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Applications Equations Graphs
McDougal Littell
Larson Boswell Kanold Stiff
Copyright 2004
Chapter 1
Chapter 3
Chapter 5
Chapter 7
Chapter 9
Chapter 11
Chapter 2
Chapter 4
Chapter 6
Chapter 8
Chapter 10
Chapter 12
Chapter 1
Connections to Algebra
1.1 Variables in Algebra
1.2 Exponents and Powers
1.3 Order of Operations
1.4 Equations and Inequalities
1.5 Problem Solving plane using Models
1.6 Tables and Graphs
1.7 And Introduction to Functions
Chapter 1.1
Chapter 1.2
Expressions Containing Exponents
An expression like 4 6 is called a power.
The exponent 6 represents the number of times the base 4 is used as a factor.
base
exponent
46 = 4 • 4 • 4 • 4 • 4 • 4
power
6 factors of 4
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
10 1
WORDS
MEANING
ten to the first power
10
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
10 1
42
WORDS
MEANING
ten to the first power
10
four to the second power or four squared
4•4
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
WORDS
MEANING
ten to the first power
10
42
four to the second power or four squared
4•4
53
five to the third power, or five cubed
5•5•5
10 1
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
WORDS
MEANING
ten to the first power
10
42
four to the second power or four squared
4•4
53
five to the third power, or five cubed
5•5•5
76
seven to the sixth power
7•7•7•7•7•7
10 1
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
WORDS
MEANING
ten to the first power
10
42
four to the second power or four squared
4•4
53
five to the third power, or five cubed
5•5•5
76
seven to the sixth power
7•7•7•7•7•7
xn
x to the nth power
x•x•x•x…x
10 1
and so on
Reading and Writing Powers
Express the meaning of the power in words and then with numbers or variables.
SOLUTION
EXPONENTIAL
FORM
WORDS
MEANING
ten to the first power
10
42
four to the second power or four squared
4•4
53
five to the third power, or five cubed
5•5•5
76
seven to the sixth power
7•7•7•7•7•7
xn
x to the nth power
x•x•x•x…x
10 1
and so on
For a number raised to the first power, you usually do not write the exponent 1.
For instance, you write 5 1 simply as 5.
Evaluating Powers
Evaluate the expression x 3 when x = 5.
SOLUTION
x3 = 53
Substitute 5 for x.
=5•5•5
Write factors.
= 125
Multiply.
The value of the expression is 125.
Evaluating an Exponential Expression
GROUPING SYMBOLS For problems that have more than one operation, it is
important to know which operations to do first.
Grouping symbols, such as parentheses ( ) or brackets [ ], indicate the order
in which the operations should be performed.
Operations within the innermost set of grouping symbols are done first. For
instance, the value of the expression (3 • 4) + 7 is not the same as the value of
the expression 3 • (4 + 7).
Evaluating an Exponential Expression
GROUPING SYMBOLS For problems that have more than one operation, it is
important to know which operations to do first.
Grouping symbols, such as parentheses ( ) or brackets [ ], indicate the order
in which the operations should be performed.
Operations within the innermost set of grouping symbols are done first. For
instance, the value of the expression (3 • 4) + 7 is not the same as the value of
the expression 3 • (4 + 7).
Multiply. Then add.
Add. Then multiply.
(3 • 4) + 7 = 12 + 7 = 19
3 • (4 + 7) = 3 • 11 = 33
Evaluating an Exponential Expression
Evaluate the expression when a = 1 and b = 2.
(a + b) 2
SOLUTION
(a + b) 2 = (1 + 2) 2
Substitute 1 for a and 2 for b.
= 32
Add within parentheses.
=3•3
Write factors.
=9
Multiply.
Evaluating an Exponential Expression
Evaluate the expression when a = 1 and b = 2.
(a 2 ) + (b 2 )
SOLUTION
(a 2 ) + (b 2 ) = (1 2 ) + (2 2 )
Substitute 1 for a and 2 for b.
=1+4
Evaluate power.
=5
Add.
Evaluating an Exponential Expression
Evaluate the expression when a = 1 and b = 2.
(a 2 ) + (b 2 )
SOLUTION
(a 2 ) + (b 2 ) = (1 2 ) + (2 2 )
Substitute 1 for a and 2 for b.
=1+4
Evaluate power.
=5
Add.
An exponent applies only to the number, variable, or expression immediately
to its left.
In the expression 2x 3, the base is x, not 2x.
In the expression (2x) 3, the base is 2x, as indicated by the parentheses.
Exponents and Grouping Symbols
Evaluate the expression when x = 4.
2x 3
SOLUTION
2x 3 = 2(4 3)
Substitute 4 for x.
= 2(64)
Evaluate power.
= 128
Multiply.
Exponents and Grouping Symbols
Evaluate the expression when x = 4.
2x 3
(2x) 3
SOLUTION
2x 3 = 2(4 3)
SOLUTION
Substitute 4 for x.
(2x) 3 = (2 • 4) 3
Substitute 4 for x.
= 2(64)
Evaluate power.
= (8) 3
Multiply within parentheses.
= 128
Multiply.
= 512
Evaluate power.
Real-Life Applications of Exponents
Exponents often are used in the formulas for area and volume. In fact, the
words squared and cubed come from the formula for the area of a square,
A = s 2, and the formula for the volume of a cube, V = s 3.
Area of Square: A = s 2
Units of area, such as square feet, ft 2, can be written using a second power.
Real-Life Applications of Exponents
Exponents often are used in the formulas for area and volume. In fact, the
words squared and cubed come from the formula for the area of a square,
A = s 2, and the formula for the volume of a cube, V = s 3.
Area of Square: A = s 2
Volume of Cube: V = s 3
Units of area, such as square feet, ft 2, can be written using a second power.
Units of volume, such as cubic centimeters, cm 3, can be written using a third power.
Making a Table
You can find the volume of cubes that have edge lengths of 1 inch, 2 inches,
3 inches, 4 inches, and 5 inches by using the formula V = s 3.
Edge, s
1
2
3
4
5
s3
13
23
33
43
53
Volume, V
1 in. 3
8 in. 3
27 in. 3
64 in. 3
125 in. 3
Chapter 1.3
Using the Order of Operations
Mathematicians have established an order of operations to evaluate an
expression involving more than one operation.
Start with operations within grouping symbols.
Then evaluate powers.
Then do multiplications and divisions from left to right.
Finally, do additions and subtractions from left to right.
Evaluating Without Grouping Symbols
Evaluate the expression when x = 4.
3x 2 + 1
SOLUTION
3x 2 + 1 = 3 • 4 2 + 1
Substitute 4 for x.
= 3 • 16 + 1
Evaluate power.
= 48 + 1
Evaluate product.
= 49
Evaluate sum.
Evaluating Without Grouping Symbols
Evaluate the expression when x = 4.
32  x 2 – 1
SOLUTION
32  x 2 – 1 = 32  4 2 – 1
Substitute 4 for x.
= 32  16 – 1
Evaluate power.
=2–1
Evaluate quotient.
=1
Evaluate difference.
Using the Order of Operations
When you want to change the established order of operations for an
expression, you must use parentheses or other grouping symbols.
ACTIVITY
Developing
Concepts
INVESTIGATING GROUPING SYMBOLS
Without grouping symbols, the expression 3 • 4 2 + 8  4 has a value of 50.
You can insert grouping symbols to produce a different value. For example:
3 • (4 2 + 8)  4 = 3 • (16 + 8)  4 = 3 • 24  4 = 72  4 = 18
Insert grouping symbols in the expression 3 • 4 2 + 8  4 to produce the
indicated values.
146
54
38
Using the Left-to-Right Rule
THE LEFT-TO-RIGHT RULE Operations that have the same priority, such as
multiplication and division or addition and subtraction, are performed using the
left-to-right rule, as shown below.
24 – 8 – 6 = (24 – 8) – 6
= 16 – 6
= 10
Work from left to right.
Using the Left-to-Right Rule
THE LEFT-TO-RIGHT RULE Operations that have the same priority, such as
multiplication and division or addition and subtraction, are performed using the
left-to-right rule, as shown below.
24 – 8 – 6 = (24 – 8) – 6
Work from left to right.
= 16 – 6
= 10
15 • 2  6 = (15 • 2)  6
= 30  6
=5
Work from left to right.
Using the Order of Operations
ORDER OF OPERATIONS
First do operations that occur within grouping symbols.
Using the Order of Operations
ORDER OF OPERATIONS
First do operations that occur within grouping symbols.
Then evaluate powers.
Using the Order of Operations
ORDER OF OPERATIONS
First do operations that occur within grouping symbols.
Then evaluate powers.
Then do multiplications and divisions from left to right.
Using the Order of Operations
ORDER OF OPERATIONS
First do operations that occur within grouping symbols.
Then evaluate powers.
Then do multiplications and divisions from left to right.
Finally, do additions and subtractions from left to right.
Using a Fraction Bar
1+2
A fraction bar can act as a grouping symbol: (1 + 2)  (4 – 1) =
4–1
7•4
7•4
=
2
8+7 –1
8 + 49 – 1
Evaluate power.
=
28
8 + 49 – 1
Simplify the numerator.
=
28
57 – 1
Work from left to right.
=
28
56
Subtract.
=
1
2
Simplify.
Evaluating Expressions with a Calculator
Many calculators use the established order of operations, but some do not.
When you enter the following in your calculator, does the calculator display
6.1 or 0.6?
10.5 – 6.3  2.1 –
1.4 Enter
SOLUTION
If your calculator uses order of operations, it will display
10.5 – 6.3 ÷ 2.1 – 1.4 = 10.5 – (6.3 ÷ 2.1) – 1.4
= 10.5 – (3) – 1.4
= 6.1
6.1 .
Evaluating Expressions with a Calculator
Many calculators use the established order of operations, but some do not.
When you enter the following in your calculator, does the calculator display
6.1 or 0.6?
10.5 – 6.3  2.1 –
1.4 Enter
SOLUTION
If your calculator uses order of operations, it will display
6.1 .
10.5 – 6.3 ÷ 2.1 – 1.4 = 10.5 – (6.3 ÷ 2.1) – 1.4
= 10.5 – (3) – 1.4
= 6.1
SOLUTION
If your calculator displays
as they are entered.
0.6 , it performs the operations
[(10.5 – 6.3) ÷ 2.1] – 1.4 = ( 4.2 ÷ 2.1) – 1.4
= 2 – 1.4
= 0.6
Chapter 1.4
Chapter 1.5
TRANSLATING VERBAL PHRASES
Verbal Phrase
Expression
The sum of six and a number
6+x
Eight more than a number
y+8
A number plus five
n+5
A number increased
increased by seven
x+7
A number decreased by nine
n–9
Ten times a number
Seven divided by a number
10 • n or (10n)
10n)
(10
(10n
7
x
USING A VERBAL MODEL
In Mathematics there is a difference between a phrase
and a sentence. Phrases translate into expressions;
sentences translate into equations or inequalities.
Phrases
Expressions
Sentences
Equations or Inequalities
USING A VERBAL MODEL
Phrase
Expression
The sum of six and a number
The sum of six and a number is
Sentence
6+x
6+x=
In this sentence, “is” says that
one quantity is equal to one another. Equation
The sum of six and a number
number is
is twelve.
twelve.
Sentence
In this sentence, the words
“is less than” indicate an inequality.
The sum of six and a number
is less than
thantwelve.
twelve.
6 + x = 12
Inequality
6 + x < 12
USING A VERBAL MODEL
Writing algebraic expressions, equations, or inequalities
that represent real-life situations is called modeling.
The expression, equation, or inequality is a
mathematical model.
Use three steps to write a mathematical model.
WRITE A
ASSIGN
WRITE AN
VERBAL MODEL.
LABELS.
ALGEBRAIC MODEL.
Writing an Algebraic Model
You and three friends are having a dim sum lunch at a Chinese
restaurant that charges $2 per plate. You order lots of plates.
The waiter gives you a bill for $25.20, which includes tax of
$1.20. Use mental math to solve the equation for how many
plates your group ordered.
SOLUTION
Understand the problem situation
before you begin. For example,
notice that tax is added after the total
cost of the dim sum plates is figured.
Writing an Algebraic Model
VERBAL
MODEL
Cost per
plate
LABELS
Cost per plate = 2
(dollars)
Number of plates = p
(plates)
•
Number of
plates
=
Bill
–
Tax
Amount of bill = 25.20 (dollars)
Tax = 1.20
(dollars)
ALGEBRAIC
MODEL
2
p = 25.20
–
1.20
2p = 24.00
p = 12
Your group ordered 12 plates of food costing $24.00.
Writing an Algebraic Model
A PROBLEM SOLVING PLAN USING MODELS
VERBAL
MODEL
Ask yourself what you need to know to solve the
problem. Then write a verbal model that will give
you what you need to know.
LABELS
Assign labels to each part of your verbal problem.
ALGEBRAIC Use the labels to write an algebraic model based on
MODEL
your verbal model.
SOLVE
Solve the algebraic model and answer the original
question.
CHECK
Check that your answer is reasonable.
Using a Verbal Model
JET PILOT A jet pilot is flying from Los Angeles, CA to Chicago, IL at a
speed of 500 miles per hour. When the plane is 600 miles from Chicago,
an air traffic controller tells the pilot that it will be 2 hours before the
plane can get clearance to land. The pilot knows the speed of the jet must
be greater then 322 miles per hour or the jet could stall.
a. At what speed would the jet have to fly to arrive in Chicago in 2 hours?
b.
Is it reasonable for the pilot to fly
directly to Chicago at the reduced
speed from part (a) or must the
pilot take some other action?
Using a Verbal Model
a. At what speed would the jet have to fly to arrive in Chicago in 2 hours?
SOLUTION
You can use the formula (rate)(time) = (distance) to write a verbal model.
VERBAL
MODEL
Speed of
jet
LABELS
Speed of jet = x
(miles per hour)
Time = 2
(hours)
Distance to travel = 600
(miles)
ALGEBRAIC
MODEL
•
Time
=
Distance to
travel
2 x = 600
x = 300
To arrive in 2 hours, the pilot would have to slow the jet down to 300 miles per hour.
Using a Verbal Model
b. Is it reasonable for the pilot to fly directly to Chicago at 300
miles per hour or must the pilot take some other action?
It is not reasonable for the pilot to
fly at 300 miles per hour, because
the jet could stall. The pilot should
take some other action, such as
circling in a holding pattern, to use
some of the time.
Chapter 1.6
Chapter 1.7
Chapter 2
Properties of Real Numbers
2.1 Real Number Line
2.2 Addition of Real Numbers
2.3 Subtraction of Real Numbers
2.4 Adding and Subtracting Matrices
2.5 Multiplication of Real Numbers
2.6 Distributive Property
2.7 Division of Real Numbers
2.8 Probability and Odds
Chapter 2.1
Chapter 2.2
Modeling Addition of Integers
To model addition problems involving positive and negative integers, you can
use tiles labeled + and – . Each + represents positive 1, and each –
represents negative 1. Combining a + with a – gives 0. You can use
algebra tiles to find the sum of –8 and 3.
1
2
3
Model negative 8 and
positive 3 using algebra tiles.
–
Group pairs of positive and
negative tiles. Count the
remaining tiles.
–
–
–
+
+
+
The remaining tiles show the
sum of –8 and 3.
–
–
–
–
–
–
–
0–8
–
+
3
–
–
–
–
Each pair has a
sum of 0.
–
+
–
–
–
–8 + 3 = –5
–
+
Adding Real Numbers
Addition can be modeled with movements on a number line.
You add a positive number by moving to the right.
Model –2 + 5.
Start at –2.
–4
–3
Move 5 units to the right.
•
–2
–1
0
1
The sum can be written as –2 + 5 = 3.
2
End at 3.
•
3
4
Adding Real Numbers
Addition can be modeled with movements on a number line.
You add a positive number by moving to the right.
Model –2 + 5.
Start at –2.
–4
–3
Move 5 units to the right.
•
–2
–1
0
1
2
End at 3.
•
3
4
The sum can be written as –2 + 5 = 3.
You add a negative number by moving to the left.
Model 2 + (–6).
Move 6 units to the left.
End at – 4.
–5
•
–4
–3
–2
–1
0
Start at 2.
1
The sum can be written as 2 + (–6) = – 4.
•
2
3
Adding Real Numbers
The rules of addition show how to add two real numbers without a number line.
RULES OF ADDITION
TO ADD TWO NUMBERS WITH THE SAME SIGN:
1
Add their absolute values.
2
Attach the common sign.
