix, so we have to apply some sort of technique to find the
inverse. One way to do this is by a matrix equation.
330 – Inverse of a Square Matrix
Matrices and Determinants
Find the inverse of
A
=
⎛1 2⎞
⎜
⎟
⎝3 7⎠
solution
In order to find the inverse of A, we will attempt to solve the matrix equation AX = I. If we can find X such that
AX = I, we know that X must be the inverse of A. We can check our result using matrix multiplication.
First, we need to set up our matrix equation, using the rectangular array form of the matrices:
AX
=
I
original matrix equation
⎛ 1 2 ⎞ ⎛ x11 x12 ⎞
⎜
⎟ ⋅⎜
⎟
⎝ 3 7 ⎠ ⎝ x21 x22 ⎠
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
rectangular array form of matrices
⎛ 1 x11 + 2 x21 1 x12 + 2 x22 ⎞
⎜
⎟
⎝ 3 x11 + 7 x21 3 x12 + 7 x22 ⎠
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
matrix multiplication
Now we can equate corresponding elements in the matrices. This gives us the following system of linear
equations:
x11 + 2x21
=
1
and
x12 + 2x22
=
0
3x11 + 7x21
=
0
and
3x12 + 7x22
=
1
corresponding system of linear equations
Since we have four equations with four unknowns, we can easily apply standard back-substitution techniques to
solve this system. From the first system, we have:
x11
=
7
x21
=
–3
and
From the second system, we find through Gaussian elimination that:
x12
=
–2
x22
=
1
and
Therefore, our inverse matrix X is:
X
=
⎛ 7 −2 ⎞
⎜
⎟
⎝ −3 1 ⎠
answer
If we want to check our answer, then we find the products AX and XA. If they both equal the product matrix I2,
then we can conclude that X = A–1.
Matrices and Determinants
Inverse of a Square Matrix – 331
AX
XA
=
⎛ 1 2 ⎞ ⋅ ⎛ 7 −2 ⎞
⎜
⎟⎜
⎟
⎝ 3 7 ⎠ ⎝ −3 1 ⎠
=
⎡ 1( 7) + 2( −3) 1( −2) + 2( 1) ⎤
⎢
⎥
⎣ 3( 7) + 7( −3) 3( −2) + 7( 1) ⎦
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
=
⎛ 7 −2 ⎞ ⎛ 1 2 ⎞
⎜
⎟ ⋅⎜
⎟
⎝ −3 1 ⎠ ⎝ 3 7 ⎠
=
⎡ 7( 1) + ( −2) ( 3) 7( 2) + ( −2) ( 7) ⎤
⎢
⎥
⎣ −3( 1) + 1( 3) −3( 2) + 1( 7) ⎦
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
check
check
Any system of linear equations can have one solution, no solution, or infinitely many solutions. If the
coefficient matrix A of a square system (i.e. a system that has the same number of equations as unknowns) is
invertible, then the system has a unique solution, defined as follows:
A System of Linear Equations with a Unique Solution
If A is an invertible matrix, the system of linear equations represented
by the matrix equation AX = B has a unique solution given by:
X
=
A–1B
332 – Inverse of a Square Matrix
solution of matrix equation AX = B
Matrices and Determinants
Advanced
To find the inverse of a square matrix.
adjoining – creating a larger matrix out of two smaller matrices.
Edmond Halley (1656-1742) did more than watch the skies for comets. He calculated the focal length of certain
optical lenses, the trajectory of artillery shells, and prepared actuarial tables for use by insurance companies to
determine life and annuity premiums and values. This actuarial work is an early form of social statistics. Halley
also published papers on trade winds, monsoons, tides, and terrestrial magnetism. These papers lay the
groundwork for scientific geophysics. Finally, he discovered that the stars are in motion relative to each other.
Halley only studied the motion of three stars - Arcturus, Procyon, and Sirius - but theorized that smaller, fainter
stars also move. This could not be proved until 150 years later, when more advanced instruments became
available.
Matrices → Inverse of a Square Matrix → Finding Matrix Inverse
In the previous thought, we used the matrix equation AX = I to determine the inverse of A. Our result was two
systems of linear equations. We can represent those systems using augmented matrices as follows:
⎛1 2
⎜
⎝3 7
1⎞
⎟
0⎠
and
⎛1 2
⎜
⎝3 7
0⎞
⎟
1⎠
augmented matrices representing
systems of linear equations
Note that these matrices both have the same coefficient matrix. Rather than solving these systems
independently, we can solve them simultaneously by adjoining the identity matrix to the coefficient matrix, as
shown below:
⎛1 2
⎜
⎝3 7
1 0⎞
⎟
0 1⎠
adjoin identity matrix to coefficient matrix
Now, in order to solve both systems simultaneously, we apply Gauss-Jordan elimination on the adjoined
matrices. The objective is to transform the coefficient matrix (the left-half) into an identity matrix. Then the
right half of the adjoined matrix will be the solutions to the systems of linear equations.
3R1 – R2
–R2
R1 – 2R2
⎛1 2
⎜
⎝3 7
1 0⎞
⎛1 2
⎜
⎝ 0 −1
1 0⎞
⎛1 2
⎜
⎝0 1
1 0⎞
⎛1 0
⎜
⎝0 1
⎟
given matrix
0 1⎠
⎟
3 1⎠
⎟
−3 1 ⎠
7 −2 ⎞
−3 1
⎟
⎠
subtract second row from 3 times row 1
multiply second row by –1
subtract 2 times second row from first row
Thus, we go from the "doubly augmented" matrix [ A : I ] to the matrix [ I : A–1]. In other words:
Matrices and Determinants
Finding Matrix Inverse – 333
A–1
=
⎛ 7 −2 ⎞
⎜
⎟
⎝ −3 1 ⎠
inverse of original matrix A
We already demonstrated in the previous thought that this is indeed the inverse of the matrix A. Based upon
this example, we can develop a procedure (or algorithm) to find the inverse of a matrix (if it has one).
Finding an Inverse Matrix
Let A be a square matrix of order n.
1.)
Write the n x 2n matrix that consists of the given matrix A on the left and the n x n identity
matrix I on the right to obtain [ A : I ]. Note that we separate the matrices A and I by a
dotted line. We call this process adjoining the matrices A and I.
2.)
If possible, row reduce A to I using elementary row operations on the entire matrix [ A : I ].
The result will be the matrix [ I : A–1 ]. If this is not possible, then A is not invertible.
3.)
Check your work by multiplying to see that AA–1 = I = A–1A.
Let's see another example of how this works.
Find the inverse of the matrix, if it exists.
⎛1 1 1⎞
⎜3 5 4⎟
⎜
⎟
⎝3 6 5⎠
solution
First, we setup a 3 x 3 matrix that consists of our given matrix above adjoined to a 3 x 3 identity matrix, as
shown below:
⎛1 1 1
⎜3 5 4
⎜
⎝3 6 5
1 0 0⎞
0 1 0⎟
⎟
0 0 1⎠
adjoin given matrix with 3 x 3 identity matrix
Next, we apply elementary row operations to transform the left hand side of the matrix into an identity matrix.
Then the right hand side of the matrix will be miraculously transformed into the inverse matrix. Start be
eliminating the "3's" in the second and third rows:
R2 – 3R1→
R3 – 3R1→
2R2 – R3→
2R3 – 3R2→
R1 – R2 – R3→
334 – Finding Matrix Inverse
⎛1 1 1
⎜0 2 1
⎜
⎝0 3 2
1 0 0⎞
−3 1 0 ⎟
⎟
−3 0 1 ⎠
⎛1 1 1
⎜0 1 0
⎜
⎝0 0 1
1
0
⎞
−3 2 −1 ⎟
⎟
3 −3 2 ⎠
⎛1 0 0
⎜0 1 0
⎜
⎝0 0 1
1
1 −1 ⎞
0
−3 2 −1 ⎟
⎟
3 −3 2 ⎠
subtract 3 times first row from second row
subtract 3 times first row from third row
subtract third row from 2 times second row
subtract 3 times second row from 2 times third row
subtract second and third rows from first row
Matrices and Determinants
Now that we have transformed the left hand side of the equation to an identity matrix, we can see that our
inverse matrix is:
−1
⎛1 1 1⎞
⎜3 5 4⎟
⎜
⎟
⎝3 6 5⎠
=
⎛ 1 1 −1 ⎞
⎜ −3 2 −1 ⎟
⎜
⎟
⎝ 3 −3 2 ⎠
answer
Naturally, we want to check to make sure that this is indeed the correct solution, so if we label our original
matrix as A and our inverse matrix as A–1, we multiply them together and see if we obtain I3:
–1
AA
=
⎛ 1 1 1 ⎞ ⎛ 1 1 −1 ⎞
⎜ 3 5 4 ⎟ ⋅ ⎜ −3 2 −1 ⎟
⎜
⎟⎜
⎟
⎝ 3 6 5 ⎠ ⎝ 3 −3 2 ⎠
multiply matrices
=
⎡ [ 1⋅ ( 1) + 1⋅ ( −3) + 1⋅ ( 3) ] [ 1⋅ ( 1) + 1⋅ ( 2) + 1⋅ ( −3) ] [ 1⋅ ( −1) + 1⋅ ( −1) + 1⋅ ( 2) ] ⎤
⎢ [ 3⋅ ( 1) + 5⋅ ( −3) + 4⋅ ( 3) ] [ 3⋅ ( 1) + 5⋅ ( 2) + 4⋅ ( −3) ] [ 3⋅ ( −1) + 5⋅ ( −1) + 4⋅ ( 2) ] ⎥
⎢
⎥
⎣ [ 3⋅ ( 1) + 6⋅ ( −3) + 5⋅ ( 3) ] [ 3⋅ ( 1) + 6⋅ ( 2) + 5⋅ ( −3) ] [ 3⋅ ( −1) + 6⋅ ( −1) + 5⋅ ( 2) ] ⎦
multiply rows of A by columns of A–1
=
⎛ 1 − 3 + 3 1 + 2 − 3 −1 − 1 + 2 ⎞
⎜ 3 − 15 + 12 3 + 10 − 12 −3 − 5 + 8 ⎟
⎜
⎟
⎝ 3 − 18 + 15 3 + 12 − 15 −3 − 6 + 10 ⎠
simplify
=
⎛1 0 0⎞
⎜0 1 0⎟
⎜
⎟
⎝0 0 1⎠
simplify
Matrices and Determinants
check
Finding Matrix Inverse – 335
336 – Finding Matrix Inverse
Matrices and Determinants
Advanced
To use a formula to find the determinant of a 2 x 2 matrix.
determinant – associated with each square matrix is a real number called the determinant of the matrix. The
number of elements in any row or column of the square matrix is called the order of the determinant.
David Hilbert (1862-1943) was a German number theorist and the most famous mathematician of his day. In
1900, he posed a series of 23 questions to the mathematical community at a conference in Paris, France. His
seventh question dealt with the irrationality of certain numbers, among which are the transcendental numbers e
and π. Hilbert was not the first to prove these two numbers are transcendental, but his genius was such that he
simplified the proofs considerably.
Matrices → Inverse of a Square Matrix → Inverse of 2 x 2 Matrix
In the previous thought, we looked at the definition of the inverse of a matrix. In a related thought, we learned
how to find the inverse of a matrix of any order n x n. Now we are going to show a shortcut method for finding
the inverse of a 2 x 2 matrix. Here we can apply a quick and convenient formula that allows us to determine not
only the inverse of a matrix, but even if the matrix has an inverse.
Basically, the formula works as follows.
The Inverse of a 2 x 2 Matrix
If A is a 2 x 2 matrix given by:
A
=
2 x 2 matrix
then A is invertible (has an inverse) if and only if ad – bc ≠ 0.
Moreover, if ad – bc ≠ 0, then the inverse is given by:
A–1
=
inverse of matrix A
This only works with 2 x 2 matrices!
Note:
The denominator ad – bc is called the determinant of the 2 x 2 matrix A. You will study determinants
and their applications in another section of the Algebra Brain.
Let's try out this formula with a couple of examples.
Find the inverse of the matrix, if it exists.
A
=
⎛2 4⎞
⎜
⎟
⎝4 8⎠
Matrices and Determinants
Inverse of 2 x 2 Matrix – 337
solution
First, we check the determinant ad – bc. If it doesn't equal zero, then we can go ahead and compute the inverse
for this matrix:
ad – bc
=
(2)(8) – (4)(4)
find determinant
=
16 – 16
simplify
=
0
simplify
Since ad – bc = 0, we can conclude that this matrix does not have an inverse.
answer
Find the inverse of the matrix, if it exists.
B
⎛ 3 7⎞
⎜
⎟
⎝ −1 2 ⎠
=
solution
Again, we first find the determinant ad – bc. If it equals zero, then we do not need to continue any further with
the problem.
ad – bc
=
3(2) – 7(–1)
find determinant
=
6+7
simplify
=
13
simplify
Since ad – bc ≠ 0, we can conclude that this matrix does have an inverse. Moreover, its inverse is given by:
B–1
=
⎛ d −b ⎞
⎟
ad − bc ⎝ −c a ⎠
inverse of 2 x 2 matrix formula
=
⎛ 2 −7 ⎞
⎟
13 ⎝ 1 3 ⎠
substitute in ad – bc = 13
=
⎛
⎜
⎜
⎜
⎝
1
1
⋅⎜
⋅⎜
−7
⎞
13 13 ⎟
⎟
1 3
⎟
13 13 ⎠
2
scalar multiplication
answer
Let's just double check to make sure that we have found the inverse. If we multiply B by its inverse B–1, we
should end up with the identity matrix I2.
338 – Inverse of 2 x 2 Matrix
Matrices and Determinants
BB–1
−7
=
⎛
⎛ 3 7 ⎞ ⋅⎜
⎜
⎟⎜
⎝ −1 2 ⎠ ⎜
⎝
⎞
⎟
⎟
1 3
⎟
13 13 ⎠
=
⎡ 3⎛ 2 ⎞ + 7⎛ 1 ⎞ 3⎛ −7 ⎞ + 7⎛ 3 ⎞ ⎤
⎢ ⎜⎝ 13 ⎟⎠ ⎜⎝ 13 ⎟⎠ ⎜⎝ 13 ⎟⎠ ⎜⎝ 13 ⎟⎠ ⎥
⎢
⎥
⎢ −1⎛⎜ 2 ⎞⎟ + 2⎛⎜ 1 ⎞⎟ −1⎛⎜ −7 ⎞⎟ + 2⎛⎜ 3 ⎞⎟ ⎥
⎣ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎝ 13 ⎠ ⎦
multiply rows of B by columns of B–1
=
⎛ 13 0 ⎞
⎜ 13
⎟
⎜
⎟
⎜ 0 13 ⎟
13 ⎠
⎝
simplify
=
⎛1 0⎞
⎜
⎟
⎝0 1⎠
simplify
check
2
13 13
matrix multiplication
A similar process can be used to show that B–1B also equals I2. Therefore, B–1 is indeed the inverse of B.
Matrices and Determinants
Inverse of 2 x 2 Matrix – 339
340 – Inverse of 2 x 2 Matrix
Matrices and Determinants
Advanced
To find the determinant of a square matrix.
determinant – associated with each square matrix is a real number called the determinant of the matrix. The
number of elements in any row or column of the square matrix is called the order of the determinant.
expanding by cofactors – calculating the determinant of a matrix by taking the sum of the entries in any row
and multiplying by the cofactors associated with that row.
Seki Gowa (1642-1708) is often seen as the Isaac Newton of Japan. He brought Japanese problem-solving to
the modern age, conquering many of the same problems Newton was working on in England. Seki Gowa and
Newton pursued parallel lives. Whereas Newton was in charge of the Mint in England, Seki Gowa held a similar
position in Japan. Newton was knighted while Seki Gowa became a samurai. Seki Gowa is credited with the
invention of determinants at the same time that they were being introduced to Europe. His students referred to
him as the "Arithmetical Sage", which is inscribed on his tomb in the Buddhist cemetery at Ushigome in Tokyo,
Japan.
Matrices → Determinants
Every square matrix (i.e. one that has the same number of rows and columns) is associated with a real number
called a determinant. Determinants have many uses and we will explore these under the thought labeled
Applications of Determinants. Where are determinants and where do they come from? Consider the following
system of linear equations:
a1x + b1y
=
c1
a2x + b2y
=
c2
system of linear equations
With a little bit of algebraic mumbo-jumbo, we can find the solution to this system of equations. It turns out
that when we solve this system for x and y, we obtain the following solution set:
x
=
c1 b 2 − c2 b 1
a1 b2 − a2 b1
and
y
a 1 c2 − a 2 c1
=
a1 b2 − a2 b1
provided that a1b2 – a2b1 ≠ 0, of course. Note that we have the same denominator in both solutions. This
denominator is called the determinant of the coefficient matrix of the linear system.
Coefficient Matrix
⎛ a1 a2 ⎞
A = ⎜
⎟
⎝ b1 b2 ⎠
Determinant
det (A)
=
a1b2 – a2b1
We can also denote the determinant of a matrix with vertical bars on either side of the matrix, as indicated in the
definition below. This should not be confused with finding the absolute value of the matrix!
Matrices and Determinants
Determinants – 341
Definition of the Determinant of a 2 x 2 Matrix
The determinant of the matrix
A
=
is given by
det (A)
=
|A|
=
=
a1b2 – a2b1
determinant of 2 x 2 matrix
A quick and easy method for remembering the formula for finding the determinant of a 2 x 2 matrix is to simply
take the difference of the products of the diagonal terms, as indicated in the diagram below:
This works very well for 2 x 2 matrices, but does not work at all for higher order matrices. In order to find the
determinant for higher order matrices, we have to make use of minors and cofactors.
Minors and Cofactors
In the previous section we looked at a convenient method for finding the determinant of 2 x 2 matrices. Now
we need a more general method for finding determinants that applies to all matrices. This process is not nearly
so simple or trivial. The basic procedure is to break a 3 x 3 or larger matrix into bite-size chunks that we can
work with. We will still need to refer to the formula for finding the determinant of a 2 x 2 matrix. Essentially,
we find several of these smaller "mini-determinants". The total determinant of the matrix is then the sum of
these mini-determinants.
For higher order matrices, it is convenient to use cofactors and minors to define the determinant.
Minors and Cofactors of a Square Matrix
If A is a square matrix, the minor Mij of the entry aij is the
determinant of the square matrix obtained by deleting the
ith row and jth column of A. The cofactor Cij of the entry aij
is given by
Cij
=
(–1)i + 1Mij
This is a somewhat confusing definition. Let's look at an example on how to find the minors and cofactors of a
3 x 3 matrix.
342 – Determinants
Matrices and Determinants
Find all the minors and the cofactors of the given matrix.
A
⎛ 1 2 3⎞
⎜ −4 5 8 ⎟
⎜
⎟
⎝ −1 −2 1 ⎠
=
solution
To find the minor M11, delete the first row and first column of A and find the determinant of the remaining
matrix:
M11
=
(5)(1) – (–2)(8)
=
21
Similarly, to find M12, we delete the first row and second column of A and find the determinant of the remaining
matrix:
M12
=
(–4)(1) – (–1)(8)
=
4
We perform the process 7 more times. The final minor M33 is found by deleting the third row and third column
of A and finding the determinant of the remaining matrix:
M33
=
(1)(5) – (–4)(2)
=
13
Through this process of eliminating rows and columns and finding determinates, we will find the following
minors of matrix A:
M11
=
–11
M12
=
4
M13
=
13
M21
=
8
M22
=
4
M23
=
0
M31
=
1
M32
=
20
M33
=
13
Now in order to find the cofactors, we multiply each of the minors above by (–1)i+j where i is the row and j is
the column of the matrix A:
C11
=
(–1)2(–11)
=
21
C12
=
(–1)3(4)
=
–4
C13
=
(–1)4(13)
=
13
C21
=
(–1)3(8)
=
–8
C22
=
(–1)4(4)
=
4
C23
=
(–1)5(0)
=
0
C31
=
(–1)4(1)
=
1
C32
=
(–1)5(20)
=
–20
C33
=
(–1)6(13)
=
13
Note:
In the sign pattern above, even positions (where i + j is even) keep the same sign as the minor, but odd
positions (where i + j is odd) change the sign of the minor.
Matrices and Determinants
Determinants – 343
Determinant of a Square Matrix
Once we have defined minors and cofactors, we can now define the determinant of an n x n square matrix.
Determinant of a Square Matrix
If A is a square matrix (of order 2 x 2 or greater), the determinant of A
is the sum of the entries in any row (or column) of A multiplied by their
respective cofactors. For instance, expanding along the first row yields
|A|
=
a11C11 + a12C12 + a13C13 + . . . + a1nC1n
Applying this definition to find a determinant is called expanding by cofactors.
Find the determinant of the given matrix.
A
⎛ 1 2 3⎞
⎜ −4 5 8 ⎟
⎜
⎟
⎝ −1 −2 1 ⎠
=
solution
First note that this is the exact same matrix that we used in the previous example. The cofactors of the elements
in the first row of the matrix were:
C11
=
–11
and
C12
=
–4
and
C13
=
13
By using the definition above to find the determinant, we expand the first row as follows:
|A|
=
a11C11 + a12C12 + a13C13
first-row expansion
=
(1)(21) + (2)(–4) + (3)(13)
substitute in numbers
=
21 – 8 + 39
simplify
=
52
simplify
answer
Although we choose to do a first row expansion, we could just as easily have down a second row expansion or
even a third column expansion. If we did that, then we choose the relevant cofactors, as indicated below:
|A|
=
a21C21 + a22C22 + a23C23
second-row expansion
=
(–8)(–4) + (5)(4) + (8)(0)
substitute in numbers
=
52
simplify
=
a13C13 + a23C23 + a33C33
third-column expansion
=
(3)(13) + (8)(0) + (1)(13)
substitute in numbers
=
52
simplify
or
|A|
As long as we multiply our cofactors by the right row or column in the matrix, we will obtain the same number
each time. This leads to a useful strategy when finding determinants. Look for the row or column with the
344 – Determinants
Matrices and Determinants
most number of zeros. Then we only have to find a limited number of cofactors. We do not need to find
cofactors of zero entries because zero times its cofactor is still zero.
Find the determinant of the following matrix.
B
⎛ 2 −1 3 ⎞
⎜ 4 0 −5 ⎟
⎜
⎟
⎝1 0 2 ⎠
=
solution
Since we have two zeros in the second column, that would be the column we want to use cofactor expansion
with. If we use the definition for finding determinants, expanding around column 2, we have:
|B|
=
b12C12 + b22C22 + b32C32
second-column expansion
=
(–1)C12 + (0)C22 + (0)C32
substitute in numbers from B
=
–C12
simplify
Now all we need to do is determine C12. To do that, we delete the first row, second column of B and evaluate
the determinant of the remaining matrix:
C12
3
⎛ 4 −5 ⎞
⎜
⎟
⎝1 2 ⎠
=
( −1) ⋅
=
(–1)(8 – (–5))
evaluate determinant of 2 x 2 matrix
=
–13
simplify
formula to find cofactor
Our determinant for our given 3 x 3 matrix is therefore:
|B|
=
–C12
=
–(–13)
substitute in for C12
=
13
answer
Matrices and Determinants
Determinants – 345
346 – Determinants
Matrices and Determinants
Advanced
To review some of the properties associated with the determinant of a matrix.
No definitions on this page.
Alan Turing (1912-1954) was highly instrumental in cracking German codes during World War II. The famous
Enigma machine was already well-understood by the English. The reason why Enigma worked so well was that
the number of possible letter combinations used to produce a code was astronomically huge. The trick to
successful decryption of codes is to run through the most combinations in the least amount of time. Turing did
this by building a special machine known as Colossus. It reduced decryption time from hours to minutes, which
gave the Allies a huge tactical advantage.
Matrices → Determinants → Properties of Determinants
Taking the determinant of a matrix is a mathematical operation. As we should already know, many
mathematical operations have properties associated with them that allow us to greatly simplify our calculations.
For example, by this time we are familiar with expanding a binomial square, such as (4 – 3x)2. We have a
theorem that tells us that when we expand this binomial expression we will get (16 – 6x + 9x2). However, in
order to derive this result, we need to use the Distributive Property of Multiplication.
In the same way, if we know about the properties of determinants outlined below, we can immediately start
simplifying the determinant without resorting to tedious calculations.
Property 1
If all the elements in a particular row or a particular column are zero, then the determinant of the
matrix is also zero.
⎛ 3 −2 4 ⎞
⎜0 0 0⎟
⎜
⎟
⎝1 5 6⎠
=
0
=
0
⎛ −2 4 ⎞ + 0⋅ ⎛ 3 4 ⎞ + 0⋅ ⎛ 3 −2 ⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎝ 5 6⎠
⎝1 6⎠
⎝1 5 ⎠
cofactor expansion along second
row
simplify by multiplying each
determinant
by zero
Property 2
If any two rows or columns of a matrix are interchanged, then the determinant of the resulting matrix is
the opposite in sign of the determinant of the original matrix.
Matrices and Determinants
Properties of Determinants – 347
⎛7 0⎞
⎛ −2 4 ⎞
⎛ −2 4 ⎞
⎜
⎟ − 8⋅ ⎜
⎟ +1 ⎜
⎟
⎝5 6⎠
⎝ 5 6⎠
⎝ 7 0⎠
cofactor expansion along first
column
=
3⋅
=
3(42) – 8(–32) + 1(–28)
use formula for determinants of 2 x
2 matrices
=
354
simplify
Now switch the first and second columns and see what we get:
⎛7 0⎞
⎛ −2 4 ⎞
⎛ −2 4 ⎞
⎜
⎟ + 8⋅ ⎜
⎟ − 1⋅ ⎜
⎟
⎝5 6⎠
⎝ 5 6⎠
⎝ 7 0⎠
cofactor expansion along second
column
=
−3⋅
=
–3(42) + 8(–32) – 1(–28)
use formula for determinants of
2 x 2 matrices
=
–354
simplify
Property 3
If two rows or columns of a matrix have corresponding elements that are equal, then the determinant of
that matrix is zero.
⎛ 4 3 7⎞
⎜ −1 2 1 ⎟
⎜
⎟
⎝ −1 2 1 ⎠
⎛ 2 1 ⎞ − 3 ⎛ −1 1 ⎞ + 7⋅ ⎛ −1 2 ⎞
⎜
⎟
⎜
⎟
⎜
⎟
⎝2 1⎠
⎝ −1 1 ⎠
⎝ −1 2 ⎠
=
4⋅
=
4(0) – 3(0) + 7(0)
use formula for determinants of
2 x 2 matrices
=
0
simplify
cofactor expansion along first row
Property 4
If each element in one row or column is multiplied by some real number constant k, then the
determinant of that matrix is also multiplied by k.
⎛ 5 −1 3 ⎞
⎜2 1 4⎟
⎜
⎟
⎝ −7 3 −4 ⎠
⎛1 4 ⎞
⎛2 4⎞
⎛ 2 1⎞
⎜
⎟ − ( −1) ⋅ ⎜
⎟ + 3⋅ ⎜
⎟
⎝ 3 −4 ⎠
⎝ −7 −4 ⎠
⎝ −7 3 ⎠
=
5⋅
=
–21
cofactor expansion along
first row
simplify
Now multiply every element in the first row by 3 and see what happens to the determinant:
348 – Properties of Determinants
Matrices and Determinants
=
3⋅ 5⋅
=
3⎢ 5⋅
⎡
⎣
⎛1 4 ⎞
⎛2 4⎞
⎛ 2 1⎞
⎜
⎟ − 3( −1) ⋅ ⎜
⎟ + 3⋅ 3⋅ ⎜
⎟
⎝ 3 −4 ⎠
⎝ −7 −4 ⎠
⎝ −7 3 ⎠
⎛ 1 4 ⎞ − ( −1) ⋅
⎜
⎟
⎝ 3 −4 ⎠
⎛ 2 4 ⎞ + 3⋅ ⎛ 2 1 ⎞ ⎤
⎜
⎟
⎜
⎟ ⎥
⎝ −7 −4 ⎠
⎝ −7 3 ⎠ ⎦
cofactor
expansion along
first row,
multiplying each
element in first
row by 3
factor out a 3
from each term
=
3(–21)
simplify
expression
inside
parentheses
from above
=
–63
this is 3 times
original
determinant
Property 5
If each and every element of one row or one column is multiplied by a real number constant k and if
the resulting products are then added to the corresponding elements of another row or column,
respectively, the resulting determinant equals the original determinant.
This property is considerably more complicated than the other four properties, so a little more explanation is in
order. In this example, we will first find the determinant normally. Then we will apply property 5 as outlined
above to our original matrix. We will multiply the second column by some real number k. We will add the
product of k and the first element of column 2 to the first element of column 1. We will add the product of k
and the second element of column 2 to the second element of column 1. We do the same procedure once more
for the product of k and the third element of column two and the third element of column 1. Again, we find the
determinant of the resultant matrix. If everything goes as planned, we should obtain the same determinant as
we did the first time.
First, find the determinant normally:
⎛ 2 −1 7 ⎞
⎜ 6 −2 5 ⎟
⎜
⎟
⎝ −6 2 1 ⎠
⎛ −2 5 ⎞
⎜
⎟ − ( −1)
⎝ 2 1⎠
⎛ 6 5⎞
⎛ 6 −2 ⎞
⎜
⎟ +7 ⎜
⎟
⎝ −6 1 ⎠
⎝ −6 2 ⎠
cofactor expansion along first
row
=
2
=
2(–24) – (–1)(36) + 7(0)
use formula for determinants
of 2 x 2 matrices
=
12
simplify
Now we are going to mess around with the first column. First, we multiply the second column by 3. Then we
add the result of this to the first column to produce a column equivalent matrix:
Matrices and Determinants
Properties of Determinants – 349
⎡ 2 + 3( −1) −1 7 ⎤
⎢ 6 + 3( −2) −2 5 ⎥
⎢
⎥
⎣ −6 + 3( 2) 2 1 ⎦
=
⎛ −1 −1 7 ⎞
⎜ 0 −2 5 ⎟
⎜
⎟
⎝ 0 2 1⎠
⎛ −2 5 ⎞
⎜
⎟ + 0+ 0
⎝ 2 1⎠
add 2 times second column to first column
=
−1
=
–1(–12)
use formula for determinants of 2 x 2 matrices
=
12
simplify
cofactor expansion along first column
From the previous example, we can see one reason why we like to have these properties around. In the final
example for Property 5, we were able to convert one column of the matrix so that it contained mostly zeros.
This made it much easier to perform the cofactor expansion since we then knew that we would end up with a lot
of zero terms, thus reducing the number of times we have to find the determinant of a 2 x 2 matrix. Anything
that reduces the workload is all right by us.
The properties are summarized below for convenient reference.
Properties of Determinants
1.)
If all the elements in a particular row or a particular column are zero, then the determinant of the
matrix is also zero.
2.)
If any two rows or columns of a matrix are interchanged, then the determinant of the resulting
matrix is the opposite in sign of the determinant of the original matrix.
3.)
If two rows or columns of a matrix have corresponding elements that are equal, then the
determinant of that matrix is zero.
4.)
If each element in one row or column is multiplied by some real number constant k, then the
determinant of that matrix is also multiplied by k.
5.)
If each and every element of one row or one column is multiplied by a real number constant k and
if the resulting products are then added to the corresponding elements of another row or column,
respectively, the resulting determinant equals the original determinant.
350 – Properties of Determinants
Matrices and Determinants
Advanced
To look at a special type of matrix.
upper triangular matrix – matrix containing nonzero elements on and above the main diagonal and zero
elements below the main diagonal.
lower triangular matrix – matrix containing nonzero elements on and below the main diagonal and zero
elements above the main diagonal.
diagonal matrix – a matrix that has nonzero elements on the main diagonal and zero elements everywhere
else.
Thales of Miletus (c.640-546 B.C.) was a famous astronomer in addition to his vocations as mathematician and
philosopher. One story reports that he once fell down a well while star-gazing. The slave girl who discovered
him commented that Thales was "so interested in the heavens that he could not see what was in front of his
feet." Thales's interest in astronomy may be a result of his travels to Babylonia, where he studied with the
Chaldean magi. He advised navigators to rely on the constellation Ursa Minor (Little Bear) for navigation rather
than Ursa Major (Great Bear). This was because Ursa Minor contains the star Polaris, which is always located
due north and it is therefore easy for navigators to find their position relative to this star.
Matrices → Determinants → Triangular Matrix
There are three types of triangular matrices. An upper triangular matrix contains nonzero elements above the
main diagonal and zero elements below the diagonal. A lower triangular matrix is just the opposite,
containing nonzero elements below the main diagonal and zero elements above the main diagonal. Finally, a
matrix that has zero elements above and below the main diagonal is said to simply be a diagonal matrix .
Upper Triangular
a 1n ⎞
⎛ a11 a12 a13
⎜
⎜ 0 a22 a23
⎜ 0 0 a
33
⎜
⎜
⎜ 0 0 0
⎝
⎟
a 2n ⎟
a 3n ⎟
⎟
⎟
a mn ⎟⎠
Lower Triangular
Diagonal
0 0
⎛⎜ a11
⎜ 0 a22 0
⎜ 0 0 a
33
⎜
⎜
⎜ 0 0 0
⎝
⎞⎟
0 ⎟
0 ⎟
⎟
⎟
a mn ⎟⎠
0
One of the neat little advantages of such a triangular matrix (whether upper, lower, or diagonal) is that the
determinant of such a matrix is simply the product of the entries along the main diagonal.
Find the determinant of the following matrix.
A
=
⎛1
⎜ −2
⎜
⎜ 14
⎜ −3
⎜
⎝2
⎞
⎟
⎟
1 8 0 0 ⎟
4 7 6 0 ⎟
⎟
2 1 6 −5 ⎠
0 0 0 0
3 0 0 0
Matrices and Determinants
Triangular Matrix – 351
solution
Normally, we would cringe in terror at the thought of finding the determinant of a 5 x 5 matrix. However, we
note that it is a lower triangular matrix in that all the entries above the main diagonal are zero. Thus, we know
that the determinant is simply the product of the entries along the main diagonal:
|A|
=
⎛1
⎜ −2
⎜
⎜ 14
⎜ −3
⎜
⎝2
⎞
⎟
⎟
1 8 0 0 ⎟
4 7 6 0 ⎟
⎟
2 1 6 −5 ⎠
0 0 0 0
3 0 0 0
determinant of matrix A
=
(1)(3)(8)(6)(–5)
product of main diagonal entries
=
–720
simplify
answer
We can check our answer by inputting this matrix into a computer program and asking it to find the determinant
for us. Sure enough, it will return the answer we got above, –720.
352 – Triangular Matrix
Matrices and Determinants
Now that we know what determinants are, we want to find applications for them in the real world. In this
section of the Algebra Brain, we will examine four common uses of determinants.
Cramer's Rule
Cramer's Rule allows us to quickly use determinants to find the solutions to systems of linear equations. At this
point, we now have four ways to solve systems of linear equations: back-substitution, elimination with
equations, elimination with matrices (a la Gaussian and Gauss-Jordan elimination), and finally Cramer's Rule.