Example:
– 4 + (–5)
– 4 + –5 = 9
–9
Adding Real Numbers
The rules of addition show how to add two real numbers without a number line.
RULES OF ADDITION
TO ADD TWO NUMBERS WITH THE SAME SIGN:
1
Add their absolute values.
2
Attach the common sign.
Example:
– 4 + (–5)
– 4 + –5 = 9
–9
TO ADD TWO NUMBERS WITH OPPOSITE SIGNS:
1
Subtract the smaller absolute value from the larger absolute value.
2
Attach the sign of the number with the larger absolute value.
Example:
3 + (–9)
–9 – 3 = 6
–6
Adding Real Numbers
The rules of addition on the previous slide will help you find sums of positive
and negative numbers. It can be shown that these rules are a consequence of
the following Properties of Addition.
PROPERTIES OF ADDITION
COMMUTATIVE PROPERTY
The order in which two numbers are added does not change the sum.
a+b=b+a
Example:
3 + (–2) = –2 + 3
ASSOCIATIVE PROPERTY
The way you group three numbers when adding does not change the sum.
(a + b) + c = a + (b + c)
Example:
(–5 + 6) + 2 = –5 + (6 + 2)
Adding Real Numbers
The rules of addition on the previous slide will help you find sums of positive
and negative numbers. It can be shown that these rules are a consequence of
the following Properties of Addition.
PROPERTIES OF ADDITION
IDENTITY PROPERTY
The sum of a number and 0 is the number.
a+0=a
Example:
–4 + 0 = –4
PROPERTY OF ZERO (INVERSE PROPERTY)
The sum of a number and its opposite is 0.
a + (–a) = 0
Example:
5 + (–5) = 0
Adding Three Real Numbers
Use a number line to find the following sum.
–3 + 5 + (– 6)
SOLUTION
Start at –3. Move 5 units to the right.
–5
• •
–4
End at – 4.
–3
–2
–1
0
1
Move 6 units to the left.
The sum can be written as –3 + 5 + (–6) = – 4.
–3 + 5 = 2
•
2
3
Finding a Sum
Find the following sums.
1.4 + (–2.6) + 3.1 = 1.4 + (–2.6 + 3.1)
= 1.4 + 0.5
= 1.9
Use associative property.
Simplify.
Finding a Sum
Find the following sums.
1.4 + (–2.6) + 3.1 = 1.4 + (–2.6 + 3.1)
= 1.4 + 0.5
Use associative property.
Simplify.
= 1.9
–
1
1
1
1
+3+
= –
+ +3
2
2
2
2
=
(– 12 + 12 ) + 3
= 0+3=3
Use commutative property.
Use associative property.
Use identity property and property of zero.
Using Addition in Real Life
SCIENCE CONNECTION Atoms are composed of electrons, neutrons, and
protons. Each electron has a charge of –1, each neutron has a charge of 0,
and each proton has a charge of +1. The total charge of an atom is the sum of
all the charges of its electrons, neutrons, and protons. An atom is an ion if it
has a positive or a negative charge. If an atom has a charge of zero, it is not
an ion. Are the following atoms ions?
Aluminum: 13 electrons, 13 neutrons, 13 protons
SOLUTION
The total charge is –13 + 0 + 13 = 0, so the atom is not an
ion. In chemistry, this aluminum atom is written as Al.
Using Addition in Real Life
SCIENCE CONNECTION Atoms are composed of electrons, neutrons, and
protons. Each electron has a charge of –1, each neutron has a charge of 0,
and each proton has a charge of +1. The total charge of an atom is the sum of
all the charges of its electrons, neutrons, and protons. An atom is an ion if it
has a positive or a negative charge. If an atom has a charge of zero, it is not
an ion. Are the following atoms ions?
Aluminum: 13 electrons, 13 neutrons, 13 protons
SOLUTION
The total charge is –13 + 0 + 13 = 0, so the atom is not an
ion. In chemistry, this aluminum atom is written as Al.
Aluminum: 10 electrons, 13 neutrons, 13 protons
SOLUTION
The total charge is –10 + 0 + 13 = 3, so the atom is an ion.
In chemistry, this aluminum atom is written as Al 3 +.
Finding the Total Profit
A consulting company had the following monthly results after comparing income
and expenses. Add the monthly profits and losses to find the overall profit or
loss during the six-month period.
SOLUTION
January
February
March
–$13,142.50
–$6783.16
–$4734.86
April
May
June
$3825.01
$7613.17
$12,932.54
With this many large numbers, you may want to use a calculator.
13142.50 +/–
+
3825.01
+
+
6783.16
7613.17
+/–
+
+
4734.86
12932.54
=
+/–
–289.8
The display is –298.8. This means the company had a loss of $298.80.
Chapter 2.3
Chapter 2.4
Chapter 2.5
Chapter 2.6
MODELING THE DISTRIBUTIVE PROPERTY
You can use algebra tiles to model algebraic expressions.
1
1 x-tile
1-tile
1
x
This 1-by-1 square tile has
an area of 1 square unit.
This 1-by-x square tile has
an area of x square units.
Model the Distributive Property using Algebra Tiles
x+2
x+2
Area = 3(x + 2)
3
3+
x
2
Area = 3(x ) + 3(2)
3
USING THE DISTRIBUTIVE PROPERTY
THE DISTRIBUTIVE PROPERTY
The product of a and (b + c):
a(b + c) = ab + ac
2(x + 5) = 2(x) + 2(5) = 2x + 10
(b + c)a = ba + ca
(x + 5)2 = (x)2 + (5)2 = 2x + 10
y(1 – y) = y(1) – y(y) = y – y 2
(1 + 5x)2 = (1)2 + (5x)2 = 2 + 10x
USING THE DISTRIBUTIVE PROPERTY
Remember that a factor must multiply each term of an expression.
Distribute the –3.
(–3)(1 + x) = (–3)(1) + (–3)(x)
= –3 – 3x
Simplify.
(y – 5)(–2) = (y)(–2) + (–5)(–2)
Distribute the –2.
= –2y + 10
–(7 – 3x) = (–1)(7) + (–1)(–3x)
= –7 + 3x
Simplify.
–a = –1 • a
Simplify.
Forgetting to distribute the negative sign when
multiplying by a negative factor is a common error.
MENTAL MATH CALCULATIONS
SOLUTION
You are shopping for CDs.
You want to buy six CDs
for $11.95 each.
6(11.95) = 6(12 – 0.05)
The mental math is easier if you
think of $11.95 as $12.00 – $.05.
Write 11.95 as a difference.
= 6(12) – property
6(0.05)
Use the distributive
to calculate the total cost
= 72 – 0.30
mentally.
Use the distributive property.
= 71.70
Find the difference mentally.
Find the products mentally.
The total cost of 6 CDs at $11.95 each is $71.70.
SIMPLIFYING BY COMBINING LIKE TERMS
Each of these terms
terms is the product of a number
number and a variable.
variable.
– x + 3y2
x isis the
–1
the
y3 is the
2.
coefficient
variable.
of
x.
coefficient
variable.
of
y
Like
Liketerms
terms have the samevariable
variable raised to the same power.
power.
y2 – x2 + 3y3 – 5 + 3 – 3x2 + 4y3 + y
The constant2 terms
3 –5 and 3 2are also
3 like terms.
x
y
x
y
SIMPLIFYING BY COMBINING LIKE TERMS
8x + 3x = (8 + 3)x
= 11x
4x2 + 2 – x2 = 4x2 – x2 + 2
= 3x2 + 2
3 – 2(4 + x) = 3 + (–2)(4 + x)
Use the distributive property.
Add coefficients.
Group like terms.
Combine like terms.
Rewrite as addition expression.
= 3 + [(–2)(4) + (–2)(x)]
Distribute the –2.
= 3 + (–8) + (–2x)
Multiply.
= –5 + (–2x)
= –5 – 2x
Combine like terms
and simplify.
Chapter 2.7
Chapter 2.8
Chapter 3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Chapter 3.1
Modeling One-Step Equations
Algebra tiles can be used to solve one-step equations.
1
x
2
x
Model the equation x + 5 = –2. A
+
+
+
+
+
=
represents the unknown value x.
–
–
To find the value of x, get the x-tile by itself on the left side of the equation.
You can undo the addition by subtracting 5. Be sure to subtract the same
amount on each side of the equation, so that the two sides stay equal.
x
+
+
+
+
+
=
–
–
–
–
–
–
–
+
+
+
+
+
3 The resulting equation shows the value of x. So, x = –7.
First add 5 zero pairs so
you can subtract 5
Addition and Subtraction Equations
You can solve an equation by using the transformations on the next slide to
isolate the variable on one side of the equation. When you rewrite an equation
using these transformations, you produce an equation with the same solutions
as the original equation. These equations are called equivalent equations.
For example, x + 3 = 5 and x = 2 are equivalent because x + 3 = 5 can be
transformed into x = 2 and each equation has 2 as its only solution.
To change, or transform, an equation into an equivalent equation, think of an
equation as having two sides that are “in balance.”
Original Equation: x + 3 = 5
Subtract 3 from both sides
to isolate x on the left.
Scale stays in balance
Simplify both sides.
Equivalent equation: x = 2
Addition and Subtraction Equations
You can use transformations to produce equivalent equations.
Any transformation you apply to an equation must keep the equation in balance.
For instance, if you subtract 3 from one side of the equation, you must also
subtract 3 from the other side of the equation.
Simplifying one side of the equation does not affect the balance, so you don’t
have to change the other side.
Addition and Subtraction Equations
You can use transformations to produce equivalent equations.
TRANSFORMATIONS THAT PRODUCE EQUIVALENT EQUATIONS
ORIGINAL EQUATION
EQUIVALENT EQUATION
Add the same number
to each side.
x–3=5
Add 3.
x=8
Subtract the same
number from each side.
x + 6 = 10
Subtract 6.
x=4
Simplify one or both
sides of the equation.
x=8–3
Simplify.
x=5
Interchange the sides.
7=x
Interchange.
x=7
Addition and Subtraction Equations
You can use transformations to produce equivalent equations.
TRANSFORMATIONS THAT PRODUCE EQUIVALENT EQUATIONS
ORIGINAL EQUATION
EQUIVALENT EQUATION
Add the same number
to each side.
x–3=5
Add 3.
x=8
Subtract the same
number from each side.
x + 6 = 10
Subtract 6.
x=4
Simplify one or both
sides of the equation.
x=8–3
Simplify.
x=5
Interchange the sides.
7=x
Interchange.
x=7
Inverse operations are operations that undo each other, such as addition and
subtraction. Inverse operations help you to isolate the variable in an equation.
Adding to Each Side
Solve the following equation by adding to each side of the equation.
x – 5 = –13
SOLUTION
x – 5 = –13
x – 5 + 5 = –13 + 5
x = –8
Write original equation.
Add 5 to each side.
Simplify.
The solution is –8. Check by substituting –8 for x in the original equation.
Adding to Each Side
Solve the following equation by adding to each side of the equation.
x – 5 = –13
SOLUTION
x – 5 = –13
x – 5 + 5 = –13 + 5
x = –8
Write original equation.
Add 5 to each side.
Simplify.
The solution is –8. Check by substituting –8 for x in the original equation.
CHECK
x – 5 = –13
Write original equation.
x – 5 = –13
–8
?
Substitute –8 for x.
–13 = –13
Solution is correct.
Simplifying First
Solve the following equation.
–8 = n – (– 4)
SOLUTION
–8 = n – (– 4)
Write original equation.
–8 = n + 4
Simplify.
–8 – 4 = n + 4 – 4
–12 = n
Subtract 4 from each side.
Simplify.
Using Linear Equations in Real-Life Problems
The equations used in this chapter are called linear equations. In a linear
equation the variable is raised to the first power and does not occur in a
denominator, inside a square root symbol, or inside absolute value symbols.
LINEAR EQUATION
x+5=9
– 4 + n = 2n – 6
NOT A LINEAR EQUATION
x2 + 5 = 9
x+3 =7
Later on you will see that linear equations get their name from the fact that their
graphs are straight lines.
Modeling a Real-Life Problem
Several record temperature changes have taken place in Spearfish, South
Dakota. On January 22, 1943, the temperature in Spearfish fell from 54˚F at
9:00 A.M. to –4˚F at 9:27 A.M. By how many degrees did the temperature fall?
SOLUTION
Verbal
Model
Labels
Temperature
at 9:27 A.M.
=
Temperature
at 9:00 A.M.
Temperature at 9:27 A.M. = – 4
Temperature at 9:00 A.M. = 54
Degrees fallen
…
=T
–
Degrees fallen
Modeling a Real-Life Problem
Several record temperature changes have taken place in Spearfish, South
Dakota. On January 22, 1943, the temperature in Spearfish fell from 54˚F at
9:00 A.M. to –4˚F at 9:27 A.M. By how many degrees did the temperature fall?
SOLUTION
…
Algebraic
Model
– 4 = 54 – T
– 4 – 54 = 54 – T – 54
–58 = –T
58 = T
Write algebraic model.
Subtract 54 from each side.
Simplify.
T is the opposite of –58.
The temperature fell by 58˚F. Check this in the original statement of the
problem.
Translating Verbal Statements
Match the real-life problem with an equation.
x – 4 = 16
x + 16 = 4
16 – x = 4
You owe $16 to your cousin. You paid x dollars back and you now owe $4.
How much did you pay back?
SOLUTION
Original amount owed (16) – Amount paid back (x) = Amount now owed (4)
Translating Verbal Statements
Match the real-life problem with an equation.
x – 4 = 16
x + 16 = 4
16 – x = 4
You owe $16 to your cousin. You paid x dollars back and you now owe $4.
How much did you pay back?
The temperature was x˚F. It rose 16˚F and is now 4˚F. What was the original
temperature?
SOLUTION
Original amount owed (16) – Amount paid back (x) = Amount now owed (4)
Original temperature (x) + Degrees risen (16) = Temperature now (4)
Translating Verbal Statements
Match the real-life problem with an equation.
x – 4 = 16
x + 16 = 4
16 – x = 4
You owe $16 to your cousin. You paid x dollars back and you now owe $4.
How much did you pay back?
The temperature was x˚F. It rose 16˚F and is now 4˚F. What was the original
temperature?
A telephone pole extends 4 feet below ground and 16 feet above ground.
What is the total length x of the pole?
SOLUTION
Original amount owed (16) – Amount paid back (x) = Amount now owed (4)
Original temperature (x) + Degrees risen (16) = Temperature now (4)
Total length (x) – Length below ground (4) = Length above ground (16)
Chapter 3.2
Solving Equations Using Multiplication and Division
Goals
Solve linear equations using multiplication and division.
Use multiplication and division equations to solve problems.
VOCABULARY
Properties of equality
The rules of algebra used to transform equations into equivalent
equations.
Ratio of a to b
If a and b are two quantities measured in the same units, then the
ratio of a to b is a/b.
Similar triangles
Two triangles are similar triangles if they have equal
corresponding angles.
Chapter 3.3
Solving Multi-Step Equations
Goals
Use two or more transformations to solve an equation.
Use multi-step equations to solve real-life problems.
USING TWO OR MORE TRANSFORMATIONS
Solving a Linear Equation
SOLUTION 1
Solve
x + 6 = –8
To isolate the
3 variable, undo the addition and then the multiplication.
1
3 x + 6 = –8
1
3 x + 6 – 6 = –8 – 6
1 x = –14
3
1
3( 3 x) = 3(–14)
x = – 42
Write original equation.
Subtract 6 from each side.
Simplify.
Multiply each side by 3.
Simplify.
Combining Like Terms First
Solve 7x – 3x – 8 = 24.
7x – 3x – 8 = 24
Write original equation.
4x – 8 = 24
Combine like terms.
4x – 8 + 8 = 24 + 8
Add 8 to each side.
4x = 32
Simplify.
4x = 32
4
4
Divide each side by 4.
x=8
Simplify.
The solution is 8. When you check the solution, substitute 8 for
each x in the original equation.
Using the Distributive Property
Solve 5x + 3(x + 4) = 28.
METHOD 1 Show All Steps
METHOD 2 Do Some Steps Mentally
5x + 3(x + 4) = 28
5x + 3x + 12 = 28
5x + 3(x + 4) = 28
5x + 3x + 12 = 28
8x + 12 = 28
8x + 12 = 28
8x + 12 – 12 = 28 – 12
8x = 16
8x 16
=
8
8
x=2
8x = 16
x=2
Multiplying by a Reciprocal First
6 (x + 3)
Solve 66 = –
5
6
= – systematically
(x + 3)
Solving 66
equations
5
is an example of deductive
6 (x + 3)
reasoning.