Use whichever method suits your fancy at the time. Cramer's Rule is ideal for use with computers because they
can find determinants of matrices many times faster than a human. As long as we input the data correctly into a
formula, then a computer program like MathCAD 2000® will do the number crunching and spit out the solution
of the linear system.
Area of a Triangle
We can also use matrices and determinants to find the area of a triangle whose coordinates are given as points in
a coordinate plane. Basically, we use a formula that takes the coordinate points as inputs into a matrix and spits
out the area as the output of the formula.
Test for Collinear Points
If we are given three points in a coordinate plane, we can determine 1 of two things. Either they are not all on a
line, in which case we can form a triangle with the three points whose area can be calculated using the formula
above, or they are all on one line. If the three points are on one line, then by definition the area enclosed by the
points must be zero. Therefore, if the determinant of the matrix used to calculate the area of a triangle equals
zero, then all three points must lie on the same line in the plane.
Two-Point Form of the Equation of a Line
Given only two points, we can use the same 3 x 3 determinant from the previous two applications to find the
equation of a line. The only difference between this application and the previous application is that instead of
using six numbers in the matrix, we use four numbers (the coordinates of our two points) and the variables x
and y as elements in the determinant of the matrix.
Matrices and Determinants
Applications of Determinants – 353
354 – Applications of Determinants
Matrices and Determinants
Advanced
To find the area of a triangle using a 3 x 3 matrix.
No definitions on this page.
John Wallis (1616-1703) was an English algebraist, who, like many of his countrymen, led a varied and
interesting life. He was a human calculating prodigy who could perform amazing mental feats, such as finding
the square root of a 53-digit number to 17 decimal places without notation. During the English Civil War he
deciphered Royalist documents for Lord Oliver Cromwell. As a reward for his services Lord Cromwell appointed
Wallis the Savilian Professor of Geometry at Oxford.
Matrices → Determinants → Applications of Determinants → Area of Triangle
One useful application of determinants is to find the area of a triangle given its three vertices. The usual
geometric formula only finds the area of a triangle if we know the base and the height (A = 1/2Bh), which is fine
for many triangles, but is more difficult to use if we have a weirdly arranged triangle, like the one below.
If we want to use the geometric formula, then we need to do a lot of calculation of distances and such. Very
tedious and not much fun. However, there is a better method. Since we know what the three vertices of the
triangle are, we can use the determinant of a 3 x 3 matrix.
We set up the matrix so that the first two columns contain the coordinates of the three vertices. The third
column contains all "1's". Then we find the area of this triangle by taking the determinant of this matrix, and
multiplying it by 1/2.
Matrices and Determinants
Area of a Triangle – 355
Area of a Triangle
The area of a triangle with vertices (x1, y1), (x2, y2) and (x3, y3) is given by
Area
=
where the symbol ± indicates that the appropriate sign should be chosen to yield a positive area.
Find the area of the triangle whose vertices are (0, 0), (2, 5), and (4, 1) as shown below.
solution
First, let's find the determinant of the matrix formed from the three vertices given above. We let
(x1, y1) = (0, 0), (x2, y2) = (2, 5), and (x3, y3) = (4, 1)
⎛⎜ x1 y1 1 ⎞⎟
⎜ x2 y2 1 ⎟
⎜ x3 y3 1 ⎟
⎝
⎠
determinant of
matrix formed from
vertices of triangle
=
=
=
⎛ 5 1 ⎞ + 0 ( −1) 3⋅
0 ( −1) ⋅ ⎜
⎟
⎝1 1⎠
2
⎛2 5⎞
⎜
⎟
⎝4 1⎠
⎛ 2 1 ⎞ + 1 ( −1) 4⋅
⎜
⎟
⎝4 1⎠
⎛2 5⎞
⎜
⎟
⎝4 1⎠
cofactor expansion
along first row
(because it has
the most zeros)
first two minors are
multiplied by zero,
so all we have
left is the third
minor
=
(2)(1) – (5)(4)
determinant of 2 x 2
matrix
=
–18
simplify
We plug this number into our formula for area. Note that we have to choose our sign so that our answer ends up
being positive. In this case, we need to choose the negative sign.
356 – Area of a Triangle
Matrices and Determinants
Area
=
formula for area
=
use result from above
=
–(–9)
choose negative sign to get a positive solution
=
9
answer
And that is all there is to finding the area of a triangle using determinants and three points.
Matrices and Determinants
Area of a Triangle – 357
358 – Area of a Triangle
Matrices and Determinants
Advanced
To solve systems of linear equations using determinants and Cramer's Rule.
No definitions on this page.
Gabriel Cramer (1704-1752), a Swiss geometer, is not as well known as other mathematicians, but he still made
important contributions to the areas of analysis, determinants, and geometry. In addition to Cramer's Rule, he
also came up with Cramer's paradox, involving an apparent contradiction in a theorem put forth by Colin
MacLaurin.
Matrices → Determinants → Applications of Determinants → Cramer's Rule
So far in algebra, we have become familiar with three basic methods of solving systems of linear equations.
The first involves solving one or more of the equations independently, then back-substituting into the rest of the
equations until we have found all of the solutions (assuming that the system has a solution). The second method
involves some sort of elimination of a variable combined with back-substitution to find the solution. The third
method uses matrices to perform the elimination process more efficiently. We can use either Gaussian
elimination or Gauss-Jordan elimination. If we use Gaussian elimination, then we have to rely on backsubstitution to finish the process. If we use Gauss-Jordan elimination, we keep working on the matrix until it is
in reduced row-echelon form, revealing the solution to the system.
All of these methods work great under certain circumstances. However, if we have access to a computer
program, there is another method that we can use to almost instantaneously find the solution to a system of
linear equation. This method is called Cramer's Rule, named after Gabriel Cramer (1704–1752). This rule uses
determinants to write the solution to a system of linear equations.
Cramer's Rule: 2 x 2 Matrices
In the previous thought about Determinants, we told you that the solution to the following linear system:
a1x + b1y
=
c1
a2x + b2y
=
c2
and
is given by:
x
=
c1 b 2 − c2 b 1
a1 b2 − a2 b1
and
y
=
a 1 c2 − a 2 c1
a1 b2 − a2 b1
as long as a1b2 – a2b1 ≠ 0. Notice that the numerators and denominators correspond to determinants of 2 x 2
matrices, as follows:
Matrices and Determinants
Cramer's Rule – 359
x
c1 b 2 − c2 b 1
=
y
a1 b2 − a2 b1
⎛ c1
⎜
⎝ c2
⎛ a1
⎜
⎝ a2
=
=
a 1 c2 − a 2 c1
a1 b2 − a2 b1
⎛ a1
⎜
⎝ a2
⎛ a1
⎜
⎝ a2
b1 ⎞
⎟
b2 ⎠
b1 ⎞
⎟
b2 ⎠
=
c1 ⎞
⎟
c2 ⎠
b1 ⎞
⎟
b2 ⎠
For both the variables x and y, we find the solution by dividing a determinant by another determinant based on
the coefficient matrix. The numerator matrix is determined by the constant terms and the coefficients of the
other variable. Above, the coefficients b1 and b2 correspond to the y-terms in the linear equation, so we use
those numbers to help us find x. Similarly, since a1 and a2 are coefficients of the x-term, we use them to help us
find y.
If we denote the determinant of the coefficient matrix by D and the other two determinants in the numerators by
Dx and Dy respectively, then we have the following:
D
=
⎛ a1 b1 ⎞
⎜
⎟
⎝ a2 b2 ⎠
and
Dx
=
⎛ c1 b 1 ⎞
⎜
⎟
⎝ c2 b 2 ⎠
and
Dy
=
⎛ a 1 c1 ⎞
⎜
⎟
⎝ a 2 c2 ⎠
which means that we can write our solution to the linear equation as:
x
=
=
⎛ c1
⎜
⎝ c2
⎛ a1
⎜
⎝ a2
b1 ⎞
⎟
b2 ⎠
b1 ⎞
⎟
b2 ⎠
Dx
y
=
=
D
⎛ a1
⎜
⎝ a2
⎛ a1
⎜
⎝ a2
c1 ⎞
⎟
c2 ⎠
b1 ⎞
⎟
b2 ⎠
Dy
D
Now that we have seen how Cramer's Rule works in the abstract sense, let's use a few numbers.
Use Cramer's Rule to solve (if possible) the system of equations.
3x + 4y = –2
5x + 3y =
4
solution
First we need to find the determinant of the coefficient matrix and make sure it does not equal to zero.
Otherwise, Cramer's Rule doesn't work.
360 – Cramer's Rule
Matrices and Determinants
D
⎛3 4⎞
⎜
⎟
⎝5 3⎠
=
determinant of coefficient matrix
=
9 – 20
formula for determinant of 2 x 2 matrix
=
–11
simplify
Since D ≠ 0, we can go ahead and apply Cramer's Rule as follows:
x
=
Dx
formula to find x
D
⎛ −2 4 ⎞
⎜
⎟
⎝ 4 3⎠
=
y
find Dx and divide by D
=
=
Dy
⎛ 3 −2 ⎞
⎜
⎟
⎝5 4 ⎠
−11
=
=
x
=
formula to find y
D
find Dy and divide by D
−11
−6 − 16
−11
−22
−11
use formula for
determinant of 2 x 2
matrix
=
simplify
=
simplify
answer
2
y
=
12 − ( −10)
−11
22
−11
–2
use formula for
determinant of 2 x 2
matrix
simplify
simplify
answer
The solution is x = 2 and y = –2. Feel free to check this in the original system.
Cramer's Rule: n x n Matrix
Now we will expand Cramer's Rule to matrices of higher order. If we have n equations and n unknowns, then
the value of each variable is given as the quotient of two determinants. The denominator is the determinant of
the coefficient matrix. The numerator is the determinant of the matrix formed by replacing the column
corresponding to the variable (being solved for) with the column representing the constants.
For example, if we have the following system of three unknowns and three equations:
a11x1 + a12x2 + a13x3
=
b1
a21x1 + a22x2 + a23x3
=
b2
a31x1 + a32x2 + a33x3
=
b3
then the solution for x1 in the system is given by:
x1
=
A1
A
=
⎛⎜ b1
⎜ b2
⎜ b3
⎝
⎛⎜ a11
⎜ a21
⎜ a31
⎝
a 12 a 13 ⎞
⎟
a 22 a 23 ⎟
a 32 a 33 ⎟⎠
a 12 a 13 ⎞
⎟
a 22 a 23 ⎟
a 32 a 33 ⎟⎠
At this point, we would probably break down and use a calculator or computer to find the determinants rather
than find them by hand. In general, we can write Cramer's Rule as:
Matrices and Determinants
Cramer's Rule – 361
Cramer's Rule
If a system of n linear equations containing n unknowns has a coefficient matrix A with a nonzero
determinant | A |, the solution of the system is given by:
x1
=
,
x2
=
,...,
xn
=
where the ith column of Ai is the column of constants in the system of equations. If the determinant of
the coefficient matrix is zero, the system has either no solution or infinitely many solutions.
Let's look at an example of how this works with a system of three equations and three unknowns. For
simplicity, we will not show the steps needed to find the determinants. If you need to go back and review how
to do that, please visit the Determinant page two thoughts up in the Algebra Brain.
Use Cramer's Rule to solve (if possible) the following system of linear equations.
4x – 2y + 3z = –2
2x + 2y + 5z = 16
8x – 5y – 2z =
4
solution
First we need to find the determinant of the coefficient matrix:
D
2
⎛ 2 5 ⎞ + 2 ( −1) 3⋅
⎜
⎟
⎝ −5 −2 ⎠
⎛ −2 3 ⎞ + 8 ( −1) 4⋅
⎜
⎟
⎝ −5 −2 ⎠
⎛ −2 3 ⎞
⎜
⎟
⎝ 2 5⎠
=
4 ( −1) ⋅
=
4(21) – 2(19) + 8(16)
find determinants of 2 x 2
matrices
=
84 – 38 – 128
simplify
=
–82
simplify
expand along first column
Since the determinant is not zero, we can apply Cramer's Rule. We find the determinant of three matrices. Dx
is formed by replacing the first column of D with the constant terms. Dy is formed by replacing the second
column of D with the constant terms. Dz is formed by replacing the third column of D with the constant terms.
362 – Cramer's Rule
Matrices and Determinants
x
=
y
=
z
=
Dx
D
Dy
D
Dz
D
=
=
=
=
=
=
−410
−82
−656
−82
164
−82
=
5
=
8
=
–2
Thus, the solution to the system of linear equations is:
x
=
5
and
y
=
8
and
z
=
–2
answer
Matrices and Determinants
Cramer's Rule – 363
364 – Cramer's Rule
Matrices and Determinants
Advanced
To use a matrix to determine if three points all lie on the same line.
No definitions on this page.
John Wallis (1616-1703) made numerous contributions to modern mathematics. He investigated an infinite
approximation to 4 / π, negative and fractional exponents, and the center of gravity in cycloids using indivisibles.
Wallis was the first to use " " to indicate infinity and introduced the mantissa in logarithms. Finally, he is the
first to graphically display the complex roots of a real quadratic equation.
Matrices → Determinants → Applications of Determinants → Test for Collinear Points
Collinear Points
Suppose we have three points that are all on the same line in a plane. By definition, the area enclosed by these
three points is equal to zero. We use three points to define the area of a triangle. Therefore, we can conclude
that if the area of a triangle given by three points in a plane is equal to zero, those three points must lie on the
same line.
Consider the three points given below:
The points (0, 3), (3, 4) and (6, 5) are clearly on the same line. If we use the formula for finding the "area"
enclosed by these points using determinants, we find that the area is:
Matrices and Determinants
Test for Collinear Points – 365
Area
=
=
=
=
=
1
⎛0 3 1⎞
⎜3 4 1⎟
⎟
2 ⎜
⎝6 5 1⎠
formula for area of
triangle
1⎡
2 ⎛4 1⎞
3 ⎛3 1⎞
4 ⎛3 4⎞ ⎤
⎢ 0 ( −1) ⋅ ⎜
⎟ + 3 ( −1) ⋅ ⎜
⎟ + 1 ( −1) ⎜
⎟ ⎥
2⎣
⎝5 1⎠
⎝6 1⎠
⎝6 5⎠ ⎦
cofactor expansion along
first row
1
[ 0 − 3( 3 − 6) + ( 15 − 24) ]
determinants of 2 x 2
matrices
( 9 − 9)
simplify
2
1
2
0
simplify again
We have just demonstrated that the "area" enclosed by three points on the same line is zero. This leads us to the
conclusion that if the determinant in the formula above is zero, all three points must lie on the same line.
Formally, we can write this as:
Test for Collinear Points
Three points (x1, y1), (x2, y2), and (x3, y3) are collinear (lie on the same line) if and only if
=
0
test for collinear points
This makes sense because if the determinant did not equal zero, then you would have a triangle.
Use a determinant to see whether the points are collinear.
⎛ 2, − 1 ⎞
⎜ 2⎟
⎝
⎠
(–4, 4)
(6, –3)
given points
solution
To see if these three points are collinear, we find the determinant of the matrix formed from these points.
⎛⎜ 2 − 1 1 ⎞⎟
2 ⎟
⎜
⎜ −4 4 1 ⎟
⎜ 6 −3 1 ⎟
⎝
⎠
=
2
2 ( −1) ⋅
⎛ 4 1⎞ ⎛ 1⎞
3 ⎛ −4 1 ⎞
4 ⎛ −4 4 ⎞
⎜
⎟ + ⎜ − ⎟ ( −1) ⋅ ⎜
⎟ + 1 ( −1) ⋅ ⎜
⎟
⎝ −3 1 ⎠ ⎝ 2 ⎠
⎝ 6 1⎠
⎝ 6 −3 ⎠
cofactor expansion along first row
=
2( 4 + 3) +
1
( −4 − 6) + ( 12 − 24)
2
determinants of 2 x 2 matrices
= –3
simplify
Since the determinant is not equal to zero, we conclude that these three points do not lie on the same line, as we
can see on the graph below.
answer
366 – Test for Collinear Points
Matrices and Determinants
Equation of a Line
We can adapt the test for collinear points to another use. If we are only given two points in a rectangular
coordinate system instead of three, we can use the same formula above to find the equation of the line described
by those two points. The only difference between the formula above and the next formula is that we replace the
first two elements of the array with x and y, representing any two points on the line.
Two-Point Form of the Equation of a Line
An equation of the line passing through the points (x1, y1) and (x2, y2) is given by
=
0
two-point form of the equation of a line
The result of using this formula is an equation of a line in standard form. Let's see how this works.
Find an equation of the line passing through the points (1, 3) and (7, 5), as shown in the figure below.
Matrices and Determinants
Test for Collinear Points – 367
solution
Applying the determinant formula for the equation of a line produces:
⎛x y 1⎞
⎜1 3 1⎟
⎜
⎟
⎝7 5 1⎠
=
0
two-point form of the equation of a line
To evaluate this determinant, we expand by cofactors along the first row to obtain the following:
2
x( −1) ⋅
⎛3 1⎞
3
⎜
⎟ + y( −1) ⋅
5
1
⎝
⎠
⎛1 1⎞
4
⎜
⎟ + 1 ( −1) ⋅
7
1
⎝
⎠
⎛1 3⎞
⎜
⎟
⎝7 5⎠
=
0
cofactor expansion along first
row
x(3 – 5) – y(1 – 7) + (5 – 21)
=
0
determinants of 2 x 2 matrices
–2x + 6y – 16
=
0
simplify
Therefore, an equation of the line is:
–2x + 6y – 16
=
0
answer
Compare this method to another two-point method for finding an equation of a line.