– 5 66 =Notice
– 5how
– each
6
6
5
solution step is based on number
properties or properties of
–55 = x + 3
equality.
–58 = x
It is easier to solve this equation
if you don’t distribute – 6 first.
5
Write original equation.
6
Multiply by reciprocal of – 5 .
Simplify.
Subtract 3 from each side.
SOLVING REAL-LIFE PROBLEMS
Using a Known Formula
SOLUTION
Use the formula to convert 95 °F
to Celsius.
9 C + 32
F
=
The Fahrenheit
5 and Celsius
scales are related by the
equation
95 = 9 C + 32
5
9
F =9 5CC + 32
63 =
5
35 = C
A body temperature of 95 °F or
lower may indicate the medical
condition called hypothermia.
What temperature on the Celsius
scale may
indicate hypothermia?
Write known
formula.
Substitute 95° for F.
Subtract 32 from each side.
5
9
Multiply by 9 , the reciprocal of
.
5
A temperature of 35°C or lower may indicate hypothermia.
Chapter 3.4
Chapter 3.5
Linear Equations and Problem Solving
Drawing a Diagram
You have learned to use a verbal model to solve real-life problems. Verbal
models can be used with other problem solving strategies, such as drawing a
diagram.
Visualizing a Problem
1
YEARBOOK DESIGN A page of your school yearbook is 8 inches by 11
2
3
inches. The left margin is
inch and the space to the right of the pictures is
4
2 7 inches. The space between pictures is 3 inch. How wide can each
8
16
picture be to fit three across the width of the page?
SOLUTION
DRAW A DIAGRAM The diagram
shows that the page width is made
up of the width of the left margin, the
space to the right of the pictures, two
spaces between pictures, and three
picture widths
Visualizing a Problem
1
YEARBOOK DESIGN A page of your school yearbook is 8 inches by 11
2
3
inches. The left margin is
inch and the space to the right of the pictures is
4
2 7 inches. The space between pictures is 3 inch. How wide can each
8
16
picture be to fit three across the width of the page?
Verbal Model
Labels
Left
margin
+
Left margin =
Space to
right of
pictures
+ 2 •
3
4
Space between pictures =
+ 3 •
(inches)
Space to right of pictures = 2
…
Space
between
pictures
7
8
3
16
(inches)
(inches)
Picture width = w
(inches)
1
2
(inches)
Page width = 8
Picture
width
=
Page
width
Visualizing a Problem
1
YEARBOOK DESIGN A page of your school yearbook is 8 inches by 11
2
3
inches. The left margin is
inch and the space to the right of the pictures is
4
2 7 inches. The space between pictures is 3 inch. How wide can each
8
16
picture be to fit three across the width of the page?
…
Algebraic
Model
3
7
3
1
+2 +2•
+3•w=8
4
8
16
2
1
When you solve for w, you will find that each picture can be 1 inches
2
wide.
Using Tables and Graphs as Checks
At East High School, 579 students take Spanish. This number has been
increasing at a rate of about 30 students per year. The number of students
taking French is 217 and has been decreasing at a rate of about 2 students
per year. At these rates, when will there be three times as many student
taking Spanish as taking French?
SOLUTION
First write an expression for the number of students taking each language after
some number of years. Then set the expression for students taking Spanish
equal to three times the expression for students taking French. Subtraction is
used in the expression for students taking French, because there is a decrease.
Verbal
Model
Number taking
Spanish now
3•
…
+
(
Rate of increase for
Spanish
Number taking
French now
–
•
Number
of years
=
Rate of decrease for
French
•
Number
of years
)
Using Tables and Graphs as Checks
At East High School, 579 students take Spanish. This number has been
increasing at a rate of about 30 students per year. The number of students
taking French is 217 and has been decreasing at a rate of about 2 students
per year. At these rates, when will there be three times as many student
taking Spanish as taking French?
…
Labels
…
Number of students taking Spanish now = 579
(students)
Rate of increase for Spanish = 30
(students per year)
Number of years = x
(years)
Number of students taking French now = 217
(students)
Rate of decrease for French = 2
(students per year)
Using Tables and Graphs as Checks
At East High School, 579 students take Spanish. This number has been
increasing at a rate of about 30 students per year. The number of students
taking French is 217 and has been decreasing at a rate of about 2 students
per year. At these rates, when will there be three times as many student
taking Spanish as taking French?
…
Algebraic
Model
579 + 30 • x = 3(217 – 2 • x)
579 + 30 x = 651 – 6 x
Use distributive property.
579 + 36 x = 651
Add 6 x to each side.
36 x = 72
x=2
Subtract 579 from each side.
Divide each side by 36.
At these rates, the number of students taking Spanish will be three times
the number of students taking French in two years.
Using Tables and Graphs as Checks
At East High School, 579 students take Spanish. This number has been
increasing at a rate of about 30 students per year. The number of students
taking French is 217 and has been decreasing at a rate of about 2 students
per year. At these rates, when will there be three times as many student
taking Spanish as taking French?
CHECK Use the information above to make a table.
For example, the number of students
taking Spanish in Year 1 is 579 + 30 = 609
Year
0 (now)
1
2
Number taking Spanish
579
609
639
Number taking French
217
215
213
3 • Number taking French
651
645
639
equal
Using a Graph as a Check
GAZELLE AND CHEETAH A gazelle can run 73 feet per second for several
minutes. A cheetah can run faster (88 feet per second) but can only sustain
its top speed for about 20 seconds before it is worn out. How for away from
the cheetah does the gazelle need to stay for it be be safe?
SOLUTION
You can use a diagram to picture the situation. If the gazelle is too far away
for the cheetah to catch it within 20 seconds, the gazelle is probably safe.
If the gazelle is closer to the
cheetah than x feet, the cheetah
can catch the gazelle.
If the gazelle is farther away
than x feet, the gazelle is in
the safety zone.
“Safety zone”
0
x
gazelle caught in 20 seconds
Using a Graph as a Check
GAZELLE AND CHEETAH A gazelle can run 73 feet per second for several
minutes. A cheetah can run faster (88 feet per second) but can only sustain
its top speed for about 20 seconds before it is worn out. How for away from
the cheetah does the gazelle need to stay for it be be safe?
To find the “safety zone,” you can first find the starting distance for which the
cheetah reaches the gazelle in 20 seconds.
To find the distance each animal runs in 20 seconds, use the formula d = r t.
Verbal Model
…
Distance
gazelle runs
in 20 sec
Gazelle’s
+ starting distance
from cheetah
=
Distance cheetah
runs in 20 sec
Using a Graph as a Check
GAZELLE AND CHEETAH A gazelle can run 73 feet per second for several
minutes. A cheetah can run faster (88 feet per second) but can only sustain
its top speed for about 20 seconds before it is worn out. How for away from
the cheetah does the gazelle need to stay for it be be safe?
…
Labels
Algebraic
Model
Gazelle’s distance = 73 • 20
(feet)
Gazelle’s starting distance from cheetah = x
(feet)
Cheetah’s distance = 88 • 20
(feet)
73 • 20 + x = 88 • 20
You find that x = 300, so the gazelle needs to stay more than 300 feet away
from the cheetah to be safe.
Using a Graph as a Check
GAZELLE AND CHEETAH A gazelle can run 73 feet per second for several
minutes. A cheetah can run faster (88 feet per second) but can only sustain
its top speed for about 20 seconds before it is worn out. How for away from
the cheetah does the gazelle need to stay for it be be safe?
CHECK You can use a table
or a graph to check your
answer. A graph helps you
see the relationships. To
make a graph, first make a
table and then plot the points.
You can just find and mark
the points for every 5
seconds.
Distance from cheetah's
starting point (feet)
Gazelle and Cheetah
2000
1500
1000
Gazelle
500
300
Cheetah
0
0
5
10
Gazelle starting
at 300 feet from
cheetah is caught
in 20 seconds.
15
Time (seconds)
20
Chapter 3.6
Solving Decimal Equations
Goals
Find exact and approximate solutions of equations
that contain decimals.
Solve real-life problems that use decimals.
 Round-off error
Round-off error is the error produced when a
decimal result is rounded in order to provide a
meaningful answer.
Chapter 3.7
Formulas and Functions
Goals
Solve a formula for one of its variables.
Rewrite an equation in function form.
 Formula
A formula is an algebraic equation that
relates two or more real-life quantities.
Chapter 3.8
Rates, Ratios, and Percents
Goals
Use rates and ratios to model and solve real-life
problems.
Use percents to model and solve real-life problems.
 Rate of a per b
If a and b are two quantities measured in
different units, then the rate of a per b is a/b
 Unit rate
A unit rate is a rate per one given unit.
Chapter 4
Chapter 4.1
Chapter 4.2
Graphing a Linear Equation
A solution of an equation in two variables x and y is an ordered pair (x, y)
that makes the equation true.
The graph of an equation in x and y is the set of all points (x, y) that are
solutions of the equation.
In this lesson you will see that the graph of a linear equation is a line.
Verifying Solutions of an Equation
Use the graph to decide whether the point lies on the graph of x + 3y = 6.
Justify your answer algebraically.
y
(1, 2)
3
1
SOLUTION
–1 O
–1
x + 3y = 6
?
1 + 3(2) = 6
76
1
3
Write original equation.
Substitute 1 for x and 2 for y.
Simplify. Not a true statement.
(1, 2) is not a solution of the equation x + 3y = 6, so it is not on the graph.
5
x
Verifying Solutions of an Equation
Use the graph to decide whether the point lies on the graph of x + 3y = 6.
Justify your answer algebraically.
y
(–3, 3)
3
1
SOLUTION
–1 O
–1
x + 3y = 6
?
–3 + 3(3) = 6
6=6
1
Write original equation.
Substitute –3 for x and 3 for y.
Simplify. True statement.
(–3, 3) is a solution of the equation x + 3y = 6, so it is on the graph.
3
5
x
Graphing a Linear Equation
In the previous example you learned that the point (–3, 3) is on the graph of
x + 3y = 6, but how many points does a graph have in all?
The answer is that most graphs have too
many points to list.
y
3
Then how can you ever graph an equation?
1
One way is to make a table or a list of a few
values, plot enough solutions to recognize a
pattern, and then connect the points.
–1 O
–1
1
3
5
Even then, the graph extends without limit to the left of the smallest input and to
the right of the largest input.
x
Graphing a Linear Equation
In the previous example you learned that the point (–3, 3) is on the graph of
x + 3y = 6, but how many points does a graph have in all?
The answer is that most graphs have too
many points to list.
y
3
Then how can you ever graph an equation?
1
One way is to make a table or a list of a few
values, plot enough solutions to recognize a
pattern, and then connect the points.
–1 O
–1
1
3
5
Even then, the graph extends without limit to the left of the smallest input and to
the right of the largest input.
When you make a table of values to graph an equation, you may want to
choose values for x that include negative values, zero, and positive values.
This way you will see how the graph behaves to the left and right of the y -axis.
x
Graphing an Equation
Use a table of values to graph the equation y + 2 = 3x.
SOLUTION
Rewrite the equation in function form by solving for y.
y + 2 = 3x
y = 3x – 2
Write original equation.
Subtract 2 from each side.
Choose a few values for x and
make a table of values.
Choose x.
Substitute to find the
corresponding y-value.
–2
y = 3(–2) – 2 = – 8
–1
y = 3(–1) – 2 = –5
0
y = 3(0) – 2 = –2
1
y = 3(1) – 2 = 1
2
y = 3(2) – 2 = 4
Graphing an Equation
Use a table of values to graph the equation y + 2 = 3x.
SOLUTION
x
y
–2
–8
–1
–5
0
1
–2
2
4
1
With the table of values
you have found the five
solutions (–2, –8), (–1, –5),
(0, –2), (1, 1), and (2, 4).
y
3

1
–3
–1
–1

Plot the points. Note that they appear to
lie on a straight line.
The line through the points is the graph of
the equation.


–6

–8
1
3x
Graphing a Linear Equation
You may remember examples of linear equations in one variable. The solution
of an equation such as 2 x – 1 = 3 is a real number. Its graph is a point on the
real number line.
The equation y + 2 = 3x in the previous example is a linear equation in two
variables. Its graph is a straight line.
GRAPHING A LINEAR EQUATION
STEP 1
Rewrite the equation in function form, if necessary.
STEP 2
Choose a few values of x and make a table of values.
STEP 3
Plot the points from the table of values. A line through these
points is the graph of the equation.
Graphing a Linear Equation
Use a table of values to graph the equation 3x + 2y = 1.
SOLUTION
1
Rewrite the equation in function form by solving for y. This will make it
easier to make a table of values.
3x + 2y = 1
2y = –3x + 1
y=– 3x+ 1
2
2
2
Choose a few values
for x and make a table
of values.
Write original equation.
Subtract 3x from each side.
Divide each side by 2.
Choose x.
–2
0
2
Evaluate y.
7
2
1
2
– 5
2
Graphing a Linear Equation
Use a table of values to graph the equation 3x + 2y = 1.
SOLUTION
With the table of values you
have found three solutions.
(
–2,
3
7
2
) ( )(
, 0,
1
2
5
, 2, –
2
)
Plot the points and draw a line
through them.
Choose x.
–2
0
2
Evaluate y.
7
2
1
2
– 5
2
y

3
2
1

O
The graph of 3x + 2y = 1 is shown
at the right.
–2
–3
–4
1
2

3
x
Using the Graph of a Linear Model
An Internet Service Provider estimates that the number of households h (in
millions) with Internet access can be modeled by h = 6.76 t + 14.9, where t
represents the number of years since 1996. Graph this model. Describe the
graph in the context of the real-life situation.
SOLUTION
Make a table of values.
Use 0  t  6, for 1996 – 2002.
From the table and the
graph, you can see that the
number of households with
Internet access is projected
to increase by about 7
million households per year.
t
0
1
2
3
4
5
6
h
14.90
21.66
28.42
35.18
41.94
48.70
55.46
Horizontal and Vertical Lines
All linear equations in x and y can be written in the form A x + B y = C. When
A = 0 the equation reduces to B y = C and the graph is a horizontal line.
When B = 0 the equation reduces to A x = C and the graph is a vertical line.
EQUATIONS OF HORIZONTAL AND VERTICAL LINES
y=b
In the coordinate plane, the
graph y = b is a horizontal line.
x=a
In the coordinate plane, the
graph x = a is a horizontal line.
Graphing y = b
Graph the following equation.
y=2
SOLUTION
y
The equation does not have x as a variable.
(–3, 2)
The y -value is always 2, regardless of the value
of x. For instance, here are some points that are
solutions of the equation:
(–3, 2), (0, 2), (3, 2)

3 (0, 2)
2
(3, 2)
y=2

1
O
1
–2
–3
–4
The graph of the equation is a horizontal line 2 units above the x-axis.
2
3
x
Graphing x = a
Graph the following equation.
x = –3
SOLUTION
y
The x-value is always –3, regardless of the
value of y.
(–3, 3)

2
x = –3
For instance, here are some points that are
solutions of the equation:
(–3, –2), (–3, 0), (–3, 3)
3


(–3, 0)
1
O
1
2
(–3, –2)
–2
–3
–4
The graph of the equation is a vertical line 3 units to the left of the y-axis.
3
x
Chapter 4.3
FINDING THE SLOPE OF A L INE
The slope m of a nonvertical line is the number of units
the line rises or falls for each unit of horizontal change
from left to right.
FINDING THE SLOPE OF A L INE
y
nonvertical line
The slope m of a nonvertical
line is
the number of units the line rises or
falls for each unit of horizontal
change from left to right.
9
7
The slanted line at the right rises
3 units for each 2 units of horizontal
change from left to right. So, the
slope m of the line is 3 .
2
Two points on a line are all that is
needed to find its slope.
rise = 5 - 2 = 3 units
3
(0, 2)
run = 2 - 0 = 2 units
1
-1
rise
slope =
=
run
(2, 5)
5
5-2
2-0
=
1
-1
3
2
3
5
7
9
x
FINDING THE SLOPE OF A L INE
y
The slope m of the nonvertical line
passing through the points (x1, y1)
and (x2, y2) is
(x2, y2)
(x1, y1)
Read y1 as “y sub one”
Read x1 as “x sub one”
(y2 - y1 )
(x2 - x1 )
x
m =
rise
run
=
change in y
change in x
= xy22 -- yx11
FINDING THE SLOPE OF A L INE
When you use the formula for the slope,
m
y2 - y1
m =
x1
y -y
rise
changexin
2 -y
= run = change in x = x2 - x1
2
1
The order of subtraction
important. You
label either point as (x1, y1)
theis numerator
andcan
denominator
and the other point as
(x2,use
y2).the
However,
both the numerator
numerator
denominator
must
same subtraction
order. and denominator
must use the same order.
CORRECT
y2 - y 1
x2 - x1
Subtraction order is the same
INCORRECT
y2 - y1
x1 - x2
Subtraction order is different
A Line with a Zero Slope is Horizontal
(-1, 2)
2) and (3,
Find the slope of a line passing through (-1,
(3, 2).
2).
y
SOLUTION
9
(x11,, yy11)) = (-1, 2) and
Let (x
(x22, y2) = (3,
(3, 2)
2)
7
m=
y2 - y1
x2 - x1
2
= 32- -(-1)
=
0
4
= 0
Rise: difference of y-values
5
rise = 2 - 2 = 0 units
Run: difference of x-values
(-1, 2)
3
(3, 2)
Substitute values.
run = 3 - ( -1) = 4 units
1
Simplify.
-1
-1
1
3
Slope is zero. Line is horizontal.
5
7
9
x
INTERPRETING SLOPE AS A RATE OF CHANGE
Slope as a Rate of Change
You are parachuting. At time tt == 00 seconds,
2500 feet
feet
seconds,you open your parachute at hh == 2500
above the ground. At t = 35 seconds,
seconds, you are at h = 2115 feet.
y
2700
a. What is your rate of change in height?
(0, 2500)
2500
b. About when will you reach the ground?
Height (feet)
2300
2100
(35, 2115)
1900
5
15
25
Time (seconds)
35
45 x
Slope as a Rate of Change
SOLUTION
a. Use the formula for slope to find the rate of change.
change. The change
change in time is
35 - 0 = 35 seconds. Subtract in the same order. The change
change in height
height is
2115 - 2500 = -385 feet.
VERBAL
MODEL
LABELS
Rate of
Change
=
Change in Height
Change in Time
Rate of Change =
m (ft/sec)
Change in Height = -- 385 (ft)
Change in Time =
ALGEBRAIC
MODEL
m =
35 (sec)
- 385
35
= -11
Your rate of change is -11 ft/sec. The negative value indicates you are falling.
Slope as a Rate of Change
b. Falling at a rate of -ll ft/sec, find the time it will take you to fall 2500 ft.
Time
Time
Time
=
=
Distance
Distance
Rate
Rate
-2500 ft
-11 ft/sec
227 sec
You will reach the ground about 227 seconds
after opening your parachute.
Chapter 5
5.1
5.2
5.3
5.4
5.5
5.6
5.7
Chapter 5.1
Chapter 5.2
Chapter 5.3
Chapter 5.4
FITTING A LINE TO DATA
8
There
Usually,
arethere
several
is no
ways
single
to find
line
the
thatbest-fitting
passes through
lineall
forthe
a given
data
set
points,
of data
so you
points.
try to
In find
this lesson,
the line
you
that will
best use
fits athe
graphical
data. This is
best-fitting line.
line.
approach.
called the best-fitting
6
4
2
–8
–6
–4
–2
0
–2
–4
–6
–8
2
4
6
Approximating a Best-Fitting Line
250
DISCUS THROWS
Write an equation of
your line.
230
220
210
200
Distance (ft)
The winning Olympic
discus throws from 1908
to 1996 are plotted in
the graph. Approximate
the best-fitting line for
these throws.
240
190
180
170
160
150
140
130
120
110
100
0
8
16
24
32
40
48
56
64
72
Years since 1900
80
88
96
104
Approximating a Best-Fitting Line
250
SOLUTION
Find the slope of the line
through these points.
(96, 230)
230
220
210
200
Distance (ft)
Find two points that lie
on the best-fitting line,
such as (8, 138) and
(96, 230))..
240
190
180
170
160
150
140
(8, 138)
130
120
110
100
0
8
16
24
32
40
48
56
64
72
Years since 1900
80
88
96
104
Approximating a Best-Fitting Line
250
y2 – y1 230 – 138
m = x – x = 96 – 8 230= 92
88
2
1
220
240
y138
= mx
+ b+ b
= 8.4
Distance (ft)
– 138
= y230
=
mx
96 – 8 + b
138
= (1.05)
= 92
1.05(8) + b
88
1.05
(96, 230)
210
200
190
180
Write slope intercept form.
Substitute 1.05 for m, 8 for x,
138 for y.
170
160
Simplify.
150
129.6 = b
140
130
Solve for b.
(8, 138)
An equation of the best-fitting line is y = 1.05 x + 129.6.
120
110
In most years, the winner of
100 the discus throw was able to
16
24
32
40
48
56
64
72
80
throw the discus farther than 0the8 previous
winner.
Years since 1900
88
96
104
DETERMINING THE CORRELATION OF X AND Y
In this scatter plot, x and y
have a positive correlation,
which means that the points
can be approximated by a line
with a positive slope.
DETERMINING THE CORRELATION OF X AND Y
In this scatter plot, x and y
have a negative correlation,
which means that the points
can be approximated by a
line with a negative slope.
DETERMINING THE CORRELATION OF X AND Y
In this scatter plot, x and y
have relatively no correlation,
which means that the points
cannot be approximated by
a line.
DETERMINING THE CORRELATION OF X AND Y
TYPES OF CORRELATION
Positive Correlation
Negative Correlation
No Correlation
Chapter 5.5
Using the Point-Slope Form
Previously you learned one strategy for writing a linear equation when given
the slope and a point on the line. In this lesson you will learn a different
strategy – one that uses the point-slope form of an equation of a line.
POINT-SLOPE FORM OF THE EQUATION OF A LINE
The point-slope form of the equation of the nonvertical line
that passes through a given point (x 1, y 1) with a slope of m is
y – y 1 = m (x – x 1).
Developing the Point-Slope Form
Write an equation of the line. Use the points (2, 5) and (x, y).
SOLUTION
You are given one point on the line.
Let (x, y) be any point on the line.
Because (2, 5) and (x, y) are two points
on the line, you can write the following
expression for the slope of the line.
m=
y–5
x–2
y–5 2
=
x–2 3
y–5=
2
(x – 2)
3
The equation y – 5 =
Use formula for slope.
2
Substitute for m.
3
Multiply each side by (x – 2).
The graph shows that the slope
2
2
is . Substitute for m in the
3
3
formula for slope.
2
(x – 2) is written in point-slope form.
3
Using the Point-Slope Form
You can use the point-slope form when you are given the slope and a point on
the line.
In the point-slope form, (x 1, y 1) is the given point and (x, y) is any other point
on the line.
You can also use the point-slope form when you are given two points on the
line.
First find the slope. Then use either given point as (x 1, y 1).
Using the Point-Slope Form
Write an equation of the line shown below.
SOLUTION
First find the slope. Use the points (x 1, y 1) = (–3, 6) and (x 2, y 2) = (1, –2).
y2 – y1
m= x –x
2
1
=
–2 – 6
1 – (–3)
= –8
4
= –2
Using the Point-Slope Form
Write an equation of the line shown below.
SOLUTION
Then use the slope to write the point-slope form. Choose either point as (x 1, y 1).
y – y 1 = m (x – x 1)
Write point-slope form.
y – 6 = –2[x – (–3)]
Substitute for m, x 1 and y 1.
y – 6 = –2(x + 3)
Simplify.
y – 6 = –2 x – 6
Use distributive property.
y = –2 x
Add 6 to each side.
Modeling a Real-Life Situation
MARATHON The information below was taken from an article that appeared
in a newspaper.
Use the information to write a linear model for optimal running pace, then
use the model to find the optimal running pace for a temperature of 80°F.
Writing and Using a Linear Model
SOLUTION
Let T represent the temperature in degrees Fahrenheit.
Let P represent the optimal pace in feet per second.
From the article, you know that the optimal running pace at 60°F is 17.6 feet per
second so one point on the line is (T 1, P 1) = (60, 17.6). Find the slope of the line.
m=
change in P
– 0.3
=
= – 0.06
change in T
5
Use the point-slope form to write the model.
P – P 1 = m (T – T 1)
Write the point-slope form.
P – 17.6 = (– 0.06)(T – 60)
Substitute for m, T1, and P1.
P – 17.6 = – 0.06T + 3.6
Use distributive property.
P = –0.06T + 21.2
Add 17.6 to each side.
Writing and Using a Linear Model
SOLUTION
Use the model P = –0.06T + 21.2 to find the optimal pace at 80°F.
P = – 0.06(80) + 21.2
= 16.4
At 80°F the optimal running pace is 16.4
feet per second.
CHECK A graph can help you check this
result. You can see that as the
temperature increases the optimal
running pace decreases.
Chapter 5.6
Chapter 5.7
Chapter 6
Chapter 6.1
Chapter 6.2
Chapter 6.3
Solving Compound Inequalities
Previously you studied four types of simple inequalities. In this lesson you
will study compound inequalities. A compound inequality consists of two
inequalities connected by and or or.
Write an inequality that represents the set of numbers and graph the inequality.
All real numbers that are greater than zero and less than or equal to 4.
SOLUTION
0<x4
–1
0
1
2
3
4
5
This inequality is also written as 0 < x and x  4.
Solving Compound Inequalities
Previously you studied four types of simple inequalities. In this lesson you
will study compound inequalities. A compound inequality consists of two
inequalities connected by and or or.
Write an inequality that represents the set of numbers and graph the inequality.
All real numbers that are greater than zero and less than or equal to 4.
SOLUTION
0<x4
–1
0
1
2
3
4
5
This inequality is also written as 0 < x and x  4.
All real numbers that are less than –1 or greater than 2.
SOLUTION
x < –1 or x > 2
–2 –1
0
1
2
3
4
Compound Inequalities in Real Life
Write an inequality that describes the elevations of the regions of Mount Rainier.
Timber region below 6000 ft
14,140 ft
Glacier and
permanent snow
field region
SOLUTION
Let y represent the approximate elevation (in feet).
Timber region:
2000  y < 6000
7500 ft
6000 ft
Alpine meadow region
Timber region
2000 ft
0 ft
Compound Inequalities in Real Life
Write an inequality that describes the elevations of the regions of Mount Rainier.
Timber region below 6000 ft
Alpine meadow region below 7500 ft
14,140 ft
Glacier and
permanent snow
field region
SOLUTION
Let y represent the approximate elevation (in feet).
Timber region:
2000  y < 6000
Alpine meadow region:
6000  y < 7500
7500 ft
6000 ft
Alpine meadow region
Timber region
2000 ft
0 ft
Compound Inequalities in Real Life
Write an inequality that describes the elevations of the regions of Mount Rainier.
Timber region below 6000 ft
Alpine meadow region below 7500 ft
14,140 ft
Glacier and permanent snow field region
SOLUTION
Let y represent the approximate elevation (in feet).
Timber region:
2000  y < 6000
Alpine meadow region:
6000  y < 7500
Glacier and permanent snow field region:
7500  y  14,410
Glacier and
permanent snow
field region
7500 ft
6000 ft
Alpine meadow region
Timber region
2000 ft
0 ft
Solving a Compound Inequality with And
Solve –2  3x – 8  10. Graph the solution.
SOLUTION
Isolate the variable x between the two inequality symbols.
–2  3 x – 8  10
Write original inequality.
6  3 x  18
Add 8 to each expression.
2x6
Divide each expression by 3.
The solution is all real numbers that are greater than or equal to 2
and less than or equal to 6.
0
1
2
3
4
5
6
7
Solving a Compound Inequality with Or
Solve 3x + 1 < 4 or 2x – 5 > 7. Graph the solution.
SOLUTION
A solution of this inequality is a solution of either of its simple parts.
You can solve each part separately.
3x + 1 < 4
or
2x – 5 > 7
3x < 3
or
2x > 12
x<1
or
x>6
The solution is all real numbers that are less than 1 or greater than 6.
–1
0
1
2
3
4
5
6
7
Reversing Both Inequality Symbols
Solve –2 < –2 – x < 1. Graph the solution.
SOLUTION
Isolate the variable x between the two inequality signs.
–2 < –2 – x < 1
Write original inequality.
0 < –x < 3
Add 2 to each expression.
0 > x > –3
Multiply each expression by –1 and
reverse both inequality symbols.
To match the order of numbers on a number line, this compound is usually
written as –3 < x < 0. The solution is all real numbers that are greater than
–3 and less than 0.
–5 – 4 –3 –2 –1
0
1
2
3
Modeling Real-Life Problems
You have a friend Bill who lives three miles from school and another friend
Mary who lives two miles from the same school. You wish to estimate the
distance d that separates their homes.
What is the smallest value d might have?
SOLUTION
Problem
Solving
Strategy
DRAW A DIAGRAM A good way to begin this problem is to draw a
diagram with the school at the center of a circle.
Bill’s home is somewhere on
the circle with radius 3 miles
and center at the school.
Mary’s home is somewhere on
the circle with radius 2 miles
and center at the school.
Modeling Real-Life Problems
You have a friend Bill who lives three miles from school and another friend
Mary who lives two miles from the same school. You wish to estimate the
distance d that separates their homes.
What is the smallest value d might have?
SOLUTION
If both homes are on the same line
going toward school, the distance is
1 mile.
Modeling Real-Life Problems
You have a friend Bill who lives three miles from school and another friend
Mary who lives two miles from the same school. You wish to estimate the
distance d that separates their homes.
What is the largest value d might have?
SOLUTION
If both homes are on the same
line but in opposite directions from
school, the distance is 5 miles.
Modeling Real-Life Problems
You have a friend Bill who lives three miles from school and another friend
Mary who lives two miles from the same school. You wish to estimate the
distance d that separates their homes.
What is the largest value d might have?
SOLUTION
If both homes are on the same
line but in opposite directions from
school, the distance is 5 miles.
Write an inequality that describes all the
possible values that d might have.
SOLUTION
The values of d can be described by the inequality 1  d  5.
Chapter 6.4
SOLVING ABSOLUTE-VALUE EQUATIONS
You can solve some absolute-value equations
using mental math. For instance, you learned that
the equation | x | 8 has two solutions: 8 and 8.
To solve absolute-value equations, you can use
the fact that the expression inside the absolute
value symbols can be either positive or negative.
Solving an Absolute-Value Equation
Solve | x  2 |  5
SOLUTION
The expression x  2 can be equal to 5 or 5.
x  2 IS
IS POSITIVE
POSITIVE
x  2 IS NEGATIVE
|| xx  22 ||  55
xx  22  5
5
|x2|5
xx  77
x  2  5
x  3
The equation has two solutions: 7 and –3.
CHECK
|72||5|5
| 3  2 |  | 5 |  5
Solving an Absolute-Value Equation
Solve | 2x  7 |  5  4
SOLUTION
Isolate the absolute value expression on one side of the equation.
2x  7 IS
IS POSITIVE
POSITIVE
2x  7 IS
IS NEGATIVE
NEGATIVE
| 2x  7 |  5  4
| 2x  7 |  5  4
| 2x  7 |  9
| 2x  7 |  9
2x  7  +9
2x  16
x8
2x  7  9
9
2x  2
TWO SOLUTIONS
x  1
Solving an Absolute-Value Equation
Recall that |xx is
| isthe
thedistance
distancebetween
betweenxxand
and0.0.IfIf x
| x 
| 8,8,then
then
any number between 8 and 8 is a solution of the inequality.
8 7 6 5 4 3 2 1
0
1
2
3
4
5
6
You can use the following properties to solve
absolute-value inequalities and equations.
7
8
SOLVING ABSOLUTE-VALUE INEQUALITIES
SOLVING ABSOLUTE-VALUE EQUATIONS AND INEQUALITIES
| ax  b |  c
means
ax  b  c
and
a x  b   c.
means valueais
| a xWhen
 b | an
c absolute
x less
b than
c
and
a x the
b   c.
a number,
inequalities are connected by and. When an absolute
means
| a xvalue
 b | iscgreater
a x  b  cthe inequalities
or
a x  bare  c.
than a number,
connected by or.
| ax  b |  c
means
ax  b  c
or
a x  b   c.
| ax  b |  c
means
ax  b  c
or
a x  b   c.
Solving an Absolute-Value Inequality
Solve | x  4 | < 3
x  4 IS POSITIVE
|x4|3
x  4  3
x7
x  4 IS NEGATIVE
|x4|3
x  4  3
x1
Reverse
inequality
symbol.
The solution is all real numbers greater than 1 and
less than
7.
This can be written as 1  x  7.