368 – Test for Collinear Points
Matrices and Determinants
Unit 6
Sequences and Probability
Sequences and Probability – 369
370 – Sequences and Probability
In this section of the Algebra Brain, we will examine sequences and probability.
In English, a sequence is simply a set of items arranged in a particular order. A list of names is often listed in
alphabetical order because this makes it easy to find a particular name on the list. In mathematics, a sequence is
a function whose domain consists of all the positive integers, listed in some specific order. Once we know the
general pattern for the sequence, it becomes a relatively simply matter to find the nth term for that sequence.
Two very common types of sequences are the arithmetic sequence and the geometric sequence.
In an arithmetic sequence, each successive term of the sequence is found by adding some number to the current
term in the sequence. The difference between two successive terms is a constant number d. For example, in the
sequence:
1, 3, 5, 7, 9, …
we add 2 to each term of the sequence to obtain the next term of the sequence.
In a geometric sequence, the ratio of two successive terms is a constant number r. In other words, to find the
next term of the sequence, we multiply the current term by r. For example, in the sequence:
1, 2, 4, 8, 16, 32, …
we multiply each term of the sequence by 2 to obtain the next term of the sequence.
We also present the Binomial Theorem in this section of the Algebra Brain. Recall that we can use a simple
formula to expand a binomial term raised to the second formula. This becomes much more difficult for
binomials raised to higher powers. The Binomial Theorem gives us a formula that allows us to find the
coefficients of a binomial term raised to the nth power.
Finally, we will investigate the basic laws of probability. Most people would agree that a coin tossed into the
air has a 1 in 2 chance of coming up heads. However, we can also use the laws of probability to determine the
likelihood of buying a defective product if we know how many defective products are in a particular shipment.
We can also determine the probability of winning the lottery, in which case we no longer need to study algebra
and can go live on a yacht in the Caribbean. In fact, random drawings are a primary use of probability models,
whether we are trying to win a million dollar jackpot or at a charity raffle.
Sequences and Probability – 371
372 – Sequences and Probability
Advanced
To understand the definition of a factorial.
factorial – for a positive integer n, the product of all the positive integers less than or equal to n.
The word "factorial" was originally used by the German mathematician Christian Kramp. He referred to the
product of factors in arithmetic progression, indicated as an|d. That is, a(a + d)(a + 2d) . . . a (a + n – 1)d = an|d.
He is also responsible for the notation n! used today to represent the product of the first n positive integers. This
corresponds directly to d = 1 in Kramp's original calculation above.
Sequences and Probability → Factorials
Some of the most important sequences in mathematics involve terms that are defined with special types of
products called factorials . Probability is one of the more common applications of factorials. The Binomial
Theorem also uses factorials to define coefficients of binomial expansion. They also play an important role in
sequences and series in calculus.
Definition of Factorial
If n is a positive integer, n factorial is defined by:
n!
=
1 · 2 · 3 · 4 · . . . · (n – 1) · n!
n factorial
As a special case, zero factorial is defined as 0! = 1.
From this definition it is easy to see that 1! = 1, 2! = (2)(1) = 2, 3! = (3)(2)(1) = 3 and so on. Also, factorials
have the same determination in the order of operations as exponents. That is, just as 2x3 and (2x)3 imply
different orders of operations, 2n! and (2n)! imply the following orders:
2n!
(2n)!
=
2(n!)
=
=
2(1 · 2 · 3 · . . . · n)
1 · 2 · 3 · . . . · n · (n + 1) · . . . · 2n
Fractions involving factorials are often very easy to simplify. We can cancel all the like factors in the
numerator and the denominator, leaving just the unlike factors.
Simplify each factorial expression:
a.)
b.)
c.)
7!
2!⋅ 6!
3!⋅ 4!
4!⋅ 5!
n!
( n − 1)!
Sequences and Probability
Factorials – 373
solution
For each of these expressions, we write out each of the factorials, then cancel like factors in the numerators and
denominators. Then we evaluate the remaining expression.
a.)
7!
2!⋅ 6!
write out factorials in long form
cancel common factors
=
b.)
3!⋅ 4!
4!⋅ 5!
2
write out factorials in long form
cancel common factors
=
1
=
5⋅ 4
1
=
c.)
n!
( n − 1)!
remaining factors
simplify
answer
20
write out factorials in long form
cancel common factors
=
=
374 – Factorials
simplify whatever is left after
our cancellations
answer
7
=
n
remaining factor
answer
Sequences and Probability
Advanced
To learn the Fundamental Counting Principle, used to determine the number of outcomes of an event.
No definitions on this page.
In Archimedes's (287-212 B.C.) work, The Sand Reckoner, he lays out a system in which he attempts to
calculate the number of grains of sand in the universe. It also contained a special notation for estimating and
expressing very large numbers. Another of his famous mathematical puzzles is known as the Cattle Problem.
The idea is to determine the number of cows and bulls of various colors, given that each cattle color is
represented in a particular ratio to the others. As it turns out, there are an infinite number of solutions, but the
Greeks had a difficult time solving the puzzle because they had no knowledge of algebra.
Sequences and Probability → Counting Principle
One of the most fundamental concepts to understanding probability is to be able to count the number of ways a
particular event can occur. Oftentimes, the outcome of one event A can affect the outcome of a subsequent
event B, so it is useful to know the likelihood of A occurring. The only way to do that is to count the number of
ways A can occur.
A certain game requires a roll of three six-sided dice (indicated by 3d6). A success on one dice roll is indicated
when the total on all three dice is "8". How many different ways can a total of "8" be obtained on 3d6?
solution
One way we can do this is to construct the following table:
one die
another die
the other die
1
1
6
2
2
4
3
3
2
4
3
1
5
2
1
The order of the numbers on the dice is unimportant, since all three are absolutely identical. The important
thing to consider is the numbers. As we can see, there are only 5 ways we can obtain a total of "8" on three
dice.
answer
In another game, the rules indicate that we need to roll a six-sided die at one point. However, if we roll a "5" or
a "6" on the die, we roll again, this time disregarding any more 5's or 6's. Assuming that we roll a "5" or a "6"
on the first roll, how many different ways can we obtain a total of "8" on the two dice rolls?
solution
Again, we construct a table to help us count the different possibilities:
first roll
second roll
5
3
6
2
Here, we can see that if we get a "5" on the first roll, we must roll a "3" in order to get a total of "8". Similarly,
if we roll a "6" on the first roll, then we must roll a "2" in order to get a total of "8". Thus, there are only 2 ways
to obtain a total of "8" on the two die rolls. answer
Sequences and Probability
Counting Principle – 375
Each of the previous examples illustrates an application of the counting principle. The first example involves
counting with replacement, meaning that each roll of the die is independent of the other rolls. The second
example, however, illustrates counting without replacement, meaning that the second die roll is dependent on
the first. You will study this type event in more detail when you look at Probability.
These examples are fairly simple. We can easily list the possible outcomes and determine which of those
outcomes fit our criteria (in our examples, we wanted a total of "8" on all the dice). When possible, listing the
outcomes is always the best approach. However, in many cases that is not feasible, because there are so many
outcomes. For example, if we wanted to know how many ways we could rearrange the letters in
"supercalifragilisticexpialidocious", we would be here the rest of our lives writing down the list of possible
combinations. Fortunately, we can take advantage of a number of techniques to find the solution without
having to list all the possible outcomes. One such method is the Fundamental Counting Principle, defined
below:
The Fundamental Counting Principle
Let E1 and E2 be two events. The first event E1 can occur in m1 different ways.
After E1 has occurred E2 can occur in m2 different ways.
The number of ways that the two events can occur is given by m1 · m2.
Note:
The Fundamental Counting Principle can be extended to three or more events. For instance, the
number of ways that three events E1, E2, and E3 can occur is m1 · m2 · m3.
Telephone numbers in the United States have ten numbers. The first three digits are the area code, and indicate
where in the United States a particular call originates. The next seven digits are the local telephone number.
Now, there are a few restrictions on which digits can be used in the local telephone number. The digits 0 and 1
are not permitted as the first digit of the local telephone number. Also, the first 3 digits of the number cannot be
"555", as that indicates a phony telephone number (you may have noticed that all telephone numbers in movies
and television begin with "555". This is so nobody calls those numbers). How many different telephone
numbers are possible within each area code?
solution
The first digit cannot be 0 or 1. That leaves 8 choices (2 – 9). The 6 digits remaining can be 0 – 9. If we apply
the Fundamental Counting Principle, then the total number of local telephone numbers is given by:
(8)(10)(10)(10)(10)(10)(10)
=
8,000,000
total number of local telephone numbers
However, we have to exclude all those numbers whose first three digits are "555", which means that we
subtract (10)(10)(10)(10) = 10,000 numbers, leaving us with:
8,000,000 – 10,000
=
7,990,000
actual number of possible local phone numbers
answer
This is actually not very many phone numbers, considering the widespread use of faxes and cell-phones, so
many businesses often have extension numbers, indicating different parts of an organization.
376 – Counting Principle
Sequences and Probability
Advanced
To find the number of combinations of n elements taken r at a time.
No definitions on this page.
The multiplicative rule of combinations, nCr, n things taken r at a time dates as far back as the Jain
mathematician Mahavira in 850 A.D. in India. Later mathematicians of the 10th and 11th centuries are
responsible for creating the combinatorial triangle known today as Pascal's triangle, the elements of which are
derived from the multiplicative rule of combinations.
Sequences and Probability → Counting Principle → Combinations
In a related thought, we discussed permutations, which refer to arranging items in a particular order. Now we
are going to talk about combinations, where order is completely irrelevant.
Basically, we want to look at a subset of a larger set. The arrangement of the subset does not matter. Most
standard card games use this principle. Consider the collectible card game Magic: The Gathering®. Each player
is dealt 7 cards from his or her deck. Once they pick up the cards, each player is free to rearrange the cards in
any order they want. There is no particular arrangement of cards, so each arrangement of the 7 cards is the
same combination. The only time the combination changes is when a player draws one card and then plays a
card from his or her hand.
In a more concrete example, consider the word SPEAR. While we can rearrange the letters in a number of
ways, the combinations PEARS and REAPS are both equivalent combinations because they contain the exact
same elements. Note, however, that if order were important then they would be different permutations. To sum
up:
Combinations
order NOT important
Permutations
order VERY important
Like permutations, we look at subsets of a larger set. That is, we can find the number of combinations of n
elements taken r at a time. For example, we could find the number of 3-letter combinations we can find from
the 5 letters in the word SPEAR. The formula for finding the number of combinations of n elements taken r at a
time is similar to the formula for the number of permutations with an extremely important difference. The
formula is given below:
Combinations of n Elements Taken r at a Time
The number of combinations of n elements taken r at a time is given by:
nCr
Note:
=
number of combinations of
n elements taken r at a time
This is the exact same formula used to compute the coefficients of a binomial expansion.
In a classroom experiment, four students out of a class of 20 are selected to participate in the experiment. How
many different combinations of four students are possible?
Sequences and Probability
Combinations – 377
solution
The order of the students chosen is irrelevant, so we know that we can go ahead and use the formula given
above, letting n = 20 and r = 4:
20C4
=
=
=
20!
( 20 − 4)! 4!
20!
16! 4!
20⋅ 19⋅ 18⋅ 17⋅ 16!
16! 4⋅ 3⋅ 2⋅ 1
number of combinations of
20 elements taken 4 at time
simplify denominator
expand a few factorials
cancel common factors
=
=
(5)(19)(3)(17)
remaining factors
=
4845
simplify
answer
Out of 20 students, there are nearly 5000 different ways of arranging them into 4-person teams.
378 – Combinations
Sequences and Probability
Advanced
To find the number of ways elements can be arranged in order.
No definitions on this page.
Permutations actually come from a study of linguistics. At the turn of the 13th century, the Arab ibn Mun'im wrote
in his Fiqh al-Hisab rules and results obtained in linguistics. These results illustrated the use of both
permutations and combinations. At the end of the 13th century, another Arab, ibn al-Banna, demonstrated the
rules for the number Pn of permutations of n elements, of combination nCr and arrangement nPr of n things taken
r at a time.
Sequences and Probability → Counting Principle → Permutations
One important application of the Fundamental Counting Principle is in determining the number of ways we can
arrange n elements (in order). An ordering of n elements is called a permutation and is defined below:
Definition of Permutation
A permutation of n different elements is an ordering of the elements
such that one element is first, one element is second, one is third, and so on.
How many permutations are possible for the letters B, R, A, I and N?
solution
To understand the number of permutations possible, we undertake the following reasoning.
We basically want to rearrange the five letters in as many different ways as possible, with the order of the letters
playing an important role. For the first letter of a permutation, we have 5 choices. For the second letter of a
permutation, we only have 4 letters to choose from. The third letter of a permutation only offers 3 choices, and
so on. We only get one possible choice for the final letter of a permutation (whichever letter we have not
used).
Thus:
First position:
Second position:
Third position:
Fourth position:
Fifth position:
Any of the 5 letters
Any of the remaining 4 letters
Any of the remaining 3 letters
Any of the remaining 2 letters
The 1 remaining letter
We use the Fundamental Counting Principle to determine the total number of permutations:
(5)(4)(3)(2)(1)
=
5!
total number of permutations
=
120
simplify factorial
answer
The result of this example can be generalized as follows:
Sequences and Probability
Permutations – 379
Permutations of n Elements
The number of permutations of n elements is given by:
n • (n – 1) . . . 4 • 3 • 2 • 1
=
n!
number of permutations of n elements
In other words, there are n! different ways that n elements can be ordered.
In many cases, we are more interested in ordering a subset of a collection of elements rather than the entire
collection. For instance, we might want to choose and order 3 elements out of a total of 8 elements. This
means that we extract any three elements out of the group of eight and order them. Then we extract a second
group of three elements (possibly the same, but not necessarily) and order those three elements. The idea is to
find as many different orders of three elements out of the group of eight elements. If we generalize this, we can
say that we are ordering r elements out of a total of n elements. This kind of ordering is called a permutation of
n elements taken r at a time, defined as follows:
Permutations of n Elements Taken r at a Time
The number of permutations of n elements taken r at time is
=
nPr
=
n(n – 1)(n – 2) . . . (n – r + 1)
number of permutations of
n elements taken rat a time
Eight students race at a swim meet. In how many ways can a group of three students out of the original eight
come in first, second, and third (assuming there are no ties)?
solution
Using the formula above, we let n = 8 (total number of students) and r = 3 (subset of larger group). Then we
get:
8P3
=
=
8!
( 8 − 3)!
8!
5!
substitute in n = 8 and r = 3
simplify denominator
cancel common factors
=
=
(8)(6)(7)
remaining factors
=
336
simplify
answer
The key thing to remember about permutations is that ORDER IS IMPORTANT! That is, if you take a look at
the permutations of the letters C, A and T taken three at a time, then the permutations (C, A, T) and (T, A, C) are
different because of the order. Note that there are three distinct elements in this set of three letters. However,
what if we have the letters K, I, W, and I? How can we distinguish between the first I and the second I? The
short answer is that we can't. The total number of permutations is given by 4P4 = 4! = 24. But some of the
permutations will be the same because there are two I's. In order to find the total number of distinguishable
permutations, we use the following formula:
380 – Permutations
Sequences and Probability
Distinguishable Permutations
Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind,
and so on, with n = n1 + n2 + n3 + . . . + nk.
Then the number of distinguishable permutations of the n objects is:
number of distinguishable permutations
In how many distinguishable ways can the letters in CALCULUS be arranged?
solution
This word has 8 letters, of which 2 are C's, 2 are L's, two are U's, 1 is an A, and 1 is an S. Thus, the number of
distinguishable ways the letters can be written is:
8!
2! 2! 2! 1! 1!
=
8⋅ 7!
8
number of distinguishable permutations
=
7!
cancel common factors
=
5040
simplify factorial
answer
There are over five thousand different ways we can arrange the letters in the word CALCULUS that are
completely distinct from every other permutation. Note that the total number of permutations (allowing for
duplicate permutations) is given by 8P8 = 8! = 40,320. And no, we are not going to list them for you.
Sequences and Probability
Permutations – 381
382 – Permutations
Sequences and Probability
Advanced
To use the Binomial Theorem to expand binomials raised to the nth power.
binomial series – a binomial expansion which contains infinitely many terms.
Although the Binomial Theorem is usually credited to the Chinese or to Isaac Newton, during the first half of the
fifteenth century, the Persian mathematician Ghiyat ad din Ghamshid ibn Mas'ud al-Kashi also did important
work on binomials. He also worked with exponential powers of whole numbers, n roots, and irrational numbers.
Sequences and Probability → Binomial Theorem
A binomial is a polynomial that has exactly two terms. If it had three terms, then it would be a trinomial. Now,
we can raise a binomial to an integer power (such as 2). When that happens, we can expand the binomial by
multiplying all its terms together. However, for any power of the binomial greater than 2, this becomes tedious
and difficult. Therefore, what we want to find is a quick method of finding the coefficients of a binomial raised
to any power higher than 2 (although the formula we come up with will work for binomials raised to any integer
power).
The key word here is integer power. The binomial theorem does not work for powers that are not integers. But
we don't usually need to worry about that in algebra.
Let's take a look at the expansion of (x + y)n for several values of n and see if we can't notice a pattern
developing among the coefficients:
(x + y)0
=
1
binomial raised to zero power
(x + y)1
=
x+y
binomial raised to first power
(x + y)2
=
x2 + 2xy + y2
binomial raised to second power
(x + y)3
=
x3 + 3x2y + 3xy2 + y3
binomial raised to third power
(x + y)4
=
x4 + 4x3y + 6x2y2 + 4xy3 + y4
binomial raised to fourth power
(x + y)5
=
x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5
binomial raised to fifth power
Here are some observations we can make based on the expansions given above:
1.