Solving an Absolute-Value Inequality
Solve
 1 | 3  6 and graph
2x + 1| 2x
IS POSITIVE
2x +the
1 ISsolution.
NEGATIVE
| 2x  1 |  3  6
| 2x  1 | 3  6
2x + 1 IS POSITIVE
| 2x|2x1 | 13|  69
2x + 1 IS NEGATIVE
1 |  69
| 2x|2x1 | 3
2x  1  9
| 2x  1 |  9
2x  10
2x  8
2x  1  9
2x  1  +9
x4
x  5
2x  10
2x  8
The solution is all real numbers greater than or equal
x4
x  5
to 4 or less than or equal to  5. This can be written as
the compound inequality
x   5 or x  4.
Reverse
2x11|  +9
| 2x
9
inequality symbol.
6 5 4 3 2 1
0
1
2
3
4
5
6
Writing an Absolute-Value Inequality
You work in the quality control
department of a manufacturing
company. The diameter of a drill bit
must be between 0.62 and 0.63 inch.
a. Write an absolute-value inequality
to represent this requirement.
b. Does a bit with a diameter of 0.623
inch meet the requirement?
Writing an Absolute-Value Inequality
The diameter of a drill bit must be between 0.62 and 0.63 inch.
a. Write an absolute-value inequality to represent this requirement.
Let d represent the diameter (in inches) of the drill bit.
Write a compound inequality.
0.62  d  0.63
Find the halfway point.
0.625
Subtract 0.625 from
each part of the
compound inequality.
0.62  0.625  d  0.625  0.63  0.625
0.005  d  0.625  0.005
Rewrite as an absolute-value inequality.
| d  0.625 |  0.005
This inequality can be read as “the actual diameter must
differ from 0.625 inch by no more than 0.005 inch.”
Writing an Absolute-Value Inequality
The diameter of a drill bit must be between 0.62 and 0.63 inch.
b. Does a bit with a diameter of 0.623 meet the requirement?
| d  0.625 |  0.005
| 0.623  0.625 |  0.005
| 0.002 |  0.005
0.002  0.005
Because 0.002  0.005, the bit does meet the requirement.
Chapter 6.5
Chapter 6.6
Chapter 6.7
Chapter 7
Systems of Linear Equations
7.1 Solving Linear Systems by Graphing
7.2 Solving Linear Systems by Substitution
7.3
Solving Linear Systems by Linear Combinations
7.4 Applications of Linear Systems
7.5 Special Types of Linear Systems
7.6 Solving Systems of Linear Inequalities
7.1 Solving Linear Systems by Graphing
Goals
Solve a system of linear equations by graphing.
Model a real-life problem using a linear system.
System of linear equations - Two or more linear
equations in the same variables form a system of linear
equations, or simply a linear system.
Solution of a system of linear equations A solution of a system of linear equations in two variables x
and y is an ordered pair (x, y) that satisfies each equation
in the system.
SOLVING A LINEAR SYSTEM USING GRAPHAND-CHECK
To use the graph-and-check method to solve a system of
linear equations in two variables, use the following steps.
 Step 1 Write each equation in a form that is easy to graph
 Step 2 Graph both equations in the same coordinate
plane
 Step 3 Estimate the coordinates of the point of
intersection
 Step 4 Check the coordinates algebraically by substituting
into each equation of the original linear system.
7.2 Solving Linear Systems by Substitution
Goals
Use substitution to solve a linear system.
Model a real-life situation using a linear system.
SOLVING A LINEAR SYSTEM BY SUBSTITUTION
 Step 1 Solve one of the equations for one of its variables .
 Step 2 Substitute the expression from Step 1 into the other
equation and solve for the other variable .
 Step 3 Substitute the value from Step 2 into the revised
equation from Step 1 and solve.
 Step 4 Check the solution in each of the original equations.
IDENTIFYING THE NUMBER OF SOLUTIONS
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
Lines intersect
one solution
x
IDENTIFYING THE NUMBER OF SOLUTIONS
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
Lines are parallel
no solution
x
IDENTIFYING THE NUMBER OF SOLUTIONS
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
y
Lines coincide
infinitely many
solutions
(the coordinates
of every point on
the line)
x
IDENTIFYING THE NUMBER OF SOLUTIONS
CONCEPT
NUMBER OF SOLUTIONS OF A LINEAR SYSTEM
SUMMARY
y
y
y
x
x
x
Lines intersect
Lines are parallel
Lines coincide
one solution
no solution
infinitely many solutions
A Linear System with No Solution
Show that this linear system
METHOD 1: GRAPHING
has no solution.
2x  y  5
2x  y  1
Equation 1
Equation 2
Rewrite each equation
in slope-intercept form.
y  –2 x  5
y  –2 x  1
Revised Equation 1
Revised Equation 2
6
Graph the linear system.
5
y  2x  5
4
The lines are parallel; they
have the same slope but
different y-intercepts. Parallel
lines never intersect, so the
system has no solution.
3
y  2x  1
5
4
3
2
2
1
1
1
0
1
2
3
4
5
A Linear System with No Solution
Show that this linear
METHOD 2: SUBSTITUTION
system has no solution.
2x  y  5
2x  y  1
Equation 1
Equation 2
Because Equation 2 can be revised
to y  –2 x  1, you can substitute
–2 x  1 for y in Equation 1.
2x  y  5
2 x  (–2 x  1)  5
15
Write Equation 1.
Substitute –2 x  1 for y.
Simplify. False statement!
Once the variables are eliminated, the statement is not true regardless
of the values of x and y. This tells you the system has no solution.
A Linear System with Many Solutions
Show that this linear system
has infinitely many solutions.
METHOD 1: GRAPHING
Rewrite each equation
in slope-intercept form.
–2x  y  3
– 4 x  2y  6
y  2x  3
y  2x  3
Equation 1
Equation 2
Revised Equation 1
Revised Equation 2
6
Graph the linear system.
–4x  2y  6
From these graphs you
can see that the equations
represent the same line.
Any point on the line is
a solution.
–2x  y  3
5
4
3
2
1
5
4
3
2
1
1
0
1
2
3
4
5
A Linear System with Many Solutions
Show
that this linear system
M
ETHOD 2: LINEAR COMBINATIONS
has infinitely many solutions.
–2x  y  3
– 4 x  2y  6
Equation 1
Equation 2
You can multiply Equation 1 by –2.
4x – 2y  – 6
– 4 x  2y  6
0  0
Multiply Equation 1 by –2.
Write Equation 2.
Add Equations. True statement!
The variables are eliminated and you are left with a statement that is true
regardless of the values of x and y. This result tells you that the linear system
has infinitely many solutions.
MODELING A REAL-LIFE PROBLEM
Error Analysis
GEOMETRY CONNECTION Two students are visiting the
mysterious statues on Easter Island in the South Pacific. To find
the heights of two statues that are too tall to measure, they tried
a technique involving proportions. They measured the shadow
lengths of the statues at 2:00 P.M. and again at 3:00 P.M.
3:00
2:00
Error Analysis
SOLUTION
They let a and b represent the heights of the two statues. Because
the ratios of corresponding sides of similar triangles are equal, the
students wrote the following two equations.
a
bb
27
30
aa = bb
=
2:00
3:00
27
30
27 ft
30
18 ft
20
20
18
18
20
30
aa = 27
b
18
20
a = 3 b
2
Error Analysis
They got stuck because the equations that they wrote are equivalent.
All that the students found from these measurements was that
a = 3 b. In other words, the height of the first statue is one and
2
one-half times the height of the second.
a
To findb the heights of the
two statues,
the
a
b
=
=
=
students need to measure the shadow
length of something else whose
height
27
30
18
20
they know
or can measure. Then they
3
a = 3 b
a =
b
can compare
those measurements 2with
2
the measurements of the statue.
Solving Systems of Linear Inequalities
The graph of a linear inequality in two variables is a half-plane. The boundary
line of the half-plane is dashed if the inequality is < or > and solid if the
inequality is  or .
Two or more linear inequalities form a system of linear inequalities or
simply a system of inequalities.
A solution of a system of a system of linear inequalities is an ordered pair
that is a solution of each inequality in the system.
The graph of a system of linear inequalities is the graph of all solutions
of the system.
A Triangular Solution Region
Graph the system of linear inequalities.
y<2
x  –1
y>x–2
Inequality 1
Inequality 2
Inequality 3
SOLUTION
Graph all three inequalities in the same coordinate plane. The graph of the
system is the overlap, or the intersection, of the three half-planes shown.
CHECK You can see from the graph
that the point (2, 1) is a solution of the
system. To check this, substitute the
point into each inequality.
1<2
2  –1
1>2–2
True.
True.
True.
A Triangular Solution Region
Graph the system of linear inequalities.
The point (0, 3) is not in the graph of the
system. Notice (0, 3) is not a solution of
inequality 1. This point is not a solution of
the system.
When graphing a system of inequalities, it is
helpful to find each corner point (or vertex).
For instance, this graph has three corner
points: (–1, 2), (–1, –3), and (4, 2).
y<2
x  –1
y>x–2
Inequality 1
Inequality 2
Inequality 3
Solution Region Between Parallel Lines
Write a system of inequalities that defines the shaded region shown.
SOLUTION
The graph of one inequality is the half-plane below
the line y = 3.
The graph of the other inequality is the half-plane
above the line y = 1.
The shaded region of the graph is the horizontal band that lies between the two
horizontal lines, y = 3 and y = 1, but not on the lines.
The system of linear
inequalities at the right
defines the shaded region.
y<3
y>1
Inequality 1
Inequality 2
A Quadrilateral Solution Region
Graph the system of linear inequalities. Label each vertex of the solution
region. Describe the shape of the region.
x0
The graph of the first inequality is
the half-plane on and to the right
of the y-axis.
A Quadrilateral Solution Region
Graph the system of linear inequalities. Label each vertex of the solution
region. Describe the shape of the region.
x0
y0
The graph of the first inequality is
the half-plane on and to the right
of the y-axis.
The graph of the second
inequality is the half-plane on and
above of the x-axis.
A Quadrilateral Solution Region
Graph the system of linear inequalities. Label each vertex of the solution
region. Describe the shape of the region.
y2
The graph of the third inequality is
the half-plane on and below the
horizontal line y = 2.
A Quadrilateral Solution Region
Graph the system of linear inequalities. Label each vertex of the solution
region. Describe the shape of the region.
y2
The graph of the third inequality is
the half-plane on and below the
horizontal line y = 2.
1
y  – 2 x + 3.
The graph of the fourth inequality is
the half-plane on and below the line
1 x + 3.
y= –
2
A Quadrilateral Solution Region
Graph the system of linear inequalities. Label each vertex of the solution
region. Describe the shape of the region.
The region that lies in all four half-planes is a quadrilateral with vertices at
(0, 2), (0, 0), (6, 0), and (2, 2).
Note that (0, 3) is not a vertex of the solution region even though two
boundary lines meet at that point.
Modeling A Real-Life Problem
You are ordering lighting for a theater so the spotlights can follow the
performers. The lighting technician needs at least 3 medium-throw spotlights and
at least 1 long-throw spotlight. A medium-throw spotlight costs $1000 and a longthrow spotlight costs $3500. The minimum order for free delivery is $10,000.
Write and graph a system of linear inequalities that shows how many mediumthrow spotlights and long-throw spotlights should be ordered to get the free
delivery.
Verbal Model
Number of medium-throws 
Number of long-throws 
Number of
mediumthrows
…
•
Price of a
mediumthrow
3
1
+
Number
of longthrows
•
Price of a
longthrow
 10,000
Modeling A Real-Life Problem
You are ordering lighting for a theater so the spotlights can follow the
performers. The lighting technician needs at least 3 medium-throw spotlights and
at least 1 long-throw spotlight. A medium-throw spotlight costs $1000 and a longthrow spotlight costs $3500. The minimum order for free delivery is $10,000.
Write and graph a system of linear inequalities that shows how many mediumthrow spotlights and long-throw spotlights should be ordered to get the free
delivery.
…
Labels
…
Number of medium-throws = x
(no units)
Number of long-throws = y
(no units)
Price of a medium-throw = 1000
(dollars)
Price of a long-throw = 3500
(dollars)
Modeling A Real-Life Problem
You are ordering lighting for a theater so the spotlights can follow the
performers. The lighting technician needs at least 3 medium-throw spotlights and
at least 1 long-throw spotlight. A medium-throw spotlight costs $1000 and a longthrow spotlight costs $3500. The minimum order for free delivery is $10,000.
Write and graph a system of linear inequalities that shows how many mediumthrow spotlights and long-throw spotlights should be ordered to get the free
delivery.
…
Algebraic
Model
x3
Inequality 1
y1
Inequality 2
1000x + 3500y  10,000
Inequality 3
Modeling A Real-Life Problem
You are ordering lighting for a theater so the spotlights can follow the
performers. The lighting technician needs at least 3 medium-throw spotlights and
at least 1 long-throw spotlight. A medium-throw spotlight costs $1000 and a longthrow spotlight costs $3500. The minimum order for free delivery is $10,000.
Write and graph a system of linear inequalities that shows how many mediumthrow spotlights and long-throw spotlights should be ordered to get the free
delivery.
The graph of the system of inequalities is
shown.
Any point in the shaded region of the
graph is a solution to the system.
A fraction of a spotlight cannot be ordered, so
only ordered pairs of integers in the shaded
region will correctly answer the problem.
Modeling A Real-Life Problem
You are ordering lighting for a theater so the spotlights can follow the
performers. The lighting technician needs at least 3 medium-throw spotlights and
at least 1 long-throw spotlight. A medium-throw spotlight costs $1000 and a longthrow spotlight costs $3500. The minimum order for free delivery is $10,000.
Will an order of 4 medium-throw spotlights and 1 long-throw spotlight be
delivered free?
The point (4, 1) is outside the solution
region, so an order of 4 medium-throw
spotlights and 1 long-throw spotlight would
not be delivered free.
Chapter 8
Exponents and Exponential Functions
8.1 Multiplication Properties of Exponents
8.2 Zero and Negative Exponents
8.3 Division properties of Exponents
8.4 Scientific Notation
8.5 Exponential Growth Functions
8.6 Exponential Decay Functions
Chapter 8.1
Multiplication Properties of Exponents
INVESTIGATING POWERS
How can you use addition to multiply exponential expressions?
1
Complete the table. To simplify an expression, expand the product.
Then count the factors.
Product
of powers
2
Expanded product
Number Product as
of factors a power
73 · 72
(7 · 7 · 7) · (7 · 7)
5
75
24 · 24
(2 · 2 · 2 · 2) · (2 · 2 · 2 · 2)
8
28
x4 · x5
(x · x · x · x) · (x · x · x · x · x)
9
x9
Add a column to your table that shows the sum of the exponents that
are in the first column. What pattern do you notice?
INVESTIGATING POWERS
How can you use multiplication to raise an exponential expression to a power?
3 Complete the table. To simplify an expression, expand the product.
Then count the factors.
Power of
Expanded product
a power
(5 2) 3
(5 2) · (5 2) · (5 2)
[(–3) 2] 2 [(–3) 2] · [(–3) 2]
(b 2) 4
4
Expanded product
Number Product as
of factors a power
(5 · 5) · (5 · 5) · (5 · 5)
6
56
[(–3) · (–3)] · [(–3) · (–3)]
4
(–3) 4
8
b8
(b 2) · (b 2) · (b 2) · (b 2) (b · b) · (b · b) · (b · b) · (b · b)
Add a column to your table that shows the product of the exponents that
are in the first column. What pattern do you notice?
Multiplying Exponential Expressions
To multiply two powers that have the same base, you add the exponents.
Here is an example.
5 factors
a2 · a3 = a · a · a · a · a = a5 = a2 + 3
2 factors 3 factors
Multiplying Exponential Expressions
To multiply two powers that have the same base, you add the exponents.
Here is an example.
5 factors
a2 · a3 = a · a · a · a · a = a5 = a2 + 3
2 factors 3 factors
To find a power of a power, you multiply the exponents. Here is an example.
3 factors
6 factors
(a 2) 3 = a 2 · a 2 · a 2 = a · a · a · a · a · a = a 6 = a 2 · 3
Multiplying Exponential Expressions
CONCEPT
SUMMARY
MULTIPLICATION PROPERTIES OF EXPONENTS
Let a and b be numbers and let m and n be positive integers.
PRODUCT OF POWERS PROPERTY
To multiply powers having the same base, add the exponents.
a m · a n = a m+n
Example: 3 2 · 3 7 = 3 2+7 = 3 9
POWER OF A POWER PROPERTY
To find the power of a power, multiply the exponents.