In each expansion there are n + 1 terms. That is, if n is even, then we have an odd number of terms
and if n is odd, then we have an even number of terms.
2.
In each expression, x and y take on symmetrical roles. The powers of x decrease by 1 in each
successive term and the powers of y increase by 1 in each successive term.
3.
The sum of the powers of x and y equals n in each term. That is, the first term has xny0, the second term
has xn–1y1, the third term has xn–2y2, and so on until we get to x0yn.
4.
The coefficients increase and decrease in a symmetric manner.
In order to determine the coefficients, we need to make use of the following theorem, which unfortunately
requires calculus to prove:
Sequences and Probability
Binomial Theorem – 383
The Binomial Theorem
In the expansion of (x + y)n
(x + y)n
=
xn + nxn – 1y + . . . + nCrxn – ryr + . . . + nxyn – 1 + yn
The coefficient nCr of xn – ryr is given by
nCr
Note:
=
As it turns out, this theorem is just a special case of a much more powerful theorem used for what are
called binomial series . Isaac Newton, who came up with the binomial theorem about 2600 years after
the Chinese, also figured out a way to expand a binomial that is raised to a rational power, such as 1/2.
However, you don't get a nice neat package like you get with the binomial theorem. Instead you get an
infinite number of terms, the coefficients of which are determined in a similar manner as described
above.
Find the binomial coefficients:
a.)
9C7
b.)
4C1
c.)
6C3
solution
a.)
9C7
=
=
=
b.)
4C1
9⋅ 8⋅ 7⋅ 6⋅ 5⋅ 4⋅ 3⋅ 2⋅ 1
( 2⋅ 1) ( 7⋅ 6⋅ 5⋅ 4⋅ 3⋅ 2⋅ 1)
9⋅ 8
write out factorials long hand
cancel common factors in numerator and denominator
2⋅ 1
9x4
simplify rational expression
=
36
answer
=
=
=
6C3
Binomial Theorem
( 9 − 7)! 7!
=
=
c.)
9!
=
4!
(4
1) ! 1 !
4.3.2.1
( 3.2.1) ( 1)
4
Binomial Theorem
write out factorials long hand
cancel common factors
1
4
answer
6!
3! 3!
384 – Binomial Theorem
Binomial Theorem
Sequences and Probability
=
6.5.4.3 !
3! 3!
write out factorials long hand
=
6.5.4
3.2.1
cancel common factorials
=
5x4
cancel common factors
=
20
answer
Each coefficient belongs in front of xn –ryr, so the answer to part (a) above would go in front of x9 – 7y7. The full
term would look like:
36x2y7
Sequences and Probability
Binomial Theorem – 385
386 – Binomial Theorem
Sequences and Probability
Advanced
To find the coefficients of a binomial expansion.
No definitions on this page.
Blaise Pascal (1623-1662) was a French geometer, probabilist, physicist, inventor, and philosopher. His most
famous invention is a mechanical calculator that automatically adds and subtracts. The basic principle behind
this device was still being used in the first half of the 20th century until modern electronics came along. The
triangle that bears his name was actually invented 600 years earlier by the Chinese. Pascal used it so creatively
in his studies on probability that the triangle was given his name. Pascal's work on binomial coefficients helped
lead Isaac Newton to his discovery of the general binomial theorem, which works for all real numbers, not just
integers.
Sequences and Probability → Binomial Theorem → Pascal's Triangle
There is a convenient method to remember the binomial coefficients. It is called Pascal's Triangle, after the
French mathematician Blaise Pascal (1623 – 1662).
The basic idea is to build a triangle out of the binomial coefficients. Each row of the triangle corresponds to a
binomial raised to a particular degree. We start with degree 0 and work our way down the triangle. The
coefficients of lower rows are formed by adding coefficients of the two numbers in the row above.
The top row is row zero because it corresponds to the binomial (x + y )0.
The first and last numbers of each row are 1.
Each of the other numbers in the rows is formed by adding the two numbers directly above it.
The middle number of row two, for example, is formed by adding the two ones in row 1.
The second number in row 4 is formed by adding the 1 and 3 directly above it in row 3.
Here are the expanded binomials of (x + y)n using Pascal’s Triangle to determine the coefficients:
Sequences and Probability
Pascal's Triangle – 387
Note:
(x + y)0
=
1
zeroth row
(x + y)1
=
1x + 1y
first row
(x + y)2
=
1x2 + 2xy + 1y2
second row
(x + y)3
=
1x3 + 3x2y + 3xy2 + 1y3
third row
(x + y)4
=
1x4 + 4x3y + 6x2y2 + 4xy3 + 1y4
fourth row
(x + y)5
=
1x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + 1y4
fifth row
(x + y)6
=
1x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + 1y6
sixth row
(x + y)7
=
1x7 + 7x6y + 21x5y2 + 35x4y3 + 35x3y4 + 21x2y5 + 7xy6 + 1y7
seventh row
The sum of the exponents of x and y always equals the degree of the binomial.
Pascal's triangle is only useful for the first half-dozen rows or so. After that, it gets a bit tedious to build the
triangle each time. However, if, for some reason, you have trouble remembering the Binomial Theorem method
of generating coefficients, then by all means use Pascal's Triangle.
Although credit for the triangle is given to Pascal, versions of the triangle were known to Eastern cultures
(particularly the Chinese) long before its "discovery" by the West.
388 – Pascal's Triangle
Sequences and Probability
Advanced
To learn the definition of probability.
No definitions on this page.
Dice are among the first examples of a random number generator. Besides gambling, dice have also been used
for foretelling the future. Indeed, one of Julius Caesar's most famous classical quotations is, "Alea jacta est,"
meaning, "The die is cast." This refers to the fact that once one has started upon a course of action, the outcome
is in the hands of Fate or Destiny, represented in the form of dice.
Sequences and Probability → Probability
Probability, ladies and gentlemen, is all about finding the likelihood of an event occurring. For example, the
probability that the Cubs win the World Series might be one in a million, depending on whom you talk to. This
means that the Cubs will only win the World Series once every million years.
Probability is always associated with uncertainty. That is, there is seldom a guarantee that an event will happen.
We can only look for the possibilities of an event occurring in terms of the possibility that the event will not
occur. In general, any happening in which the result is uncertain is called an experiment. The possible results
of the experiment are called outcomes. The set of all possible outcomes of an experiment is called the sample
space. Any sub-collection of the sample space is called an event.
Let's look at a simple example, a six-sided die (abbreviated as 1d6). When we toss 1d6 into the air, the sample
space (i.e. the total number of the outcomes) can be represented by all the numbers 1 through 6. However, the
event, or the number that we actually get, is just one of the numbers 1 through 6. In other words, if we roll a
"5", then we have:
S
=
{1, 2, 3, 4, 5, 6}
sample space
E
=
{5}
event
The probability that we roll a "5" is given by counting the number of times that this occurs in the sample space
and dividing that number by the total number of outcomes. In this case, the number "5" only occurs 1 time in
the sample space, which contains 6 outcomes. Therefore, the probability of rolling a "5" is given by:
P(5)
=
1
6
probability of rolling a "5" on 1d6
Suppose we now roll 2d6 (two six-sided dice) and add the number together. The smallest number we can
get is "2" and the largest number is "12". Therefore, the sample space includes all numbers between 2 and
12. However, we now have a new wrinkle. Some numbers will occur more than once in the sample space.
For instance, we can roll a 1 and a 4 to get 5 or we can roll a 2 and a 3 to get 5. Or we can reverse the
order: 4 + 1 = 5 or 3 + 2 = 5. One way to get a better picture of what is going on is with a chart, listing
each die roll and the total on 2d6:
Sequences and Probability
Probability – 389
die
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
Notice that there are actually 36 outcomes in the sample space. The number "5" occurs 4 times in the sample
space, so the probability of rolling a total of 5 on 2d6 is therefore:
4
=
P(5)
36
1
=
9
probability of rolling 5 on 2d6
simplify fraction
In general, the number of outcomes in event E is denoted by n(E). The number of outcomes in the sample space
is denoted by n(S). The probability that event E will occur is given by n(E) / n(S). This is summarized in the
following theorem:
The Probability of an Event
If an event E has n(E) equally likely outcomes and the sample space S has n(S)
equally likely outcomes, then the probability of event E is:
P(E)
=
probability of event E
Since the number of outcomes of an event E must be less than or equal to the number of outcomes in the sample
space, the probability of an event occurring is ALWAYS between 0 and 1. That is,
0
≤
P(E)
≤
1
If P(E) = 0, then the event E cannot occur and E is called an impossible event. If P(E) = 1, then event E
absolutely has to occur, and E is called a certain event.
When determining the number of outcomes in a sample space, it is often very helpful to apply the Fundamental
Counting Principle.
A certain game requires a roll of 3d6. In order for it to be a "critical success" (i.e. extremely good), the total of
all three dice must be 5 or less. What is the probability of rolling 5 or less on 3d6?
solution
First, we need to know the total number of outcomes. In order to do that, we use the Fundamental Counting
Principle. Since each die has 6 outcomes, we conclude that the total number of outcomes is given by
n(S)
=
(6)(6)(6)
number of outcomes in sample space
=
216
simplify
Next, we need to determine how many of the outcomes are 5 or less. The smallest number we can roll on 3d6 is
3, which only occurs once. Here is the list of ways the dice can land:
390 – Probability
Sequences and Probability
first die
second die
third die
total
1
1
1
3
2
1
1
4
1
2
1
4
1
1
2
4
2
2
1
5
2
1
2
5
1
2
2
5
3
1
1
5
1
3
1
5
1
1
3
5
When we count the columns, we see that there are 10 ways in which we can achieve a sum of 5 or less on 3d6.
Thus, n(E) = 10, and the probability of rolling 5 or less on 3d6 is given by:
P(5 or less)
=
=
10
216
5
108
probability of rolling 5 or less on 3d6
simplify fraction
answer
When we convert this fraction to percentile form, we find that we only have about a 4.6% chance of rolling a
"5" on 3d6.
Sequences and Probability
Probability – 391
392 – Probability
Sequences and Probability
Advanced
To find the probability of independent events.
independent events – the occurrence of one event has no impact whatsoever on the occurrence of another
event.
Jakob Bernoulli (1645-1705) contributed significantly to the field of probability theory. One of his theorems,
dubbed "The Law of Large Numbers" by Simeon-Denis Poisson, states that if one makes a very large number of
independent trials - flipping a coin, for example - then the observed ratio of successes to failures for a given
event - obtaining "heads" in a coin flip - will be very close to the theoretical probability for that event on each
individual trial. In other words, flipping a coin 1000 times should yield roughly 500 instances of "heads", since
500/1000 = 1/2, which is the probability of obtaining "heads" on an individual trial.
Sequences and Probability → Probability → Independent Events
Two events are said to be independent events when the occurrence of one event has no impact whatsoever on
the occurrence of the other event. For instance, rolling a "6" on 1d6 has no effect on future rolls of the same
die.
In general, we can use a very simple formula for finding the probability of independent events. Simply multiply
the probabilities of each independent event together.
Probability of Independent Events
If A and B are independent events, the probability that both A and B will occur is given by:
P(A and B)
=
P(A) · P(B)
probability of independent events
We have 6 dice, a four-sider (1d4), a six-sider (1d6), an eight-sider (1d8), a ten-sider (1d10), a twelve-sider
(1d12) and a twenty-sider (1d20). What is the probability that we roll a "3" on all six dice if we roll them all
together?
solution
Each die can be considered to be an independent event. The probability of rolling a "3" on 1d4 is P(3 on d4) =
1
/4. The probability of rolling a "3" on 1d6 is P(3 on d6) = 1/6. We can easily find the probabilities for each of
the other dice in a similar manner. The probability for rolling a "3" on all six dice, then, is the product of each
of the independent probabilities:
P(3 on all 6 dice)
=
=
⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞⎛ 1 ⎞
⎜ 4 ⎟ ⎜ 6 ⎟ ⎜ 8 ⎟ ⎜ 10 ⎟ ⎜ 12 ⎟ ⎜ 20 ⎟
⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠
1
460800
probability of rolling a "3" on all six dice
simplify
answer
Thus, there is only a 1 in 460,800 chance that we will roll a "3" on all six dice when we roll them all together.
Sequences and Probability
Independent Events – 393
394 – Independent Events
Sequences and Probability
Advanced
To find the probability of the union of two events.
mutually exclusive – sets with no common elements.
Daniel Bernoulli (1700-1782) was a Swiss mathematician and physician. He is most famous for his theories on
hydrodynamics. Thanks to Daniel Bernoulli, we understand how an airplane is able to fly. He also made
significant contributions to the field of probability. Among his hypotheses is the supposition that as a person's
material fortune increases geometrically, his or her moral fortune increase arithmetically. Thus Bernoulli drew a
connection between economics and ethics, describing both in mathematical terms.
Sequences and Probability → Probability → Union of Two Events
In the previous thought, we looked at the probability of a single event occurring, independently of other events.
Now we want to look at what happens to probability if we have two or more events occurring.
Two events are called mutually exclusive if the events A and B have no outcomes in common. Oftentimes we
express this notion in the terminology of sets. That is, if A and B are mutually exclusive then
P(A
=
B)
0
probability of the intersection of sets A and B
All this means is that A and B have nothing in common, so the probability that we will find an event that
occurs in both the sets A and B is zero. For example, if we roll 2 six-sided dice (2d6), then the probability of
rolling a 5 has nothing whatsoever to do with the probability of rolling a 10. They are mutually exclusive
events. That is, there is no way possible that we can roll a 5 AND a 10. However, there is a distinct
possibility of rolling a 5 OR a 10. In that case, to find the probability, all we need to do is add the
probability of A to the probability of B.
Let's see what the possibilities of rolling any number between 2 and 12 on 2d6 are again:
die
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10
11
6
7
8
9
10
11
12
From the chart, we can see that the probability of rolling a 5 on 2d6 is 4 out of 36 possibilities. The probability
of rolling a 10 is 3 out of 36 possibilities. Therefore, the probability of rolling a 5 or a 10 is simply:
P(5 or 10)
=
=
4
36
7
36
+
3
36
probability of mutually exclusive events
simplify
answer
Simply adding probabilities is only good for when we have mutually exclusive events. When events are not
mutually exclusive, we have to tack on another term into our little sum. For instance, if we want to find the
probability of drawing an ace or a diamond from a deck of cards, we have to take into account that 1 of the
cards is both an ace and a diamond. Therefore, the total probability is determined by subtracting the number of
Sequences and Probability
Union of Two Events – 395
elements in the larger set that fit both criteria. This keeps us from counting an element twice when we are
determining the probability of an event.
In general, we can summarize the probability of the union of two events in the following theorem:
Probability of the Union of Two Events
If A and B are events in the same sample space, the probability of A or B occurring is given by
P(A U B)
=
P(A) + P(B) – P(A
B)
probability of A or B
If A and B are mutually exclusive events, then
P(A U B)
=
P(A) + P(B)
probability of mutually exclusive events
One card is selected from a standard 52-card deck. What is the probability that it is a number
card or a club? A number card includes all those with a digit between 2 and 10.
solution
The deck has 13 clubs, so the probability of selecting a club (event A) is P(A) = 13/52. The deck also contains 36
number cards (9 of each suit), so the probability of selecting a number card (event B) is P(B) = 36/52. We also
have to take into account that some of the number cards are also clubs (9 of them, to be exact), thus P(club and
number card) = 9/52. To find the net probability of drawing a club or a number card, we have to subtract the
redundant event of drawing a club and a number card. In other words, these events are not mutually exclusive,
so we have to use the first formula given in the theorem above:
P(A U B)
=
P(A) + P(B) – P(A B)
probability of non-mutually exclusive events
P(club or number card)
=
P(club) + P(number card) – P(club and number card)
events for this situation
P(club or number card)
=
=
=
13
36
9
+
−
52 52 52
substitute in our probabilities for each event
40
52
combine like fractions
10
13
reduce fraction to lowest terms
answer
The probability of drawing a club or a number card from standard 52-card deck is 10 out of 13.
396 – Union of Two Events
Sequences and Probability
Advanced
To find the probability of the complement of a given event.
complement of an event – the collection of outcomes in the sample space that are not part of a given event A.
If a given event is denoted by A, then its complement is denoted by A'.
With few exceptions, the use of probability in quality control in manufacturing is fairly new, dating back to 1920.
One exception involves the accuracy of guns and the wear of gun barrels in late 19th-century France. Another
exception is the use of a statistical model for small samples to maintain the quality of the fermentation process at
the Guinness Brewery in Dublin, Ireland. A man named W. S. Gosset (a.k.a. "Student") developed this quality
control process in the early 20th century.
Sequences and Probability → Probability → Union of Two Events → Complementary Events
In the previous thought we discussed the basics of probability. Recall that an event is defined as the subset of
the sample space, containing none, some, or all of the elements within that sample space. But what do we call
the elements of the sample space that are not a part of the subset defining the event? Do they have any
meaning?
As it turns out, sometimes it is more useful to know the probability of finding an outcome in the sample space
that is not part of a given event. We call the collections of outcomes in the sample space that are not part of
event A the complement of the event A. We denote the complement as A'.
Suppose we roll a six-sided die (1d6). The sample space for this experiment is S = {1, 2, 3, 4, 5, 6}. One
event we might want to look at is the possibility of rolling an even number. Then we would denote this
as A = {2, 4, 6}. The complement of this event would then be all the odd numbers in the sample space:
A' = {1, 3, 5}. The probability of rolling an even number is P(A) = 1/2. Similarly, the probability of
rolling an odd number is P(A') = 1/2. The probability of rolling an even or an odd number has to be 1,
because we MUST do one or the other. Therefore,
P(A or B)
=
1
1
=
=
2
probability of rolling a number from 1 to 6
+
1
2
P(A) + P(A')
sum of P(even) and P(odd)
substitute in P(A) and P(A')
If we rearrange this a little bit, we can see:
P(A) + P(A')
=
1
sum of probability of event and its
complement is ALWAYS equal to 1
P(A')
=
1 – P(A)
solve for P(A')
Formally, we can define the probability of a complement as follows:
Sequences and Probability
Complementary Events – 397
Probability of a Complement
Let A be an event and let A' be its complement. If the probability of A is P(A),
then the probability of its complement is:
=
P(A')
1 – P(A)
probability of a complement
On some occasions, it is actually quicker to find the probability of an event by first finding its complement and
then subtracting the complement from 1.
A manufacturer determines that one of its products has a failure rate of about 1 in 1500. That means that the
probability the product fails is 1/1500. A retailer orders a shipment of 75 units of the product. What is the
probability that at least one of the units is defective?
solution
One way to solve this problem is find the probability of having 1 defective unit, then find the probability of
having 2 units defective, and so on for all 75 units in the order. This is the hard way.
The easy way is to find the probability that all units are perfect. Then we subtract this value from 1. The
probability that any given unit is perfect is 1499/1500. Since each unit’s perfection is independent of every other
unit’s perfection, they are independent events and the probability that all 75 units are perfect is given by:
P(A)
=
⎛ 1499 ⎞
⎜
⎟
⎝ 1500 ⎠
≈
0.951
75
probability of independent events
decimal approximation
Now we subtract this number from 1 to find the probability that at least 1 unit is defective:
P(A')
=
1 – P(A)
probability of a complement
≈
1 – 0.951
substitute in P(A) = 0.951
≈
0.049
simplify
answer
Thus, there is an approximately 4.9% probability that at least 1 unit of the shipment is defective.
398 – Complementary Events
Sequences and Probability
Advanced
To define a sequence of numbers.
term (sequence) - an element in the range of a sequence.
finite sequence - a sequence whose domain consists of the first n integers only.
recursive (sequence) - each term of a sequence depends on the value of the previous term of the sequence.
The Italian mathematician Fibonacci was one of the leading mathematicians of the Middle Ages. He made
significant contributions to the fields of algebra, arithmetic, and geometry, but is most famous for his sequence of
numbers (1, 1, 2, 3, 5, 8, 13, ...). The Fibonacci sequence is actually the solution to an obscure problem that
appeared in his book Liber Abaci. Interestingly, Fibonacci numbers (i.e. numbers that are members of his
sequence) appear regularly in nature, particularly in flowers such as trillium, wildrose, bloodroot, cosmos,
columbine, lily blossoms, and iris, each of which has a number of petals equal to a Fibonacci number.
Sequences and Probability → Sequences
Information is often arranged in some sort of meaningful order. In fact, random information is often classified
as "noise" because our minds are not able to process it. Just about everything we encounter in the real world is
a result of an attempt to form order from chaos. Usually, when we arrange pieces of information, we place
them in a sequence, so that there is a first member, second member, third member, and so on.
In mathematics, a sequence is a very special type of function (and yes, it is a function in every sense of the
word). The domain of a sequence consists of positive integers (1, 2, 3, 4, etc.). Sequence notation is a little
different than our usual function notation f(x). Instead, we denote members of a sequence using subscripts, as
shown below:
Definition of a Sequence
An infinite sequence is a function whose domain is the set of positive integers.
The function values
a1, a2, a3, … , an, …
terms of a sequence
are the terms of the sequence.
If the domain consists of the first n positive integers only, the sequence is a finite sequence.
Note:
Sometimes it is useful to begin the subscripts of a sequence with 0 rather than 1. In that case, the terms
of the sequence become:
a0, a1, a2, a3, …
Let's look at a couple of examples of sequences to show a little bit what they are like:
Sequences and Probability
Sequences – 399
Find the first four terms for each sequence given below:
a.)
an
=
3n + 4
b.)
an
=
5 + (–1)n
c.)
an
=
( −1)
n
2n − 1
solution
a.)
The first four terms of the sequence given by an = 3n + 4 are:
a1
=
3(1) + 4
=
7
first term
a2
=
3(2) + 4
=
10
second term
a3
=
3(3) + 4
=
13
third term
a4
=
3(4) + 4
=
16
fourth term
answer
Notice that the difference between two consecutive terms is 3. This is an example of an arithmetic sequence.
b.)
The first four terms of the sequence given by an = 5 + (–1)n are:
a1
=
5 + (–1)1
=
4
first term
a2
=
5 + (–1)2
=
6
second term
a3
=
5 + (–1)3
=
4
third term
a4
=
5 + (–1)4
=
6
fourth term
answer
Notice that we are alternating between 4 and 6 in our list of terms. If we continue this sequence out to
infinity, the pattern will continue to repeat itself.
c.)
The first four terms of this sequence are as follows:
a1
=
a2
=
a3
=
a4
=
( −1)
1
2( 1) − 1
( −1)
2
2( 2) − 1
( −1)
=
3
2( 3) − 1
( −1)
=
=
4
2( 4) − 1
=
–1
1
second term
3
−
1
7
first term
1
5
third term
fourth term
answer
Just looking at the first few terms of a sequence is not enough to fully determine the pattern of a sequence. In
order to fully define a unique sequence, the nth term must be given! Consider the following two sequences, both
of which have the same first three terms:
400 – Sequences
Sequences and Probability
1
1 1 1 1
, , , , ...., , ....
n
2 4 8 16
2
and
6
1 1 1 1
, , , , ....,
2
2 4 8 15
( n + 1) n − n + 6
(
)
After the third term, these two sequences start to look very different from each other. If we are just given the
first few terms, all we can do is find an apparent nth sequence. If we had just been given the first three terms of
the first sequence above, we would probably have agreed that the apparent nth term was 1/(2n). We would
never have considered that the second nth term would be a possibility as well.
Write an expression for the apparent nth term (an) of the given sequence.
a.)
3, 7, 11, 15, 19, …
b.)
1+
1
1
,1 +
1
2
1
,1 +
3
,1 +
1
4
,1 +
1
5
, ....
solution
a.)
n:
1
2
3
4
5
...n
Terms:
3
7
11
15
19
. . . an
Apparent pattern: Each term is one less than 4 times n, which implies that:
=
4n – 1
answer
n
1
2
Terms:
1+
an
b.)
1
1+
1
3
1
2
1+
4
1
3
1+
5
1
4
1+
...n
1
5
. . . an
Apparent pattern: Each term is the sum of 1 and 1/n, which implies that:
an
=
1+
1
n
answer
Finally, some sequences are defined recursively. That is, each term of the sequence depends on the previous
term of the sequence. In order to find the terms of the sequence, you need the first few terms of the sequence as
well as the general pattern for the recursion. One of the most famous examples of a recursive sequence is the
Fibonacci Sequence:
a0 = 1, a1 = 1, ak = ak – 2 + ak – 1
Sequences and Probability
where k ≥ 2
Sequences – 401
The first six terms of the Fibonacci Sequence are:
Note:
a0
=
1
0th term
a1
=
1
first term
a2
=
a0 + a1
=
1+1
=
2
second term
a3
=
a1 + a2
=
1+2
=
3
third term
a4
=
a2 + a3
=
2+3
=
5
fourth term
a5
=
a3 + a4
=
3+5
=
8
fifth term
This is an example of a sequence that starts with a0 instead of a1.
402 – Sequences
Sequences and Probability
Advanced
To learn about arithmetic sequences and nth partial sums.
arithmetic sequence – a sequence whose consecutive terms all have a common difference separating them.
common difference – the difference between any two consecutive terms in an arithmetic sequence is a
constant, represented by d.
nth partial sum – the sum of the first n terms of an infinite sequence.
Archytus of Tarentum (c.428-c.350 B.C.) is credited as being the inventor of the arithmetic mean. He also
invented the simple pulley and the screw. Besides being a well-known mathematician, he was also a brilliant
military strategist and statesman. Supposedly, troops under his command never lost a battle. He served as
general of the citizen army of Tarentum for seven years, despite the fact that local law prohibited anyone from
holding the post for more than one year.
Sequences and Probability → Sequences → Arithmetic Sequence
In the previous thought, we introduced you to the basic concept of a sequence. There are an infinite number of
sequences out there. Some of them can be classified into two different types: arithmetic or geometric.
Simply put, an arithmetic sequence is one whose terms all have a common difference separating them. That
is, if we take any two terms and subtract them from each other, we will obtain the same number, regardless of
which terms we use.
For example, the sequence
2, 4, 6, 8, 10, . . .
is arithmetic because the difference between any two consecutive terms is 2.
More formally, we can define an arithmetic sequence as:
Definition of an Arithmetic Sequence
A sequence is arithmetic if the differences between consecutive terms are the same.
Thus, the sequence
a1, a2, a3, . . . , an, . . .
is arithmetic if there is a number d such that
a2 – a1
=
d
and
a3 – a2
=
.
.
.
=
d
and
d
and so on
an + 1 – an
The number d is the common difference of the arithmetic sequence.
Now, the definition by itself is not all that helpful. If we see a sequence, we can use the above definition to help
us determine if it is arithmetic or not (just find the difference of consecutive terms; if it is always the same, then
Sequences and Probability
Arithmetic Sequence – 403
the sequence is arithmetic). Once we know that a particular sequence is arithmetic, we usually want to find the
nth term of that sequence. To do that, we need the following theorem.
The nth Term of an Arithmetic Sequence
The nth term of an arithmetic sequence has the form
an
=
dn + c
nth term of arithmetic sequence
where d is the common difference between consecutive terms and c = a1 – d.
Find a formula for the nth term of the arithmetic sequence whose common difference is 5 and whose first term
is 3.
solution
We are told the sequence is arithmetic. Furthermore, we know that the common difference is 5 and the
first term is 3. This implies that the second term is 3 + 5 = 8, the third term is 8 + 5 = 13, the fourth term
is 13 + 5 = 18, and so on. The nth term must be in the form an = dn + c, where d in this case happens to
be 5.
Because a1 = 3, it follows that:
c
=
a1 – d
formula to find c
=
3–5
substitute in a1 = 3 and d = 5
=
–2
simplify
Thus, the formula for the nth term is:
an
=
dn + c
nth term of arithmetic sequence
an
=
5n – 2
substitute in d = 5 and c = –2
answer
Once we have determined the nth term of an arithmetic sequence, it is extremely easy to find any other term of
the sequence, provided that we know the previous term. If the nth term is given by
an
=
dn + c
nth term of arithmetic sequence
then the (n + 1)th term is given by
an + 1
=
an + d
(n + 1)th term of arithmetic sequence
This is known as a recursive formula because the (n + 1)th term is dependent on the nth term. Similarly, the
(n + 2)th term is dependent on the (n + 1)th term, and so on.
Finally, there is a very simple formula to find the sum of a finite arithmetic sequence. All we need to know is n,
the number of terms, as well as the first and last terms of the finite arithmetic sequence. Then we can find the
sum of the sequence using the following formula:
404 – Arithmetic Sequence
Sequences and Probability
The Sum of a Finite Arithmetic Sequence
The sum of a finite arithmetic sequence with n terms is given by
=
Sn
Note:
sum of a finite arithmetic sequence
This formula only works with arithmetic sequences. We use a different formula to find the sum of a
finite geometric sequence. Other types of finite sequences also have different formulas to find the
sum.
The sum of the first n terms of an infinite sequence is called the nth partial sum.
Find the sum of all integers from 1 to 150.
solution
The integers from 1 to 150 form an arithmetic sequence that has 150 terms. Thus, we can use the formula given
above to find the sum of these 150 terms:
=
S150
=
=
1 + 2 + 3 + 4 + . . . + 149 + 150
n
2
(a1 + an )
150
2
sum of integers from 1 to 150
sum of a finite arithmetic sequence
( 1 + 150)
substitute in n = 150, a1 = 1, and a150 = 150
=
75(151)
simplify
=
11325
simplify
answer
Find the sum of the first 32 terms of the arithmetic sequence
5, 18, 31, 44, 57, 70, . . .
solution
First, we need to know what the 32nd term is for this sequence. That means we need the nth term, which takes
the form an = dn + c. In this case, taking the difference of consecutive terms reveals that the common difference
d = 13. We are given that a1 = 5, which means that c = a1 – d = 5 – 13 = –8. Therefore, the nth term of the
sequence is:
an
=
13n – 8
nth term of the sequence
When we substitute in n = 32, we find that the 32nd term is:
a32
=
13(32) – 8
substitute in n = 32
=
408
32nd term of our arithmetic sequence
Sequences and Probability
Arithmetic Sequence – 405
Now that we know the 32nd term (408) and the first term (5), we can go ahead and apply the formula for
finding the sum of a finite arithmetic sequence:
Sn
=
S32
=
n
2
(a1 + an )
32
2
( 5 + 408)
sum of a finite arithmetic sequence
substitute in n = 32, a1 = 5 and a32 = 408
=
16(413)
simplify
=
6608
simplify
answer
406 – Arithmetic Sequence
Sequences and Probability
Advanced
To learn about the definition of geometric sequences and how to calculate the sum of a geometric
sequence.
geometric sequence – a sequence with a constant ratio between successive terms.
common ratio – a ratio which is the same between all consecutive terms in a geometric sequence.
nth partial sum – the sum of the first n terms of a sequence.
Archytus of Tarentum (c.428-c.350 B.C.) developed the geometric mean, the arithmetic mean, and the harmonic
mean. His most famous mathematical achievement was to develop an elegant solution to the problem of
doubling a cube, i.e. enlarging a cube according to a given ratio. He solved it by inventing a new type of threedimensional curve through the intersection of a cylinder, a cone, and a torus (or doughnut-shape).
Sequences and Probability → Sequences → Geometric Sequence
In the thought Arithmetic Sequence, we defined an arithmetic sequence as one whose terms can be defined in
terms of a common difference between consecutive terms. Now we will look at another major type of sequence:
the geometric sequence, whose terms are defined in terms of a common ratio. That is, if we take any two
consecutive terms and divide them into each other, we will obtain a constant ratio r. Here is an example of a
geometric sequence:
2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 4096, . . .
geometric sequence
If we take any two consecutive terms of this sequence and divide one term by the other, we will obtain the
common ratio of 2 (or 1/2, depending on which number gets divided into the other).
More specifically, we can define a geometric sequence as follows:
Definition of a Geometric Sequence
A sequence is geometric if the ratios of consecutive terms are the same.
=
r
and
=
r
and
r
with r ≠ 0
.
.
.
=
The number r is called the common ratio of the sequence.
All this definition can do for us is tell us if a sequence is geometric or not. To determine if a sequence is
geometric, we take any two consecutive terms and find their quotient. If we obtain the same quotient for any
two consecutive terms, then we have found the common ratio and the sequence is geometric.
Sequences and Probability
Geometric Sequence – 407
Usually, however, we also like to be able to find some formula for a sequence that will allow us to find the nth
term of the geometric sequence.
The nth Term of a Geometric Sequence
The nth term of a geometric sequence has the form:
a1rn – 1
=
an
nth term of a geometric sequence
where r is the common ratio of consecutive terms of the sequence.
Thus, every geometric sequence can be written in the form:
a1, a1r, a1r2, a1r3, . . . , a1rn – 2, a1rn – 1, . . .
Note:
first n terms of a geometric sequence
The subscript of the a-term and the superscript of r must add up to n. Thus, the 5th term of a
geometric sequence can be written in the following ways:
a1r4
=
a2r3
a3r2
=
=
a4r
=
a5r0
=
a5
equivalent expressions for the
5th term of a geometric sequence
This is another example of recursive thinking, where the next term of the sequence is dependent on the
preceding term of the sequence.
Determine whether the sequence is geometric. If it is, find the common ratio and find the 9th term of the
sequence.
5, 1, 0.2, 0.04, . . .
solution
To determine if the sequence is geometric, we divide consecutive terms and see if we obtain a constant number
each time:
a2
a1
a3
a2
a4
a3
=
=
=
1
5
0.2
1
0.04
0.2
=
0.2
divide second term by first term
=
0.2
divide third term by second term
=
0.2
divide fourth term by third term
Since we obtain the same number each time we divide consecutive terms in this sequence, we conclude that it is
geometric. Furthermore, the common ratio is r = 0.2 = 1/5.
Once we have the common ratio, we can find the nth term of this geometric sequence, since we also know the
first term is 5. This means that the nth term of the sequence has the form:
an
=
5(0.2)n – 1
nth term of a geometric sequence with
a1 = 5 and r = 0.2
a25
=
5(0.2)9 – 1
substitute in n = 9
a25
=
5(0.2)8
simplify exponent
=
0.0000128
simplify
408 – Geometric Sequence
answer
Sequences and Probability
Sum of a Geometric Sequence
Geometric sequences work quite differently than arithmetic sequences, so we need to take extra care that we are
handling them correctly. For one thing, we can find a finite sum for a finite geometric sequence, but we can, in
some cases, find a finite sum for an infinite geometric sequence. It is very important that we know if the
geometric sequence we are working with is infinite or finite.