(a m) n = a m · n
Example: (5 2) 4 = 5 2 · 4 = 5 8
POWER OF A PRODUCT PROPERTY
To find the power of a product, find the power of each factor and multiply
(a · b) m = a m · b m
Example: (2 · 3) 6 = 2 6 · 3 6
Using the Product of Powers Property
Use the product of powers property to simplify the expression.
53 · 56= 53 + 6
= 59
Using the Product of Powers Property
Use the product of powers property to simplify the expression.
53 · 56= 53 + 6
= 59
x2 · x3 · x4 = x2 + 3 + 4
= x9
Using the Power of a Power Property
When you use the power of a power property, it is the quantity within the
parentheses that is raised to the power, not the individual terms.
Correct
(a + 1) 3 = (a + 1)(a + 1)(a + 1)
Incorrect
(a + 1) 3 = a 3 + 1 3
Using the Power of a Power Property
When you use the power of a power property, it is the quantity within the
parentheses that is raised to the power, not the individual terms.
Correct
(a + 1) 3 = (a + 1)(a + 1)(a + 1)
Incorrect
(a + 1) 3 = a 3 + 1 3
Use the power of a power property to simplify the expression.
( y 2) 4 = y 2 · 4
= y8
Using the Power of a Power Property
When you use the power of a power property, it is the quantity within the
parentheses that is raised to the power, not the individual terms.
Correct
(a + 1) 3 = (a + 1)(a + 1)(a + 1)
Incorrect
(a + 1) 3 = a 3 + 1 3
Use the power of a power property to simplify the expression.
( y 2) 4 = y 2 · 4
= y8
[(a + 1) 2 ] 5 = (a+1) 2 · 5
= (a+1) 10
Using the Power of a Product Property
Use the power of a product property to simplify the expression.
(4yz) 3 = (4 · y · z)3
Identify factors.
= 43 · y3 · z3
Raise each factor to a power.
= 64y 3z 3
Simplify.
Using the Power of a Product Property
Use the power of a product property to simplify the expression.
(4yz) 3 = (4 · y · z)3
Identify factors.
= 43 · y3 · z3
Raise each factor to a power.
= 64y 3z 3
Simplify.
– (2w) 2= – (2 · w) 2
Identify factors.
= – (2 2 · w 2)
Raise each factor to a power.
= – 4w 2
Simplify.
Using All Three Properties
Simplify (4x 2y) 3 · x 5.
SOLUTION
(4x 2y) 3 · x 5 = 4 3 · (x 2) 3 · y 3 · x 5
Power of a product
= 64 · x 6 · y 3 · x 5
Power of a power
= 64x 11y 3
Product of powers
Using Powers in Real Life
Some farmers use center-pivot irrigation, which is a sprinkler system that
revolves around a center pivot. The large irrigation circles help farmers to
conserve water, maximize crop yield, and reduce cost of pesticides.
Find the ratio of the area of the larger irrigation circle to the area of the
smaller irrigation circle.
SOLUTION
(2r) 2
Ratio =
r 2
 · 22 · r2
=  · r2
 · 4 · r2
=  · r2
4
= 1
Using Powers in Real Life
Some farmers use center-pivot irrigation, which is a sprinkler system that
revolves around a center pivot. The large irrigation circles help farmers to
conserve water, maximize crop yield, and reduce cost of pesticides.
Write a general statement about doubling the radius of an irrigation circle.
SOLUTION
Doubling the radius of an irrigation circle makes the area four times as large.
Chapter 8.2
Zero and Negative Exponents
Chapter 8.3
Division Properties of Exponents
Chapter 8.4
Scientific Notation
Chapter 8.5
Exponential Growth Functions
WRITING EXPONENTIAL GROWTH MODELS
A quantity is growing exponentially if it increases by the same
percent in each time period.
EXPONENTIAL GROWTH MODEL
C is the initial amount.
t is the time period.
y = C (1 + r)t
(1 + r) is the growth factor, r is the growth rate.
The percent of increase is 100r.
Finding the Balance in an Account
COMPOUND INTEREST You deposit $500 in an account that pays 8% annual
interest compounded yearly. What is the account balance after 6 years?
SOLUTION
METHOD 1 SOLVE A SIMPLER PROBLEM
Find the account balance A1 after 1 year and multiply by the growth factor to
find the balance for each of the following years. The growth rate is 0.08, so
the growth factor is 1 + 0.08 = 1.08.
A1 = 500(1.08) = 540
Balance after one year
A2 = 500(1.08)(1.08) = 583.20
Balance after two years
A3 = 500(1.08)(1.08)(1.08) = 629.856
Balance after three years
A6 =
Balance after six years
•
•
•
500(1.08) 6 793.437
•
•
•
Finding the Balance in an Account
COMPOUND INTEREST You deposit $500 in an account that pays 8% annual
interest compounded yearly. What is the account balance after 6 years?
SOLUTION
METHOD 2 USE A FORMULA
Use the exponential growth model to find the account balance A. The growth
rate is 0.08. The initial value is 500.
EXPONENTIAL GROWTH MODEL
is the
initial
amount.
C 500
is the
initial
amount.
6tisisthe
period.
thetime
time
period.
A6y==500
C (1
(1++0.08)
r)t 6
(1(1++0.08)
thegrowth
growth factor,
factor, 0.08
is the
growthrate.
rate.
r) isisthe
r is the
growth
6  793.437
The)percent
of increase
is 100r.
A6 = 500(1.08
Balance
after 6 years
Writing an Exponential Growth Model
A population of 20
rabbits is released
into a wildlife region.
The population
triples each year for
5 years.
Writing an Exponential Growth Model
A population of 20 rabbits is released into a wildlife region.
The population triples each year for 5 years.
a. What is the percent of increase each year?
SOLUTION
The population triples each year, so the growth factor is 3.
1+r = 3
So, the growth rate r is 2 and the percent of increase each
year is 200%.
Reminder: percent increase is 100r.
Writing an Exponential Growth Model
A population of 20 rabbits is released into a wildlife region.
The population triples each year for 5 years.
b. What is the population after 5 years?
SOLUTION
After 5 years, the population is
P = C(1 + r) t
Exponential growth model
= 20(1 + 2) 5
Substitute C, r, and t.
= 20 • 3 5
Simplify.
= 4860
Evaluate.
There will be about 4860 rabbits after 5 years.
Help
GRAPHING EXPONENTIAL GROWTH MODELS
A Model with a Large Growth Factor
Graph the growth of the rabbit population.
Make a table of values, plot the points in a coordinate
plane, and draw a smooth curve through the points.
SOLUTION
t 0
P 20
1
2
3
60
180
540
4
5
1620 4860
6000
Population
5000
4000
P = 20 ( 3 ) t
3000
Here, the large
growth factor of 3
corresponds to a
rapid increase
2000
1000
0
1
2
3
4
Time (years)
5
6
7
Chapter 8.6
Exponential Decay Functions
WRITING EXPONENTIAL DECAY MODELS
A quantity is decreasing exponentially if it decreases by the same
percent in each time period.
EXPONENTIAL DECAY MODEL
C is the initial amount.
t is the time period.
y = C (1 – r)t
(1 – r ) is the decay factor, r is the decay rate.
The percent of decrease is 100r.
Writing an Exponential Decay Model
From 1982 through 1997, the purchasing power
of a dollar decreased by about 3.5% per year. Using 1982 as the base
for comparison, what was the purchasing power of a dollar in 1997?
COMPOUND INTEREST
SOLUTION
Let y represent the purchasing power and let t = 0 represent the year
1982. The initial amount is $1. Use an exponential decay model.
y = C(1 – r) t
Exponential decay model
= (1)(1 – 0.035) t
Substitute 1 for C, 0.035 for r.
= 0.965 t
Simplify.
Because 1997 is 15 years after 1982, substitute 15 for t.
y = 0.96515
Substitute 15 for t.
0.59
The purchasing power of a dollar in 1997 compared to 1982 was $0.59.
GRAPHING EXPONENTIAL DECAY MODELS
Graphing the Decay of Purchasing Power
Graph the exponential decay model in the previous example.
Use the graph to estimate the value of a dollar in ten years.
SOLUTION
Help
Make a table of values, plot the points in a coordinate
plane, and draw a smooth curve through the points.
t 0
1
2
3
4
5
6
7
8
9
10
y 1.00 0.965 0.931 0.899 0.867 0.837 0.808 0.779 0.752 0.726 0.7
y = 0.965t
0.8
(dollars)
Purchasing Power
1.0
0.6
0.4
0.2
0
1
2
3
Your dollar of today
will be worth about 70
cents in ten years.
4
5
6
7
8
Years From Now
9
10
11
12
GRAPHING EXPONENTIAL DECAY MODELS
CONCEPT
EXPONENTIAL GROWTH AND DECAY MODELS
SUMMARY
EXPONENTIAL GROWTH MODEL
y = C (1 + r)t
EXPONENTIAL DECAY MODEL
y = C (1 – r)t
An exponential model y = a • b t represents exponential
b > 1 and
exponential (1
decay
0 <decay
b < 1.factor,
(1 growth
+ r) is theifgrowth
factor,
– r) isifthe
Ct is
is the
the initial
time period.
amount.
C)
(0, C) rate.
r is the(0,growth
rate.
r is the decay
1+r>1
0<1–r<1
Chapter 9
Quadratic Equations and Functions
9.1 Solving Quad. Equations by Finding Sq. Roots
9.2 Simplifying Radicals
9.3 Graphing Quad. Functions
9.4 Solving Quad. Equations by Graphing
9.5 Solving Quad. Equations by the Quadratic Formula
9.6 Applications of the Discriminant
9.7 Graphing Quad. Inequalities
9.8 Comparing Linear, Exponential, & Quad. Models
Chapter 9.1
Chapter 9.2
Chapter 9.3
Chapter 9.4
SOLVING QUADRATIC EQUATIONS GRAPH ICALLY
SOLVING QUADRATIC EQUATIONS USING GRAPHS
The solution of a quadratic equation in one variable x can be solved
or checked graphically with the following steps:
STEP 1
Write the equation in the form ax 2 + bx + c = 0.
STEP 2
Write the related function y = ax 2 + bx + c.
STEP 3
Sketch the graph of the function y = ax 2 + bx + c.
The solutions, or roots, of ax 2 + bx + c = 0 are the x-intercepts.
Checking a Solution Using a Graph
Solve
1 2
x = 8 algebraically.
2
Check your solution graphically.
SOLUTION
1 2
x = 8
2
CHECK
Write original equation.
x 2 = 16
Multiply each side by 2.
x= 4
Find the square root of each side.
Check these solutions using a graph.
Checking a Solution Using a Graph
CHECK Check these solutions using a graph.
1
Write the equation in the form ax 2 + bx + c = 0
1 2
x =8
2
1 2
x –8=0
2
2
Rewrite original equation.
Subtract 8 from both sides.
Write the related function y = ax2 + bx + c.
y = 1 x2 – 8
2
Checking a Solution Using a Graph
CHECK Check these solutions using a graph.
2
Write the related function
y = ax2 + bx + c.
–4, 0
y = 1 x2 – 8
2
3
1
Sketch graph of y = x2 – 8.
2
The x-intercepts are  4, which
agrees with the algebraic solution.
4, 0
Solving an Equation Graphically
Solve x 2 – x = 2 graphically.
Check your solution algebraically.
SOLUTION
1
Write the equation in the form ax 2 + bx + c = 0
x2 – x = 2
x2 – x – 2 = 0
2
Write original equation.
Subtract 2 from each side.
Write the related function y = ax2 + bx + c.
y = x2 – x – 2
Solving an Equation Graphically
2
Write the related function y = ax2 + bx + c.
y = x2 – x – 2
3
Sketch the graph of the function
y = x2 – x – 2
From the graph, the x-intercepts
appear to be x = –1 and x = 2.
– 1, 0
2, 0
Solving an Equation Graphically
From the graph, the x-intercepts
appear to be x = –1 and x = 2.
– 1, 0
CHECK
You can check this by substitution.
Check x = –1:
Check x = 2:
x2 – x = 2
x2 – x = 2
?
?
(–1) = 2
22 – 2 = 2
1+1=2
4–2=2
(–1) 2 –
2, 0
USING QUADRATIC MODELS IN REAL LIFE
Comparing Two Quadratic Models
A shot put champion performs an experiment for your math class.
Assume that both times he releases the shot with the same initial
speed but at different angles. The path of each put can be modeled
by one of the equations below, where x represents the horizontal
distance (in feet) and y represents the height (in feet).
Initial angle of 35º: y = – 0.010735x 2 + 0.700208x + 6
Initial angle of 65º: y = – 0.040330x 2 + 2.144507x + 6
Which angle results in a farther throw?
Comparing Two Quadratic Models
SOLUTION
Begin by graphing both models in the same coordinate
plane. Use only positive x-values because x represents
the distance of the throw.
Which angle results in a farther throw?
Comparing Two Quadratic Models
PARABOLIC MOTION An angle of 45 º produces the maximum
distance when an object is propelled from ground level. An angle
smaller than 45 º is better when the shot is released above the
ground. If the shot is released from 6 feet above the ground, a 43 º
angle produces a maximum distance of 74.25 feet.
45º
65º
43º
35º
Chapter 9.5
Using the Quadratic Formula
In this lesson you will learn how to use the quadratic formula to solve
any quadratic equation.
THE QUADRATIC FORMULA
The solutions of the quadratic equation a x 2 + b x + c = 0 are
x=
–b 
b2 – 4 a c
2a
when a  0 and b 2 – 4 a c ≥ 0.
You can read this formula as:
x equals the opposite of b, plus or
minus the square root of b squared
minus 4 a c, all divided by 2 a.
Using the Quadratic Formula
Solve x 2 + 9x + 14 = 0.
SOLUTION
Use the quadratic formula.
1x 2 + 9x + 14 = 0
x=
x=
x=
– 9b 
9b 2 – 4( 1a )(14
c)
2( a1 )
Identify a = 1, b = 9, and c = 14.
Substitute values in the
quadratic formula.
–9 
81 – 56
2
Simplify.
–9 
25
Simplify.
2
The equation has
two solutions:
–9 + 5
= –2
x=
2
–9  5
x=
2
–9 – 5
= –7
x=
2
Using the Quadratic Formula
Check the solutions to x 2 + 9x + 14 = 0 in the original equation.
Check x = –2 :
Check x = –7 :
x 2 + 9x + 14 = 0
?
(–2) 2 + 9(–2) + 14 = 0
?
x 2 + 9x + 14 = 0
?
(–7) 2 + 9(–7) + 14 = 0
?
4 + –18 + 14 = 0
49 + –63 + 14 = 0
0=0
0=0
Finding the x-Intercepts of a Graph
Find the x-intercepts of the graph of y = –x 2 – 2 x + 5.
SOLUTION
The x-intercepts occur when y = 0.
y = –x 2 – 2 x + 5
Write original equation.
0 = –1x 2 – 2 x + 5
Substitute 0 for y, and
identify a = 1, b = –2, and c = 5.
–(–2
b)
x=
x=
x=
2
2
(–2
b ) 2 – 4(–1
a )( 5c )
2(–1
a)
4 + 20
–2
24
–2
Substitute values in the
quadratic formula.
The equation has
two solutions:
x=
Simplify.
x=
Solutions
x=
2+
24
 –3.45
–2
2
24
–2
2–
24
 1.45
–2
Finding the x-Intercepts of a Graph
Check your solutions to the equation y = –x 2 – 2 x + 5 graphically.
Check y  –3.45 and y  1.45.
You can see that the graph shows
the x-intercepts between –3 and –4
and between 1 and 2.
Using Quadratic Models in Real Life
Problems involving models for the dropping or throwing of an object are called
vertical motion problems.
VERTICAL MOTION MODELS
OBJECT IS DROPPED:
h = –16 t 2 + s
OBJECT IS THROWN:
h = –16 t 2 + v t + s
h = height (feet)
t = time in motion (seconds)
s = initial height (feet)
v = initial velocity (feet per second)
In these models the coefficient of t 2 is one half the acceleration due to gravity.
On the surface of Earth, this acceleration is approximately 32 feet per second
per second.
Remember that velocity v can be positive (object moving up), negative (object
moving down), or zero (object not moving). Speed is the absolute value of velocity.
Modeling Vertical Motion
BALLOON COMPETITION
You are competing in the Field Target Event at a hot-air balloon festival. You
throw a marker down from an altitude of 200 feet toward a target. When the
marker leaves your hand, its speed is 30 feet per second. How long will it take
the marker to hit the target?
SOLUTION
Because the marker is thrown down, the initial velocity is v = –30 feet per
second. The initial height is s = 200 feet. The marker will hit the target when the
height is 0.
v = –30, s = 200, h = 0.
Modeling Vertical Motion
BALLOON COMPETITION
SOLUTION
v = –30, s = 200, h = 0.
h = –16 t 2 + v t + s
Choose the vertical motion model for a thrown object.
h = –16 t 2 + (–30)
v t +200
s
Substitute values for v and s into the
vertical motion model.
0h = –16 t 2 – 30t + 200
Substitute 0 for h. Write in standard form.
t=
t=
–(–30
b )
30 
(–30
b ) 2 – 4(–16
a )(200
c )
2(–16
a )
13,700
–32
t  2.72 or –4.60
Substitute values in the quadratic formula.
Simplify.
Solutions
Modeling Vertical Motion
BALLOON COMPETITION
SOLUTION
t  2.72 or –4.60
As a solution, –4.60 does not make sense in the context of the problem.
Therefore, the weighted marker will hit the target about 2.72 seconds after it
was thrown.
Chapter 9.6
Chapter 9.7
Chapter 9.8
Chapter 10
Chapter 10.1
MODELING ADDITION OF POLYNOMIALS
Algebra tiles can be used to model polynomials.
+
–
1
–1
These 1-by-1 square
tiles have an area of
1 square unit.
+
–
+
–
x
–x
x2
–x 2
These 1-by-x rectangular
tiles have an area of x
square units.
These x-by-x rectangular
tiles have an area of x 2
square units.
MODELING ADDITION OF POLYNOMIALS
You can use algebra tiles to add the polynomials x 2 + 4x + 2 and 2 x 2 – 3x – 1.
1
Form the polynomials x 2 + 4x + 2 and 2 x 2 – 3x – 1 with algebra tiles.
x2
+
4x
+ 2
+
+
+
+
+
+
+
–
2 x2
+
+
– 1
3x
–
–
–
–
MODELING ADDITION OF POLYNOMIALS
You can use algebra tiles to add the polynomials x 2 + 4x + 2 and 2 x 2 – 3x – 1.
2
To add the polynomials, combine like terms. Group the x 2-tiles, the x-tiles,
and the 1-tiles.
x 2 + 4x + 2
+
+
+
+
+
+
+
+
2x 2 – 3x – 1
+
+
–
+
+
+
+
+
+
+
+
+
=
+
–
–
–
–
–
–
–
MODELING ADDITION OF POLYNOMIALS
You can use algebra tiles to add the polynomials x 2 + 4x + 2 and 2 x 2 – 3x – 1.
2
3
To add the polynomials, combine like terms. Group the x 2-tiles, the x-tiles,
and the 1-tiles.
2 + 4x + 2
xFind
and remove the zero pairs. +
+
+ sum is
+ 3x+2 + x+ + 1.
+
The
+
2x 2 – 3x – 1
+
+
–
+
+
+
+
+
+
+
+
+
=
+
–
–
–
–
–
–
–
Adding and Subtracting Polynomials
An expression which is the sum of terms of the form a x k where k is a
nonnegative integer is a polynomial. Polynomials are usually written in
standard form.
Standard form means that the terms of the polynomial are placed in descending
order, from largest degree to smallest degree.
Polynomial in standard form:
2 x 3 + 5x 2 – 4 x + 7
Leading coefficient
Degree
Constant term
The degree of each term of a polynomial is the exponent of the variable.
The degree of a polynomial is the largest degree of its terms. When a
polynomial is written in standard form, the coefficient of the first term is
the leading coefficient.
Classifying Polynomials
A polynomial with only one term is called a monomial. A polynomial with two
terms is called a binomial. A polynomial with three terms is called a trinomial.
Identify the following polynomials:
Degree
Classified by
degree
Classified by
number of terms
6
0
constant
monomial
–2 x
1
linear
monomial
3x + 1
1
linear
binomial
–x 2 + 2 x – 5
2
quadratic
trinomial
4x 3 – 8x
3
cubic
binomial
2 x 4 – 7x 3 – 5x + 1
4
quartic
polynomial
Polynomial
Adding Polynomials
Find the sum. Write the answer in standard format.
(5x 3 – x + 2 x 2 + 7) + (3x 2 + 7 – 4 x) + (4x 2 – 8 – x 3)
SOLUTION
Vertical format: Write each expression in standard form. Align like terms.
5x 3 + 2 x 2 – x + 7
3x 2 – 4 x + 7
3
2
+ – x + 4x
–8
4x 3 + 9x 2 – 5x + 6
Adding Polynomials
Find the sum. Write the answer in standard format.
(2 x 2 + x – 5) + (x + x 2 + 6)
SOLUTION
Horizontal format: Add like terms.
(2 x 2 + x – 5) + (x + x 2 + 6) = (2 x 2 + x 2) + (x + x) + (–5 + 6)
= 3x 2 + 2 x + 1
Subtracting Polynomials
Find the difference.
(–2 x 3 + 5x 2 – x + 8) – (–2 x 2 + 3x – 4)
SOLUTION
Use a vertical format. To subtract, you add the opposite. This means you
multiply each term in the subtracted polynomial by –1 and add.
–2 x 3 + 5x 2 – x + 8
– –2 x 3
+ 3x – 4
No change
Add the opposite
–2 x 3 + 5x 2 – x + 8
+ 2 x3
– 3x + 4
Subtracting Polynomials
Find the difference.
(–2 x 3 + 5x 2 – x + 8) – (–2 x 2 + 3x – 4)
SOLUTION
Use a vertical format. To subtract, you add the opposite. This means you
multiply each term in the subtracted polynomial by –1 and add.
–2 x 3 + 5x 2 – x + 8
– –2 x 3
+ 3x – 4
–2 x 3 + 5x 2 – x + 8
+ 2 x3
– 3x + 4
5x 2 – 4x + 12
Subtracting Polynomials
Find the difference.
(3x 2 – 5x + 3) – (2 x 2 – x – 4)
SOLUTION
Use a horizontal format.
(3x 2 – 5x + 3) – (2 x 2 – x – 4) = (3x 2 – 5x + 3) + (–1)(2 x 2 – x – 4)
= (3x 2 – 5x + 3) – 2 x 2 + x + 4
= (3x 2 – 2 x 2) + (– 5x + x) + (3 + 4)
= x 2 – 4x + 7
Using Polynomials in Real Life
You are enlarging a 5-inch by 7-inch photo by a scale factor of x and mounting
it on a mat. You want the mat to be twice as wide as the enlarged photo and 2
inches less than twice as high as the enlarged photo.
Write a model for the area of the mat around the photograph as a function of the
scale factor.
Use a verbal model.
Verbal Model
Area of mat = Total Area –
Labels
…
Area of
photo
7x
Area of mat = A
(square inches)
5x
Total Area = (10x)(14x – 2)
(square inches)
10x
Area of photo = (5x)(7x)
(square inches)
14x – 2
SOLUTION
Using Polynomials in Real Life
You are enlarging a 5-inch by 7-inch photo by a scale factor of x and mounting
it on a mat. You want the mat to be twice as wide as the enlarged photo and 2
inches less than twice as high as the enlarged photo.
Write a model for the area of the mat around the photograph as a function of the
scale factor.
SOLUTION
…
= 140x 2 – 20x – 35x 2
5x
= 105x 2 – 20x
10x
14x – 2
Algebraic
Model
7x
A = (10x)(14x – 2) – (5x)(7x)
A model for the area of the mat around the photograph as a function of the
scale factor x is A = 105x 2 – 20x.
Chapter 10.2
Multiplying Polynomials
You have already learned how to multiply a polynomial by a monomial by using
the distributive property.
(3x)(2x 2 – 5x + 3)
= (3x)(2x 2) + (3x)(–5x) + (3x)(3) = 6x 3 – 15x 2 + 9x
In the following slide, you will see how an area model illustrates the
multiplication of two binomials
Multiplying Polynomials
ACTIVITY
Developing
Concepts
INVESTIGATING BINOMIAL MULTIPLICATION
The rectangle shown at the right has a width of (x + 2) and a height of (2x + 1)
1
What is the area of each part of the
rectangle?
x
2
x
1
1
x2
x
x
Find the product of (x + 2) and (2 x + 1) by
adding the areas of the parts to get an
expression for total area.
2x + 1
x
x2
x
x
1
x
1
1
(x + 2)(2x + 1) = x 2 + x 2 + x + x + x + x + x + 1 + 1
= 2x 2 + 5x + 2
x+2
Multiplying Polynomials
Another way to multiply two binomials is to use the distributive property twice.
First
use (a + b)(c + d) = a(c + d) + b(c + d).
Then
use a(c + d) = ac + ad and b(c + d) = bc + bd.
This shows that (a + b)(c + d) = ac + ad + bc + bd.
This property can also be applied to the binomials of the form a – b or c – d.
Using the Distributive Property
Find the product (x + 2)(x – 3).
SOLUTION
(x + 2)(x – 3) = x(x – 3) + 2(x – 3)
(b + c) a = ba + ca
= x 2 – 3x + 2x – 6
a (b – c) = ab – ac
= x2 – x – 6
Combine like terms.
Multiplying Polynomials
When you multiply two binomials, you can remember the results given by the
distributive property by means of the FOIL pattern.
Multiply the First, Outer, Inner, and Last terms.
Product of
First Terms
(3x + 4)(x + 5)
= 3x 2
Product of
Outer Terms
+ 15x
= 3x 2 + 19x + 20
Product of
Inner Terms
+ 4x
+ 20
Product of
Last Terms
Multiplying Binomials Using the FOIL Pattern
Find the product (3x – 4)(2x + 1).
SOLUTION
First, Outer, Inner, and Last.
F
(3x – 4)(2x + 1)
O
I
L
= 6x 2 + 3x – 8x – 4
Mental math.
= 6x 2 – 5x – 4
Simplify.
Multiplying Polynomials Vertically
To multiply two polynomials that have three or more terms, remember that each
term of one polynomial must be multiplied by each term of the other polynomial.
Use a vertical or a horizontal format. Write each polynomial in standard form.
Find the product (x – 2)(5 + 3x – x 2).
SOLUTION
Align like terms in columns.
– x 2 + 3x + 5
Standard form
x–2
Standard form
2x 2 – 6x – 10
–2(– x2 + 3x + 5)

–x 3 + 3x 2 + 5x
–x 3 + 5x 2 – x – 10
x(– x2 + 3x + 5)
Combine like terms.
Multiplying Polynomials Horizontally
Find the product (4x 2 – 3x – 1)(2x – 5).
SOLUTION
Multiply 2x – 5 by each term of 4x 2 – 3x – 1.
(4x 2 – 3x – 1)(2x – 5) = 4x 2(2x – 5) – 3x(2x – 5) – 1(2x – 5)
= 8x 3 – 20x 2 – 6x 2 + 15x – 2x + 5
= 8x 3 – 26x 2 + 13x + 5
Using Polynomials in Real Life
The diagram below shows the basic dimensions for a window. The glass
portion of the window has a height-to-width ratio of 3 : 2. The framework adds
6 inches to the width and 10 inches to the height.
Write a polynomial expression that represents the total area
of the window, including the framework.
SOLUTION
5
Use a verbal model.
3x
Verbal Model
Labels
Total area =
Height
Width
·
of window
of window
5
Total area = A
(square inches)
3
…
Height of window = 3x + 10
(inches)
Width of window
(inches)
= 2x + 6
2x
3
Using Polynomials in Real Life
The diagram below shows the basic dimensions for a window. The glass
portion of the window has a height-to-width ratio of 3 : 2. The framework adds
6 inches to the width and 10 inches to the height.
Write a polynomial expression that represents the total area
of the window, including the framework.
5
SOLUTION
3x
…
Algebraic
Model
A = (3x + 10) · (2x +
6)
= 6x 2 + 18x + 20x + 60
Area model
5
FOIL pattern
3
= 6x 2 + 38x + 60
Combine like terms.
2x
3
Using Polynomials in Real Life
The diagram below shows the basic dimensions for a window. The glass
portion of the window has a height-to-width ratio of 3 : 2. The framework adds
6 inches to the width and 10 inches to the height.
Find the area when x = 10, 11, 12, 13, and 14.
SOLUTION
You can evaluate the polynomial expression 6x 2 + 38x + 60 by substituting
x-values. For instance, to find the total area of the window when x = 10,
substitute 10 for x.
A = 6x 2 + 38x + 60
= 6(10) 2 + 38(10) + 60
= 600 + 380 + 60
= 1040
The areas for all five x-values are
listed in the table.
x (in.)
10
11
12
13
14
A (in. 2)
1040
1204
1380
1568
1768
Chapter 10.3
Chapter 10.4
Chapter 10.5
Chapter 10.6
FACTORING A QUADRATIC TRINOMIAL
To factor quadratic polynomials whose leading factor is not 1:
Find the factors of a (m and n).
Find the factors of c (p and q).
The sum of the outer and
inner products (mq + pn) is b.
a = mn
b = mq + pn
ax2 + bx + c = (mx + p)(nx + q)
c = pq
FACTORING A QUADRATIC TRINOMIAL
a = mn
c = pq
b = mq + pn
22 = 3 • 4 + 5 • 2
6=3•2
6x2 + bx + 20 = (3x + 5)(2x + 4)
c = pq
One Pair of Factors for a and c
Factor 2x2 + 11x + 5
SOLUTION
Test combinations of factors for
a
c
1 and 2
1 and 5
Try a = 1 • 2 and c = 1 • 5
(1x + 1) (2x + 5) = 2x2 + 7x + 5
Not correct.
Try a = 1 • 2 and c = 5 • 1
(1x + 5) (2x + 1) = 2x2 + 11x + 5
Correct.
A Common Factor for a, b, and c
Factor 6x 2 + 2x – 8
SOLUTION
Begin by factoring out the common factor 2.
6x 2 + 2x – 8
The correct 2factorization is
2 (3x – x – 4)
=
6x 2 – 2x – 8 = 2(x + 1)(3x – 4).
Now factor 3x 2 – x – 4 by testing possible factors of a and c.
FACTORS OF a AND c
a = 1 • 3 and c = (2)(–2)
PRODUCT
When
you factor, you canCORRECT?
stop+testing
(x – 2)(3x
2) = 3x2once
– 4x –you
4 find
No
the correct factorization.
2
(x + 2)(3x – 2) = 3x + 4x – 4
No
a = 1 • 3 and c = (–4)(1)
(x – 4)(3x + 1) = 3x2 – 11x – 4
No
a = 1 • 3 and c = (1)(–4)
(x + 1)(3x – 4) = 3x2 – x – 4
Yes
a = 1 • 3 and c = (–2)(2)
SOLVING QUADRATIC EQUATIONS BY FACTORING
Writing a Quadratic Model
A cliff diver jumps from a ledge 48 feet above the ocean with an
initial upward velocity of 8 feet per second. How long will it take
until the diver enters the water?
Use
a vertical
motion
Let hv = 0.
8 and s = 48.
When
the diver
entersmodel.
the water,
Solve the resulting equation for t.
2
h = – 16t + vt + s
Vertical motion model
0 = – 16t 2 + 8t + 48
Write quadratic model.
2
= – 16t + 28t + 48
Substitute values.
Factor out – 8.
0 = (– 8)(2t – t – 6)
SOLUTION
0 = (– 8)(t – 2)(2t + 3)
The solutions are 2 and –
Factor.
3
.
2
Negative values do not make sense, so the only reasonable
solution is t = 2. It will take the diver 2 seconds.
Chapter 11
Chapter 11.1
Chapter 11.2
Chapter 11.3
USING DIRECT AND INVERSE VARIATION
DIRECT VARIATION
The variables x and y vary directly if, for a constant k,
y
= k,
x
or
y = kx,
k  0.
USING DIRECT AND INVERSE VARIATION
INDIRECT VARIATION
The variables x and y vary inversely, if for a constant k,
k
= y,
x
or
xy = k,
k  0.
USING DIRECT AND INVERSE VARIATION
MODELS FOR DIRECT AND INVERSE VARIATION
DIRECT VARIATION
y = kx
k>0
INVERSE VARIATION
y =
k>0
k
x
Using Direct and Inverse Variation
When x is 2, y is 4. Find an equation that relates x and y
in each case.
x and y vary directly
SOLUTION
y
= k
x
4
= k
2
2 = k
Write direct variation model.
Substitute 2 for x and 4 for y.
Simplify.
y
An equation that relates x and y is = 2, or y = 2x.
x
Using Direct and Inverse Variation
When x is 2, y is 4. Find an equation that relates x and y
in each case.
x and y vary inversely
SOLUTION
xy = k
Write inverse variation model.