The Sum of a FINITE Geometric Sequence
The sum of the geometric sequence
a1, a1r, a1r2, a1r3, a1r4, . . . , a1rn – 1
with common ratio r ≠ 1 is given by
=
Sn
Note:
sum of a finite geometric sequence
This is called the nth partial sum of the geometric sequence.
Find the sum.
9
∑
2
k− 1
=1
k
solution
Rather than write out each term, it is simpler to apply the formula just given. Here we are raising the number 2
to the power of k – 1, where k is the index of summation. Thus, the first term of the geometric sequence is:
a1
=
21 – 1
=
1
first term of the geometric sequence
Then our formula becomes:
9
∑
k
Note:
k−1
2
=
⎛ 1 − rn ⎞
a 1⎜
⎟
⎝ 1−r ⎠
sum of a finite geometric sequence
n is the upper limit of summation
n=9
=
1⎜
⎛ 1 − 29 ⎞
⎟
⎝ 1−2 ⎠
substitute in r = 2, n = 9 and a1 = 1
=
511
simplify using a calculator
answer
=1
We have to be careful with our indices of summation. If we had k = 0, then we have to adjust the
formula for the nth partial sum. For instance, if our example had begun with k = 0, the sum would have
been:
9
∑
k
k−1
2
=
=0
Sequences and Probability
1
2
9
+
∑
k
k−1
2
=
511.5
=1
Geometric Sequence – 409
The formula for the nth partial sum of a finite geometric sequence can, depending on the value of r, produce a
formula for the sum of an infinite geometric sequence. More specifically, if | r | < 1, then it can be shown that rn
becomes arbitrarily close to zero as n increases without bound. In calculus terms, we are talking about limits.
If we have a common ratio r such that | r | < 1, then we can apply the following formula to find the sum of an
infinite geometric sequence:
The Sum of an INFINITE Geometric Sequence
If | r | < 1, the infinite geometric sequence
a1, a1r, a1r2, a1r3, a1r4, . . . , a1rn – 1, arn, . . .
has the sum
=
S
sum of an infinite geometric sequence
Notice that this formula is completely independent of the number of terms we have in our geometric sequence.
All we need to know is the first term, which is usually easy to figure out, and the common ratio, which again
can be determined without too much difficulty.
Find the sum.
∞
∑
4 ⎛⎜
=0
n
1⎞
n
⎟
⎝ 4⎠
solution
First of all notice that the index of summation starts at 0, not 1, so we have to make a slight adjustment to our
summation notation:
∞
∑
n
=0
1
4 ⎛⎜ ⎞⎟
⎝ 4⎠
∞
n
=
4+
∑
n
=1
4 ⎛⎜
1⎞
⎟
⎝ 4⎠
n
rewrite summation notation so that
lower index starts at n = 1
Also notice that | r | = 1/4 < 1, which means that this infinite geometric sequence does, in fact, have a finite sum.
Now we need to find a1, so we let n = 1:
a1
=
4 ⎛⎜
=
1
1⎞
1
⎟
⎝ 4⎠
410 – Geometric Sequence
let n = 1
simplify
Sequences and Probability
We are ready to apply the formula for the sum of an infinite geometric sequence now. We know that a1 = 1 and
that r = 1/4. Don't forget the extra "4" in front of our summation notation:
S
=
=
1
4+
1−
4+
4+
sum of this infinite geometric series
4
1
4
4
=
1
−
1
convert terms in denominator to LCD form
4
1
simplify denominator
3
4
=
=
S
=
4+
12
3
4
flip fraction in denominator upside down
3
+
16
3
Sequences and Probability
4
3
convert all terms to LCD form
simplify
answer
Geometric Sequence – 411
412 – Geometric Sequence
Sequences and Probability
Advanced
To learn a mathematical shortcut for adding many numbers together.
summation notation – a symbol that indicates that we are taking the sum of a group of numbers.
sigma notation – a Greek symbol indicating that we are taking the sum of a group of numbers, synonymous
with summation notation.
index of summation – variable indicating the position of a number in the summation.
upper limit of summation – the largest value of the index of summation.
lower limit of summation – the smallest value of the index of summation.
convergent (sequence) – infinite sequence whose sum yields a finite number.
series – the summation of the terms of a sequence (finite or infinite).
infinite series – a series that contains an infinite number of terms in the summation.
Aristarchus of Samos (c.320-c.230 B.C.), Greek astronomer and geometer, was among the first, if not the first, to
propose the Earth traveled around the sun. He also suggested the Earth rotates on its own axis. Both ideas
were ridiculed at the time because they contradicted visual evidence and Aristotelian thought. Aristarchus
attempted to utilize trigonometric methods to find the distance between the Earth and the sun, but he lacked the
right tools for the job. His estimate for the size of the sun is off by a factor of around 400.
Sequences and Probability → Sequences → Summation Notation
Whenever we have to add a lot of numbers together, especially the terms of a finite sequence, we use a special
notation to indicate that we are taking the sum of a group of numbers. It is called summation notation or
sigma notation because it involves the use of the Greek upper case letter sigma, written as .
Definition of Summation Notation
The sum of the first n terms of a sequence is represented by:
=
a1 + a2 + a3 + . . . + an
summation notation
where i is called the index of summation, n is the upper limit of summation,
and 1 (or sometimes 0) is the lower limit of summation.
Summation is read as: "The sum of the values of a sub i from i equals 1 to i equals n is…."
While we use the letter i for the index of summation in the definition, we can use any letter we want for the
index of summation. Other common letters for the index of summation include j and k. Also the lower limit of
summation does not have to be 1. Many times it is more convenient to start at 0. We can use any positive
integer we need for the lower limit of summation.
Sequences and Probability
Summation Notation – 413
Let's see how we can evaluate a given sum:
Find the sum.
5
a.)
∑
( 2⋅ i + 1)
=1
i
4
b.)
∑
k
=1
6
c.)
∑
j
10
=3
1
j
solution
a.)
To find the sum, we need to add 5 terms together. The first term is (2(1) + 1), the second term is
(2(2) + 1) and so on.
5
∑
( 2⋅ i + 1)
=
(2(1) + 1) + (2(2) + 1) + (2(3) + 1) + (2(4) + 1) + (2(5) + 1)
=1
i
add five terms together
b.)
=
3 + 5 + 7 + 9 + 11
simplify each term
=
35
simplify
answer
Note that the argument of the summation is not dependent on the index of summation. This means that
we have four terms, each of which is simply "10".
4
∑
k
c.)
10
=
10 + 10 + 10 + 10
add four tens together
=
40
simplify
answer
=1
The argument in this case is dependent on the index of summation. Also note that the index starts at
"3" instead of "1", so the first term of the summation is 1/3, the second term is 1/4, and so on:
6
∑
j
=3
1
j
=
=
414 – Summation Notation
1
3
+
19
20
1
4
+
1
5
+
1
6
add four terms together
simplify
answer
Sequences and Probability
Summations have some important properties that we can take advantage of. For example, if we have the sum of
two summations with identical indices of summation and identical upper limits of summation, then we can
combine them into a single summation expression.
Properties of Summation
1.)
=
2.)
=
3.)
=
c is any constant
Proving any of these properties is a matter of applying the Associative Property of Addition, the Commutative
Property of Addition, and the Distributive Property of Multiplication Over Addition. If we take a look at the
first property, we start with expanding the summation on the left so that we have the sum of all its terms. Then
we factor out the "c" since it is common to all the terms. Finally, we can reduce the terms inside parentheses
into summation notation.
n
∑
i
ca i
=
ca1 + ca2 + ca3 + . . . + can
expand summation
=
c(a1 + a2 + a3 + . . . + an)
factor out the "c" from each term
=
⎛ n
⎞
c⎜
ai ⎟
⎜
⎟
⎝i = 1 ⎠
collapse remaining terms into a summation
proof
=1
∑
Sometimes two sums can have two different summation notations, depending on their lower and upper indices
of notation. Consider the following three sums:
4
∑ 3(2 )
i
i
4
3⋅
=
=1
∑
i
i
=
3⋅
5
i
∑
i
∑ ( i−1)
=2
3(21 + 22 + 23 + 24)
=
90
3
=0
32
=
=1
3
∑ 3(2i+1)
i
2
i+ 1
2
=
3(21 + 22 + 23 + 24)
=
90
=
3(21 + 22 + 23 + 24)
=
90
=0
5
=
3⋅
∑
i
i− 1
2
=2
Even though each summation has different lower and upper limits, the summation still produces the exact same
result in all three cases.
So far we have been discussing the sums of finite sequences. The sum of a finite sequence must be, by
definition (and common sense), a finite number. One of the more amazing discoveries in mathematics is that
for some infinite sequences, the sum of an infinite number of terms is also a finite number. Infinite sequences
Sequences and Probability
Summation Notation – 415
whose sums yield a finite number are said to converge to that number. In the case of the sum of an infinite
sequence, the upper index of summation is always written as (infinity). For example, we can use calculus
to demonstrate that the sum of all the terms of an infinite sequence whose nth term is 1 / 2n (with n beginning
at 0) is:
∞
∑
i
Note:
=0
1
n
=
1+
=
2
2
1
2
+
1
4
+
1
8
+
1
16
+ ....
sum of a convergent infinite sequence
a finite number
The summation of the terms of a sequence (finite or infinite) is called a series. The summation
∞
∑
i
ai
=
a1 + a2 + a3 + a4 + . . .
infinite series
=1
is called an infinite series. Infinite series have important uses in calculus. You will study infinite
sequences and series in excruciating detail when you take calculus.
416 – Summation Notation
Sequences and Probability
Appendix
Appendix – 417
418 – Appendix
Advanced
Every single letter of the Roman alphabet (the one used in English) has a use in algebra. Here is a list of some
of the most common uses of letters in algebra:
Note:
These are the most common conventional meanings for these letters. A test question may want you to
solve a problem that has e(b) = t(q) + 2 just to confuse you. It is important to understand the context in
which each letter appears. The letter "m", for example, only means slope in the context of lines. If the
problem wants you to find out how much an object weighs, then m most likely indicates mass.
Any and all of these letters can be used to indicate either a constant term or a variable.
a
b
c
d
These four letters are often used to represent constant terms, such as a = 1, b = 2, etc…
d is also used in many problems to represent distance
e
Usually reserved for the special number e ≈ 2.7182818285
Used to define the natural logarithmic function
f
Most often used in function notation y = f(x)
g
If f is already being used for a function, then we use g to represent a second function
h
Used in function notation if f and g are already taken
Also denotes height in many types of problems
Used in conic sections to describe the x-component of translation of the conic
i
Reserved to represent the complex number √(–1)
Also used in vector notation to describe the x-component of a vector
Used with matrices to indicate a specific row of a matrix, as in [aij]
j
Used in electronics to represent the complex number √(–1)
Used in vector notation to describe the y-component of a vector
Sometimes used to represent a function if f, g, and h are already being used
Used with matrices to indicate a specific column of a matrix, as in [aij]
k
Used in conic sections to describe the y-component of translation of the conic
Used in vector notation to describe the z-component of the vector
l
Most often used to describe the length dimension, as in a rod of length l (or L)
In general, l is avoided because it looks too similar to the numeral one (1 or l, which is which?)
m
In science, used to describe the mass dimension, as in a rock of mass m
Also the slope of a line
Used with matrices to indicate the number of rows of matrix A
n
Used as a constant term along with a, b, c, and d
Often used as a subscript, such as xn, or a superscript, such as xn, where n is a positive integer
(or 0)
Used with matrices to indicate the number of columns of matrix A
o
Generally avoided as it creates too much confusion with 0
O (capitalized) is often used to mark the origin of a graph
p
P (capitalized) is used to mark a point on a graph
p (lower case) is used to represent the numerator of a fraction: p / q
In a parabola, p is the distance between the vertex and the focus, as well as the distance
between the vertex and the directrix
Appendix A
Algebra Alphabet – 419
q
Q (capitalized) is used to mark another point on a graph if P is already being used
q (lower case) is used to represent the denominator of a fraction: p / q
r
Often stands for the radius of a circle or other round object (cylinder, sphere, etc…)
s
Used to denote arc length in the formula s = r (arc length of a sector of a circle), where is
measured in radians
t
Most often used to denote time
Used in parametric equations as the parameter of the equation
u
Used as a generic variable, much like x
v
Used as a second generic variable if u is already taken, sort of like y
w
Used as a third generic variable if u and v are already taken, sort of like z
Also denotes width
x
The granddaddy of generic variables, often called the independent variable, can be used to
represent any quantity
Also denotes the horizontal distance and direction from the origin on a graph
y
The second most popular generic variable, usually dependent on x: y = f(x)
Also denotes the vertical distance and direction from the origin on a graph
z
If x and y are already in use, then z is a third variable that is usually dependent on both x and y
On a graph, the z-axis is perpendicular to both the x- and y-axes
In general, we use capital letters (A, B, C, etc) in geometry to denote points or vertices of polygons. We use
them in conjunction with matrices to represent a matrix A, B, C, etc. Elements within the matrix are denoted
with lower case letters (a11, b12, etc.). The matrix X is used to indicate a matrix whose elements are all
unknowns and represents the solution to a system of linear equations.
420 – Algebra Alphabet
Appendix A
Glossary
Algebra Glossary (Advanced) – 421
422 – Algebra Glossary (Advanced)
Advanced
–A–
absolute value – the absolute value of a number is its distance from 0 on a number line.
additive identity (matrices) – a matrix O such that A + O = A, provided the matrices A and O have the same
dimensions.
additive inverse – the opposite of a number. If the number a is negative, then the additive inverse of a is a
positive number.
adjoining (matrices) – creating a larger matrix out of two or more smaller matrices.
algebraic expression – a combining of numbers, variables and operations in a way that stands for a number.
arithmetic combination – the sum, difference, product, or quotient of two or more functions.
arithmetic sequence – a sequence whose consecutive terms all have a common difference separating them.
asymptote – a line approached by the graph of a function such that the graph never touches or crosses the line.
asymptotes (hyperbola) – the two lines that are approached by the points on the branches of a hyperbola as the
points get farther from the foci. The hyperbola never touches the asymptotes.
augmented matrix – a matrix derived from a system of linear equations, where the last column of the matrix
represents the constant terms of the system.
axis of symmetry – the line passing through both the focus and the vertex.
–B–
base – the repeated factor in a power. In a2, a is the base.
bell-shaped curve – the graph of a normal distribution.
binomial series – a binomial expansion which contains infinitely many terms.
bounded – any interval which has a finite upper and lower boundary.
–C–
center (circle) – the point in the plane from which all other points on the circle are defined.
circle – a plane curve consisting of all points at a given distance from a fixed point in the plane.
coefficient – the numerical factor of a monomial expression.
coefficient matrix – the matrix derived from the coefficients of the variables in a linear system. It does not
include any of the constant terms of the system.
columns – vertical lines of numbers or variables in a matrix.
common difference – the difference between any two consecutive terms in an arithmetic sequence is a
constant, represented by d.
common logarithmic function – logarithmic function that uses base 10.
common ratio – a ratio which is the same between all consecutive terms in a geometric sequence.
complement of an event – the collection of outcomes in the sample space that are not part of a given event A.
If a given event is denoted by A, then its complement is denoted by A'.
Algebra Glossary (Advanced) – 423
complex conjugate – the complex conjugate of a + bi is a – bi and vice versa. The product of complex
conjugates is a real number.
complex number – any number, real or imaginary, of the form a + bi, where a and b are real numbers and
i = √(–1).
composition of functions – the function that results from first applying one function, then another; denoted by
the symbol o.
conditional equation – an equation that is true for only a few members in the domain.
conic section
algebraic – plane curves represented by second-degree polynomial equations.
geometric – the planar shapes that result from intersecting a cone with a plane at various angles.
conjugate axis – a line perpendicular to the transverse axis and passing through the center of the hyperbola.
consistent – a system of linear equations that has at least one solution.
constant – a numerical symbol that never changes its values.
constant function – a function in which f(x) is simply equal to a real (or complex) number.
constant term – term in an expression that does not contain any variables.
convergent (sequence) – infinite sequence whose sum yields a finite number.
cube root – a cube root of a number c is the solution of the equation x3 = c.
–D–
denominator – lies below the line of division in a fraction. This is the divisor. If the numerator equals 1, then
the denominator determines in how many parts the unit is to be divided.
dependent variable – a variable whose value(s) always depend on the values of other variable(s).
determinant – associated with each square matrix is a real number called the determinant of the matrix. The
number of elements in any row or column of the square matrix is called the order of the determinant.
diagonal matrix – matrix that has nonzero elements on the main diagonal and zero elements everywhere else.
difference of focal radii – the absolute value of the difference of the distances from a point on a hyperbola to
the two foci of the hyperbola.
directrix – a line associated with a parabola such that the distance from it to any point on the parabola is equal
to the distance from that point to the focus.
distance – a measure of how far apart two things are from each other.
domain (of a function) – the set of values which are allowable substitutions for the independent variables.
–E–
eccentricity – a measure of the ovalness of an ellipse.
eccentricity (hyperbola) – a measure of the flatness or roundedness of the branches of a hyperbola.
elementary row operations – interchanging, multiplying, and adding rows of a matrix to find solutions for
each of the variables in the linear system represented by the matrix.
ellipse – the set of points P in a plane which satisfy PF1 + PF2 = d, where F1 and F2 (its foci) are any two fixed
points and d (its focal constant) is a constant with d > F1F2.
424 – Algebra Glossary (Advanced)
entry (matrix) – the object in a particular row and column of a matrix.
equivalent equations – equations having the same solution set over a given domain.
evaluate – to find the value of an expression for a given value.
expanding by cofactors – calculating the determinant of a matrix by taking the sum of the entries in any row
and multiplying by the cofactors associated with that row.
exponent – in a power, the number of times the base occurs as a factor. In the expression 23, 3 is the exponent
and tells us that 2 occurs as a factor three times.
exponential function – a function with the independent variable as an exponent. A function with an equation
of the form y = abx.
extracting square roots – the process of solving a quadratic equation by isolating the x2 term on one side,
constant terms on the other sides, and then finding the square root of both sides.