(2)(4) = k
Substitute 2 for x and 4 for y.
8=k
Simplify.
An equation that relates x and y is xy = 8, or y = 8 .
x
Comparing Direct and Inverse Variation
Compare the direct variation model and the inverse
variation model you just found using x = 1, 2, 3, and 4.
SOLUTION
Make a table using y = 2x and y = x8 .
x
1
2
3
4
Direct Variation: k > 0.
As x increases by 1,
y increases by 2.
y = 2x
2
4
6
8
Inverse Variation: k > 0.
As x doubles (from 1 to 2),
y is halved (from 8 to 4).
y = x8
8
4
8
3
2
Comparing Direct and Inverse Variation
Compare the direct variation model and the inverse
variation model you just found using x = 1, 2, 3, and 4.
SOLUTION
Plot the points and then connect the
points with a smooth curve.
Direct Variation: the graph
for this model is a line passing
through the origin.
Inverse Variation: The graph
for this model is a curve that
gets closer and closer to the
x-axis as x increases and
closer and closer to the y-axis
as x gets close to 0.
Direct
y = 2x
Inverse
y= 8
x
USING DIRECT AND INVERSE VARIATION IN REAL LIFE
Writing and Using a Model
BICYCLING A bicyclist tips
the bicycle when making turn.
The angle B of the bicycle
from the vertical direction is
called the banking angle.
banking angle, B
Writing and Using a Model
BICYCLING The graph below shows a model for the relationship
between the banking angle and the turning radius for a bicycle
traveling at a particular speed. For the values shown, the banking
angle B and the turning radius r vary inversely.
Banking angle (degrees)
r
turning radius
Turning Radius
banking angle, B
Writing and Using a Model
Banking angle (degrees)
r
turning radius
Turning Radius
banking angle, B
Find an inverse variation model that relates B and r.
Use the model to find the banking angle for a turning radius of 5 feet.
Use the graph to describe how the banking angle changes as the
turning radius gets smaller.
Find an inverse variation model
that relates B and r.
SOLUTION
From the graph, you can see that
B = 32° when r = 3.5 feet.
k
B= r
k
32 =
3.5
112 = k
The model is B =
Banking angle (degrees)
Writing and Using a Model
Turning Radius
Write direct variation model.
Substitute 32 for B and 3.5 for r.
Solve for k.
112
, where B is in degrees and r is in feet.
r
Use the model to find the banking
angle for a turning radius of 5 feet.
Banking angle (degrees)
Writing and Using a Model
Turning Radius
SOLUTION
Substitute 5 for r in the model you just found.
112
= 22.4
B=
5
When the turning radius is 5 feet, the banking angle is about 22°.
Writing and Using a Model
Use the graph to describe how
the banking angle changes as
the turning radius gets smaller.
SOLUTION
As the turning radius gets smaller,
smaller, the
thebanking
bankingangle
angle
becomes greater.
greater.The
Thebicyclist
bicyclistleans
leansatatgreater
greaterangles.
angles.
Notice that the increase in the banking angle becomes
more rapid when the turning radius is small.
Chapter 11.4
Chapter 11.5
Chapter 11.6
Chapter 11.7
Chapter 11.8
Solving Rational Equations
A rational equation is an equation that contains rational expressions. The next
two examples show the two basic strategies for solving a rational equation.
Cross Multiplying
Cross multiplying can be used to solve a rational equation.
Solve
y
5
=
.
3
y+2
SOLUTION
y
5
=
y+2
3
5(3) = y ( y + 2)
15 = y 2 + 2y
Write original equation.
Cross multiply.
Simplify.
0 = y 2 + 2y – 15
Write in standard form.
0 = ( y + 5)( y – 3)
Factor right side.
If you set each factor equal to 0, you see that the solutions are –5 and 3.
Check both solutions in the original equation.
Multiplying by the LCD
Cross multiplying can be used only for equations in which each side is a single
fraction. The second method, multiplying by the LCD, works for any rational
equation. Multiplying by the LCD always leads to an equation with no fractions.
2 1
4
Solve x + 3 = x .
SOLUTION
2 1
4
+
=
x 3
x
2
1
4
3x • x + 3x • 3 = 3x • x
6 + x = 12
x=6
CHECK
2 1 ? 4
6 + 3 = 6
4
4
=
6
6
The LCD is 3x.
Multiply each side by 3x.
Simplify.
Subtract 6 from each side.
Substitute 6 for x.
Simplify.
Factoring to Find the LCD
Solve
–4
–10
+1= 2
.
y–3
y + y – 12
SOLUTION
The denominator y 2 + y – 12 factors as ( y + 4)( y – 3), so the LCD is ( y + 4)( y – 3).
Multiply each side of the equation by ( y + 4)( y – 3).
–4
–10
• ( y + 4)( y – 3) + 1 • ( y + 4)( y – 3) = 2
• ( y + 4)( y – 3)
y–3
y + y – 12
–10( y + 4)( y – 3)
–4( y + 4)( y – 3)
+ ( y + 4)( y – 3) =
( y + 4)( y – 3)
y–3
–4( y + 4) + ( y 2 + y – 12) = –10
Factoring to Find the LCD
Solve
–4
–10
+1= 2
.
y–3
y + y – 12
SOLUTION
The denominator y 2 + y – 12 factors as ( y + 4)( y – 3), so the LCD is ( y + 4)( y – 3).
Multiply each side of the equation by ( y + 4)( y – 3).
–4( y + 4) + ( y 2 + y – 12) = –10
–4 y – 16 + y 2 + y – 12 = –10
y 2 – 3y – 28 = –10
y 2 – 3y – 18 = 0
( y – 6)( y + 3) = 0
The solutions are 6 and –3. Check both solutions in the original equation.
Writing and Using a Rational Equation
BATTING AVERAGE You have 35 hits in 140 times at bat. Your batting
average is 35 = 0.250. How many consecutive hits must you get to increase
140
your batting average to 0.300?
SOLUTION
If your hits are consecutive, then you must get a hit each time you
are at bat, so “future hits” is equal to “future times at bat.”
Past hits + Future hits
Verbal
Model
Batting average =
Labels
Batting average = 0.300
(no units)
Past hits = 35
(no units)
Future hits = x
(no units)
Past times at bat = 140
(no units)
Future times at bat = x
(no units)
…
Past times at bat + Future times at bat
Writing and Using a Rational Equation
BATTING AVERAGE You have 35 hits in 140 times at bat. Your batting
average is 35 = 0.250. How many consecutive hits must you get to increase
140
your batting average to 0.300?
SOLUTION
…
Algebraic
Model
35 + x
0.300 =
140 + x
0.300(140 + x) = 35 + x
42 + 0.3x = 35 + x
7 = 0.7x
10 = x
Write algebraic model.
Multiply by LCD.
Use distributive property.
Subtract 0.3x and 35 from each side.
Divide each side by 0.7.
You need to get a hit in each of your next 10 times at bat.
Graphing Rational Functions
The inverse variation models you previously graphed are a type of rational
function.
A rational function is a function of the form
f(x) =
polynomial
.
polynomial
Graphing Rational Functions
The inverse variation models you previously graphed are a type of rational
function.
A rational function is a function of the form
f(x) =
polynomial
.
polynomial
In this lesson you will learn to graph rational functions whose numerators
and denominators are first-degree polynomials. Using long division, such a
function can be written in the form
y=
a
+ k.
x–h
Graphing Rational Functions
RATIONAL FUNCTIONS WHOSE GRAPHS ARE HYPERBOLAS
a
+k
x–h
is a hyperbola whose center is (h, k).
The graph of the rational function y =
The vertical and horizontal lines through the
center are the asymptotes of the hyperbola.
An asymptote is a line that the graph
approaches. While the distance between
the graph and the line approaches zero, the
asymptote is not part of the graph.
Graphing a Rational Function
Sketch the graph of y = 1 + 2.
x
1
+ 2. You can see that the
x–0
center is (0, 2). The asymptotes can be drawn as dashed lines through the
SOLUTION
Think of the function as y =
center. Make a table of values. Then plot the points and connect them with
two smooth branches.
x
y
–4
1.75
–2
1.5
–1
1
– 0.5
0
0
undefined
0.5
4
1
3
2
2.5
4
2.25
If you have drawn your graph correctly, it should be
symmetric about the center (h, k). For example, the
points (0.5, 4) and (–0.5, 0) are the same distance
from the center (0, 2) but in opposite directions.
Graphing a Rational Function When h is Negative
Sketch the graph of y =
SOLUTION
–1
+ 1. Describe the domain.
x+2
For this equation, x – h = x + 2, so h = –2. The graph is a
hyperbola with center at (–2, 1). Plot this point and draw a horizontal and a
vertical asymptote through it. Then make a table of values, plot the points,
and connect them with two smooth branches.
x
y
–6
1.25
–4
1.5
–3
2
–2.5
3
–2
undefined
–1.5
–1
–1
0
0
0.5
2
0.75
The domain is all real numbers except x = –2.
Rewriting before Graphing
Sketch the graph of y =
SOLUTION
2x + 1
.
x+2
Begin by using long division to rewrite the rational function.
2
x + 2 2x + 1
2x + 4
–3
Quotient
y=2+
Remainder
–3
–3
, or y =
+2
x+2
x+2
The graph is a hyperbola with center
at (–2, 2). Plot this point and draw a
horizontal and a vertical asymptote
through it. Make a table of values, plot
the points, and connect them with two
smooth branches.
When making a table of values, include
values of x that are less than h and
values of x that are greater than h.
Chapter 12
Chapter 12.1
Chapter 12.2
Chapter 12.3
Chapter 12.4
MODELING COMPLETING THE SQUARE
Use algebra tiles to complete a perfect square trinomial.
Model the expression x 2 + 6x.
Arrange the x-tiles to
form part of a square.
To complete the square,
add nine 1-tiles.
You have completed the square.
x2
x x x x x x
x
x
x
1
1
1
1
1
1
1
1
1
x2 + 6x + 9 = (x + 3)2
SOLVING BY COMPLETING THE SQUARE
To complete the square of the expression x2 + bx, add the
square of half the coefficient of x.
x2 + bx +
2
( ) (
b
2
=
x+ b
2
2
)
Completing the Square
What term should you add to x2 – 8x so that the result is a
perfect square?
SOLUTION
The coefficient of x is –8, so you should add
to the expression.
–8
2
x – 8x +
2
2
–8
, or 16,
( 2)
2
( )=x
2
– 8x + 16 = (x – 4)2
Completing the Square
Factor 2x2 – x – 2 = 0
SOLUTION
2x2 – x – 2 = 0
Write original equation.
2x2 – x = 2
Add 2 to each side.
x2 – 1 x = 1
Divide each side by 2.
2
x2 –
2
( )
1
1
x+ –
2
4
=1+
1
16
(
1 1
Add –
2 • 2
to each side.
2
1 2
1
= –
, or
) ( 4)
16
Completing the Square
x2 –
2
( )
1
1
x+ –
2
4
(
1
x–
4
x–
2
)
=1+
(
1 1
Add –
2 • 2
1
16
x=
16
Write left side as a fraction.
17
4
1

4
1
The solutions are
+
4
) ( 4)
to each side.
17
=
16
1
= 
4
2
1 2
1
= –
, or
Find the square root of each side.
17
4
Add
1 to each side.
4
1
17
–
 1.28 and
4
4
17
 – 0.78.
4
Completing the Square
1
The solutions are
+
4
CHECK
1
17
–
 1.28 and
4
4
17
 – 0.78.
4
You can check the solutions on a graphing calculator.
CHOOSING A SOLUTION METHOD
Investigating the Quadratic Formula
Perform the following steps on the general quadratic equation
ax2 + bx + c = 0 where a  0.
ax2 + bx = – c
x2
Subtract c from each side.
bx
–c
+ a+ = a
2
Divide each side by a.
2
( )
( )
b
–c + b
x
+
=
( 2a) a 4a
bx
b
x2 + +
a
2a
–c
= a +
2
b
2a
2
2
(
b
x+
2a
2
)
Add the square of half the coefficient
of x to each side.
–
4ac + b 2
=
2
4a
Write the left side as a perfect square.
Use a common denominator to express
the right side as a single fraction.
CHOOSING A SOLUTION METHOD
Investigating the Quadratic Formula
(
b
x+
2a
2
)
2
–
4ac
+
b
=
2
4a

b
x+
=
2a
2
b  4ac
2a
 b2  4ac b
x=
–
2a
2a
x=
2
–b  b  4ac
2a
Use a common denominator to express
the right side as a single fraction.
Find the square root of each side.
Include ± on the right side.
Solve for x by subtracting the same
term from each side.
Use a common denominator to express
the right side as a single fraction.
Chapter 12.5
Chapter 12.6
THE DISTANCE AND MIDPOINT FORMULAS
Investigating Distance:
1
Plot A(2,1) and B(6,4) on a coordinate plane. Then
draw a right triangle that has AB as its hypotenuse.
2
Find and label the coordinates of the vertex C.
3
Find the lengths of the legs of
4
Use the Pythagorean theorem to find AB.
AB
5
ABC.
Remember:
16 + 9 =
C (6, 1)
x
16 + 9 = c
AB = 5
c2
y4B –3 y1A
A (2, 1)
x6B –4 x2A
a2 + b2 = c2
42 + 32 = c2
B (6, 4)
y
25 = c
Finding the Distance Between Two Points
The steps used in the investigation can be used to develop a general formula
for the distance between two points A(x 1, y 1) and B(x 2, y 2).
Using the Pythagorean theorem
B (x 2, y 2 )
y
a2 + b2 = c2
You can write the equation
(x 2 – x 1) 2 + ( y 2 – y 1) 2 = d 2
d
A (x 1, y 1 )
Solving this for d produces the
distance formula.
THE DISTANCE FORMULA
The distance d between the points (x 1, y 1) and (x 2, y 2) is
d =
(x 2 – x 1) 2 + ( y 2 – y 1) 2
y2 – y1
x2 – x1
C (x 2, y 1 )
x
Finding the Distance Between Two Points
Find the distance between (1, 4) and (–2, 3).
SOLUTION
To find the distance, use the distance formula.
d =
(x 2 – x 1) 2 + ( y 2 – y 1) 2
Write the distance formula.
=
2
(x–2
y 1) 2
2 –– x11) + ( y32 – 4
Substitute.
=
10
 3.16
Simplify.
Use a calculator.
Applying the Distance Formula
A player kicks a soccer ball that is 10 yards from a sideline and 5 yards from a
goal line. The ball lands 45 yards from the same goal line and 40 yards from
the same sideline. How far was the ball kicked?
SOLUTION
The ball is kicked from the point (10, 5),
and lands at the point (40, 45). Use the
distance formula.
d =
=
(40 – 10) 2 + (45 – 5) 2
900 + 1600 =
2500 = 50
The ball was kicked 50 yards.
Finding the Midpoint Between Two Points
The midpoint of a line segment is the point on the segment that is equidistant
from its end-points. The midpoint between two points is the midpoint of the line
segment connecting them.
THE MIDPOINT FORMULA
The midpoint between the points (x 1, y 1) and (x 2, y 2) is
(
x1 + x2
2
,
y1 + y2
2
)
Finding the Midpoint Between Two Points
Find the midpoint between the points (–2, 3) and (4, 2). Use a graph to check
the result.
SOLUTION
(
–2 + 4
2
,
Remember, the midpoint formula is
3+2
2
The midpoint is
) =(
(
1,
5
2
2
2
).
,
5
2
) =(
1
,
5
2
(
)
x1 + x2
2
,
y1 + y2
2
).
Finding the Midpoint Between Two Points
Find the midpoint between the points (–2, 3) and (4, 2). Use a graph to check
the result.
CHECK
(–2, 3)
From the graph, you can see that the
5
point 1 ,
appears halfway
2
between (–2, 3) and (4, 2). You can also
(
(1, 5 )
2
(4, 2)
)
use the distance formula to check that
the distances from the midpoint to each
given point are equal.
Applying the Midpoint Formula
You are using computer software to design a video game. You want to place
a buried treasure chest halfway between the center of the base of a palm
tree and the corner of a large boulder. Find where you should place the
treasure chest.
SOLUTION
1 Assign coordinates to the locations of the
two landmarks. The center of the palm tree
is at (200, 75). The corner of the boulder is at
(25, 175).
(25, 175)
(112.5, 125)
(200, 75)
2 Use the midpoint formula to find the point that
is halfway between the two landmarks.
(
25 + 200 175 + 75
, 2
2
) (
=
225 250
2 , 2
)
= (112.5, 125)
Chapter 12.7