–F–
factorial – for a positive integer n, the product of all the positive integers less than or equal to n.
finite sequence – a sequence whose domain consists of the first n integers only.
focal radius – a line connecting a point and a focus.
foci (hyperbola) – the two points from which the difference of distances to a point on the hyperbola is constant.
focus (plural foci) – fixed points from which the sum of distance to a point on the conic is a constant.
focus (parabola) – the point along with the directrix from which a point on a parabola is equidistant.
function – an association of exactly one object from one set (the range) with each object from another set (the
domain). A relationship in which different ordered pairs have different first coordinates.
–G–
geometric sequence – a sequence with a constant ratio between successive terms.
greater than – one value is larger than a compared value.
greater than or equal to – one value is larger than or the same as a compared value.
–H–
horizontal asymptote – a horizontal line approached by the graph of a function such that the graph never
touches or crosses the horizontal line.
–I–
identity – an equation whose sides are equivalent expressions.
identity matrix – a matrix that consists of 1's on its main diagonal and 0's elsewhere.
imaginary number – a number that is the square root of a negative real number.
inclusive or – in the statement a or b, the answer may be a, it may be b, and the answer could also be both a
and b.
inconsistent – a system of linear equations that has no solution.
independent events – the occurrence of one event has no impact whatsoever on the occurrence of another
event.
Algebra Glossary (Advanced) – 425
independent variable – in a formula, a variable upon whose value other variables depend.
index (radical) – a real number placed above and to the left of a radical symbol to indicate what root is sought.
index of summation – variable indicating the position of a number in the summation.
infinite series – a series that contains an infinite number of terms in the summation.
integers – the set of whole numbers and their negative counterparts.
interval – any subset of real numbers.
irrational number – any number that cannot be expressed as a terminating or repeating decimal.
inverse function – the functions f and g are inverses of each other if f(g(x)) = x in the domain of g and
g(f(x)) = x for all x in the domain of f.
inverse properties – one property that can be used to undo another property, and vice versa.
invertible – a matrix with an inverse.
–J–
–K–
–L–
leading 1 – the first nonzero entry of a row that does not consist entirely of zeros.
less than – one value is smaller than a compared value.
less than or equal to – one value is smaller than or the same as a compared value.
like terms – terms that have exactly the same variables to the same powers.
linear equation – an equation in which each term is either a constant or a monomial of degree 1.
linear function – a polynomial function of the first degree.
logistics curve – a growth curve described by the equation below:
a
y
− ( x− c)
=
1 + be
d
lower limit of summation – the smallest value of the index of summation.
lower triangular matrix – matrix containing nonzero elements on and below the main diagonal and zero
elements above the main diagonal.
–M–
main diagonal entries – matrix entries that form a diagonal line from the top left corner to the lower right.
major axis – the line segment joining the two foci of an ellipse and which has endpoints at the vertices of the
ellipse.
matrix – a rectangular array of elements used to facilitate the study of problems in which the relation between
these elements is fundamental.
426 – Algebra Glossary (Advanced)
minor axis – line segment perpendicular to the major axis, passing through the center of the ellipse.
multiplicative inverse – the reciprocal of a number.
multiplicity of a root r – for a root r of a polynomial equation P(x) = 0, the highest power of x – r that appears
as a factor of P(x).
mutually exclusive – sets with no common elements.
–N–
natural logarithm – logarithms using base e.
natural numbers – all positive integers.
negative real zeros – negative real x-values for which the function equals zero.
nonsingular – a matrix with an inverse. This is synonymous with invertible.
normal distribution – the continuous function that the binomial distribution approaches as n, the number of
trials, increases without bound.
nth partial sum – the sum of the first n terms of a sequence.
numerator – signifies the number of parts of the denominator that is taken. In a fraction, the numerator lies
above the line representing division.
–O–
one-to-one correspondence – every point on the number line corresponds to exactly one real number.
origin – the point 0 on the real number line.
–P–
parabola – the set of all points in the plane of a line l and a point F not on l whose distance from F equals its
distance from l.
parallel lines – two lines are parallel if there is a plane in which they both lie and they do not cross at any point
within the plane.
partial fraction decomposition – the method by which we transform one rational function into the sum of two
or more simpler rational functions.
perpendicular lines – two straight lines in a plane, the intersection of which forms right angles.
point-slope form – one form of a linear equation derived from knowing one point on the line and its slope.
positive real zeros – positive real x-values for which the function equals zero.
power – a product of equal factors. The repeated factor is the base. A positive exponent tells the number of
times the base occurs as a factor.
principle nth root of a – the nth root that has the same sign as a.
–Q–
Quadratic Formula – an important formula that can be used to solve any quadratic equation for the independent
variable.
quadratic function – a polynomial function of the second degree.
Algebra Glossary (Advanced) – 427
–R–
radical symbol – an operator symbol that indicates we are taking the nth root of a real number, variable, or
algebraic expression.
radicand – the quantity under a radical symbol.
radius – the distance between the center and a point on the circle.
range – the set of values of the function evaluated at points contained within the domain.
rational function – a function defined by a simplified rational expression.
rational number – any number that can be expressed as a terminating or repeating decimal.
real number – a rational or an irrational number, which can be represented by a finite or infinite decimal.
real number line – a line with numbers evenly spaced throughout the line.
recursive (sequence) – each term of a sequence depends on the value of the previous term of the sequence.
reduced row-echelon form – an augmented matrix that has 1's along the main diagonal of the coefficient
matrix and zeros in every entry above and below the main diagonal of the coefficient matrix.
repeated roots (or zeros) – a solution to a polynomial equation that occurs more than once.
row-echelon form (linear system) – a linear system in a stair-step form with the leading coefficients of each
line being 1.
row-echelon form (matrices) – a matrix that has a stair-step pattern with leading coefficients of 1 along the
main diagonal and zeros below the main diagonal.
row-equivalent – two matrices are said to be row-equivalent when one matrix can be obtained from the other
matrix by applying elementary row operations.
rows – horizontal lines of numbers and variables in a matrix.
–S–
scalar – a real number.
scalar multiplication – the product of a real number and a matrix.
second-degree polynomial equation – a polynomial which takes the form ax2 + bx + c = 0, where a ≠ 0.
semi-major axis – line segment that is half the length of the major axis. Its endpoints are the center of the
ellipse and one of the endpoints.
semi-minor axis – a line segment half the length of the minor axis, perpendicular to the semi-major axis.
series – the summation of the terms of a sequence (finite or infinite).
sigma notation – a Greek symbol (Σ) indicating that we are taking the sum of a group of numbers, synonymous
with summation notation.
sigmoidal curve – another name for a logistics growth curve.
singular – a matrix without an inverse.
slant asymptote – an asymptote to a rational function that occurs when the degree of the numerator is greater
than the degree of the denominator.
solution – a replacement of the variable(s) in an algebraic sentence that makes the sentence true.
428 – Algebra Glossary (Advanced)
solve – to find the result by the use of certain given data, previously known facts or methods, and newly
observed relations.
square root – a square root of a number b is the solution of the equation x2 = b.
summation notation – a symbol that indicates that we are taking the sum of a group of numbers.
–T–
term – a constant number of a variable or the product of numbers and variables.
term (sequence) – an element in the range of a sequence.
transverse axis – the line passing through both the vertices and the foci of a hyperbola.
–U–
unbounded – an interval that has infinity as one or both of its endpoints.
upper limit of summation – the largest value of the index of summation.
upper triangular matrix – matrix containing nonzero elements on and above the main diagonal and zero
elements below the main diagonal.
–V–
variable – a symbol which can represent any one of a set of numbers or other objects (including other algebraic
expressions).
variable term – term in an expression that contains variables.
variation in sign – two consecutive coefficients in a polynomial function having opposite signs.
vertical asymptote – a vertical line approached by the graph of a function such that the graph never touches or
crosses the vertical line.
vertices (ellipse) – endpoints of the major axis of an ellipse.
vertex – the intersection of a parabola with its axis of symmetry.
–W–
whole numbers – all positive integers and zero.
–X–
–Y–
–Z–
zero matrix – a matrix in which all the elements are zero.
Algebra Glossary (Advanced) – 429
430 – Algebra Glossary (Advanced)
Works Consulted
Dolciani, Mary P., et al. Algebra and Trigonometry: Structure and Method, Book 2. Boston: Houghton
Mifflin Company, 1988.
Fair, Jan, and Sadie C. Bragg. Algebra 1. Englewood Cliffs, NJ: Prentice Hall, 1990.
Ifrah, George. The Universal History of Numbers. New York: John Wiley & Sons, Inc., 1991.
Larson, Roland E., Robert P. Hostetler, and Bruce H. Edwards. Calculus with Analytic Geometry, Fourth
Edition. Lexington, MA: D.C. Heath and Company, 1990.
Larson, Roland E., and Robert P. Hostetler. Algebra and Trigonometry, Fourth Edition. Boston: Houghton
Mifflin Company, 1997.
Larson, Ron, and Robert P. Hostetler. Algebra and Trigonometry, Fifth Edition. Boston: Houghton Mifflin
Company, 2001.
McConnell, John W., et al. The University of Chicago School Mathematics Project: Algebra, Second Edition.
Glenview, IL: Scott, Foresman, Addison, Wesley, 1998.
Pappas, Theoni. The Magic of Mathematics. San Carlos, CA: Wide World Publishing/Tetra, 1994.
Senk, Sharon L, et al. The University of Chicago School Mathematics Project: Advanced Algebra, Second
Edition. Glenview, IL: Scott, Foresman, Addison, Wesley, 1998.
Stewart, James. Calculus, Fourth Edition. Pacific Grove, CA: Brooks/Cole Publishing Company, 1999.
Young, Robyn V, ed. Notable Mathematicians: From Ancient Times to the Present. Detroit: Gale Research,
1998.
Works Consulted – 431
432 – Works Consulted
Index
Index – 433
Index
Page numbers in boldfaced type indicate a page containing a formula or definition.
–A–
absolute value
distance between points
properties of
absolute value inequality
solving an
algebraic expression
asymptote(s)
horizontal
of a rational function
slant
vertical
33
34
33
63
63
5
151
152
153
158
152
– D – (cont'd)
two-point form of equation of
line
cofactors of
minors of
properties of
of a square matrix
Distance Formula
distinguishable permutation
domain of a function
367
342
342
347 – 349,
350
344
22
381
99, 103
–E–
–B–
basic rules of algebra
negation, properties of
binomial theorem
7, 8
8
383, 384
–C–
Cartesian plane
Distance Formula
Midpoint Formula
change-of-base formula
circle
translation of
cofactors of determinants
combination(s)
completing the square
complex conjugate(s)
complex number(s)
addition of
conjugate of
division of
equality of
multiplication of
subtraction of
composition of functions
conic section(s)
circle
translation of
eccentricity
of an ellipse
of a hyperbola
ellipse
eccentricity of
translation of
hyperbola
asymptotes of
eccentricity of
translation of
parabola
translation of
translation of conic sections, general
counting principle
Cramer's Rule
–D–
Descartes's Rule of Signs
determinant(s)
applications of
area of a triangle
Cramer's Rule
test for collinear points
434 – Index
21
22
24
195
263, 264, 265
283
342
377
257, 258
146
143, 144
145
146
146
144
146
146
227
263, 264, 265
283
273
273
278
271, 272
273
289
275, 276
276, 294
278
293
267, 268
285
281, 282
375, 376
359 – 361,
362
133
341, 342
353
355, 356
359 – 361,
362
365, 366
eccentricity
of an ellipse
of a hyperbola
elementary row operations
ellipse
eccentricity of
translation of
equality
properties of
equation(s)
exponential
of a line
in general form
in one variable
in point-slope form
in slope-intercept form
in two-point form
logarithmic
quadratic
completing the square
extracting square roots
factoring
Quadratic Formula
solutions of
exponent(s)
properties of
scientific notation
exponential decay model
exponential equation
exponential function(s)
inverse properties of
exponential growth model
Extended Distributive Property
extracting square roots
273
273
278
319
271, 272
273
289
9
203, 204
59
42
55
53, 54
56
203, 204
249
257, 258
255, 256
253
259, 260
260
11, 12
12
210
203, 204
187
203
209
114
255, 256
–F–
factorial(s)
Factor Theorem
factoring a quadratic equation
fraction(s)
partial
properties of
function(s)
combinations, arithmetic
composition of
domain of
horizontal shift of
implied domain of
inverse
finding
horizontal line test
polynomial
955
125
672
15
161, 164
15, 16
99
225, 226
227
99, 103
232
103
235, 236
239
243, 244
105
– F – (cont'd)
addition of
Descartes's Rule of Signs
division of
factoring patterns, special
Intermediate Value Theorem
multiplication of
product patterns, special
Rational Zero Test
real zeros of
subtraction of
range of
rational
asymptote of
graphing
reflections in coordinate axes
transcendental
exponential
inverse properties of
logarithmic
graphs of
natural base e
vertical line test
vertical shift of
Fundamental Theorem of Algebra
linear factorization theorem
–G–
Gaussian elimination
Gaussian model
Gauss-Jordan elimination
graphing rational functions
111
133
121, 122
117
131
113
115
137
129
111
99, 103
149
153
155
229, 230
185
187
203
191
192 – 193
199 – 201
101
231, 232
107
107
323
213
327
155
–H–
horizontal
horizontal
horizontal
horizontal
asymptote
line (in a plane)
line test
shift of a function
–I–
identity matrix
independent events, probability of
index of summation
inequality (inequalities)
absolute value
solving an
linear
combined (double)
double (combined)
polynomial
finding test intervals
properties of
rational
Intermediate Value Theorem
inverse function(s)
finding
horizontal line test
152
47, 59
243, 244
232
303
393
413
63
63
61
62
62
65
66
17, 18
69
131
235, 236
239
243, 244
–J–
–K–
–L–
line(s) in the plane
horizontal
parallel
perpendicular
slope of
vertical
linear equation
in one variable
45
47, 59
163
163
45
152
41
42
– L – (cont'd)
generating equivalent equations
fractional expressions
linear inequality
combined (double)
double (combined)
logarithm(s)
change-of-base formula
properties of
logarithmic equation
logarithmic
graphs of
logarithmic model
logistics growth model
lower limit of summation
–M–
matrix (matrices)
addition of
properties of
elementary row operations
Gaussian elimination
Gauss-Jordan elimination
identity
inverse of
finding matrix inverse
2 x 2 matrix properties of
linear systems in
multiplication of
properties of
properties of matrix operations
reduced row-echelon form
representation of
row-echelon form
scalar multiplication of
properties of
triangular matrix
method of elimination
Midpoint Formula
minors of a determinant
–N–
natural base e
42
43
61
62
62
191
195
192, 197,
200
203, 204
191
192 – 193
215
219
413
299
305
313
319
323
327
303
329
333, 334
337
317
309
315
313, 315
321
301
321
307
313
351
83, 84
24
342
199 – 201
–O–
–P–
parallel lines
partial fractions
linear factors, distinct
linear factors, repeated
mixed factors
quadratic factors, distinct
quadratic factors, repeated
solving basic equation
Pascal's Triangle
permutation(s)
distinguishable
perpendicular lines
point-slope form of the equation of a
line
polynomial function(s)
addition of
Descartes's Rule of Signs
division of
factoring patterns, special
Intermediate Value Theorem
multiplication of
product patterns, special
Rational Zero Test
real zeros of
51
161, 164
165
173
181
169
177
164
387
379, 380
381
51
55
105
111
133
121, 122
117
131
113
115
137
129
Index – 435
– P – (cont'd)
subtraction of
polynomial inequality
finding test intervals
probability
of a complement
of an event
of independent events
of the union of two events
–Q–
quadratic(s)
quadratic equation(s)
completing the square
extracting square roots
factoring
Quadratic Formula
solutions of
Quadratic Formula
–R–
radical(s)
nth root of a number
principle nth root
properties of
rational function(s)
asymptote of
graphing
rational inequality
Rational Zero Test
real number(s)
ordering of
bounded intervals
unbounded intervals
Remainder Theorem
111
65
66
389, 390
397, 398
390
393
395, 396
247
249
257,
255,
253
259,
260
259,
436 – Index
transcendental function(s)
exponential
inverse properties of
logarithmic
graphs of
natural base e
triangular matrix
two-point form of the equation of a line
–U–
upper limit of summation
258
256
260
260
–V–
vertical asymptote
vertical line
185
187
203
191
192 – 193
199 – 201
351
56
413
152
48, 59
–w–
–x–
27
27
27
29
149
153
155
69
137
31
35
36
37
127
–S–
sequence(s)
arithmetic
nth term of
sum of
geometric
finite sum of
infinite sum of
nth term of
slant asymptote
slope of a line in the plane
slope-intercept form of the equation
of a line
substitution method
summation
index of
lower limit of
notation
properties of
upper limit of
synthetic division
Factor Theorem
Remainder Theorem
systems
graphical approach
graphical interpretation of
solutions
of inequalities
sketching the graph
of linear equations
method of elimination
multivariable
Gaussian elimination
substitution method
–T–
399
403
404
405
407
409
410
408
158
45
53, 54
73, 74
413
413
413
413
415
413
123
125
127
77
79, 80
93
94
71
83,
87
88
73, 74
–Y–
–Z–
zero
properties of
Zero Factor Property
19
19
19