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Writing Mathematics With Style
An (algebraic) expression is any combination of numbers, variables,
exponents, mathematical symbols, and mathematical operations, but never
include equal signs or inequality signs!
An equation is a mathematical statement that two quantities are
equal. Therefore, an equation consists of two expressions separated by an
equal sign.
An inequality is a mathematical statement expressing an order
relationship.
Therefore, an inequality consists of two (or more)
expressions separated by one of the inequality signs. If more than one
inequality sign are used, they must be in the same direction!
Classify the following as an expression, equation, inequality, or nonsense:
1) 2x 5
9) | 2x – 3 | > 1
2) 2x 5 = 1
10) a < b < c
3) y – z ≤ 9
11) 2 – 3(4 + x)2 – 3x(x – 3)
4) y – z + 9
12) 2 – 3(4 + x)2 = 3x(x – 3)
1 2
5) x = y
1
2
6) x ≥ y
7) 2 ≤ x ≥ 3
8) 2 ≤ x + 3
13) x2 = x + 6
14) x2 ≥ x + 6
15) x2 ≤ x + 6
16) x2 ≤ x ≤ 6
17) x2 ≥ x ≤ 6
Simplifying Expressions:
You usually simplify expressions by performing a series of
mathematical operations to obtain another expression. To express the fact
that the new expression is equal to the old expression, use the equal sign
(=). Always arrange problems vertically, if possible. In the examples
below, start with the given expression, and simplify to the answer.
Example 1: Simplify: GivenExpression
GivenExpression = expression1
= expression2
= expression3
= answer
Example 2:
=
=
=
=
GivenExpression
expression1
expression2
expression3
answer
Simplify 16 + 8 ÷ 4 • 2 + 4
16 + 8 ÷ 4 • 2 + 4
=
=
=
=
16 + 2 • 2 + 4
16 + 4 + 4
20 + 4
24
Example 3:
Simplify (x – y)(x + y – 2)
=
=
(x – y)(x + y – 2)
x2 + xy – 2x – xy – y2 + 2y
x2 – 2x – y2 + 2y
DO NOT:
1. Use bars under equations or expressions to show addition
or subtraction.
2. Place extraneous calculations or comments in the body of a
simplification or a solution.
Solving Equations and Inequalities:
You usually solve equations and inequalities by performing a series
of mathematical operation to obtain another equation or inequality. To
express the fact that the new equation is implied by the old equation,
sometimes we use the implication sign ( ), although it is acceptable if
problems are arranged neatly. If possible, arrange problems vertically. In
the examples below, start with the given equation, and solve.
Example 1: Simplify: GivenEquation
Either form given below is acceptable.
GivenEquation
equation1
equation2
equation3
answer
GivenEquation
equation1
equation2
equation3
answer
Example 2: Solve 2x + 1 = 3x – 5
2x + 1
2x + 1 – 3x – 1
–x
x
=
=
=
=
3x – 5
3x – 5 – 3x – 1
–6
6
Example 3: Solve 2x + 1 ≤ 3x – 5
2x + 1
2x + 1 – 3x – 1
–x
x
≤
≤
≤
≥
3x – 5
3x – 5 – 3x – 1
–6
6
Types of Numbers
Natural Numbers (Counting Numbers) N
N = {1, 2, 3, 4, 5, ...}
Whole Numbers W
W = {0, 1, 2, 3, 4, 5, ...}
Integers Z
Z = {..., –4, –3, –2, –1, 0, 1, 2, 3, 4, ...}
Rational Numbers Q
a
Q = { b | a, b,
Z , b ≠ 0}
Irrational Numbers I
Numbers that can be written as an infinite nonrepeating decimal
Real Numbers R
Any number that is rational or irrational (R = Q ≈ I)
Real Number Line
Visualize a line with equally spaced markers each of which is
associated with the integers. If the integers have their natural order, then
the real numbers can be visualized as points on the line.
-3
-2
-1
0
1
2
3
We notice that
1. Every real number corresponds to a unique point on the line.
2. Every point on the line corresponds to a unique real number.
This is why the set of real numbers is sometimes referred to as the
real number line.
Absolute Value
|x|
=
 x

–x
if x ≥ 0
if x < 0
The absolute value of a real number is equal to its (positive) distance
to the origin!
Rule for Order of Operations
(Please Excuse My Dear Aunt Sally)
1.
Parentheses: Simplify all groupings first.
2.
Exponents: Calculate exponential powers and radicals.
3.
Multiplication and Division:
4.
Perform all multiplications and
divisions as they occur from left to right.
Addition
S
and
ubtraction: Perform all additions and
subtractions as they occur from left to right.
Evaluate the following expressions and note the use of the equal signs
because we use mathematics writing style:
Ex 1: (4 – 6)2 + 6(–4) + 5
Ex 2:
(4 – 6)2 + 6(–4) + 5
(P)
= (–2)2 + 6(–4) + 5
6 + 24 ÷ 3 • 2 + 3 16
6 + 24 ÷ 3 • 2 + 3 16
(E)
= 6 + 24 ÷ 3 • 2 + 3 • 4
(E) = 4 + 6(–4) + 5
(D)
= 6+8•2+3•4
(M)
= 4 – 24 + 5
(M) = 6 + 16 + 3 • 4
(S)
= –20 + 5
(M) = 6 + 16 + 12
(A) = –15
Ex 3: 2 [ 5 + 2 ( 6 + 3 – 4) ]
2 [ 5 + 2 ( 6 + 3 – 4) ]
(A)
= 22 + 12
(A)
= 34
Ex 4: 10 + 12 ÷ 4 + 2 • 3
10 + 12 ÷ 4 + 2 • 3
= 2 [ 5 + 2 ( 9 – 4) ]
= 10 + 3 + 2 • 3
= 2 [ 5 + 2 ( 5) ]
= 10 + 3 + 6
= 2 [ 5 + 10 ]
= 13 + 6
= 2 [ 15 ]
= 19
= 30
Types of Intervals
Interval
Notation
(a, b)
[a, b]
[a, b)
(a, b]
(a, ∞)
Graph
a
b
a
b
a
a
(–∞, b]
b
a
(–∞, b)
[a, ∞)
b
b
a
b
Algebraic
Notation
Interval
Description
a<x<b
Open, finite
a≤x≤b
Closed, finite
a≤x<b
Half–open, finite
a<x≤b
Half–open, finite
x>a
Open, infinite
x<b
Open, infinite
x≥a
Closed, infinite
x≤b
Closed, infinite
If your answer is composed of two (or more) distinct intervals, then the
algebraic form of your answer must contains the conjunction 'OR'.
Example 1: Describe algebraically the following intervals:
a)
-3
b)
c)
d)
–3 < x < 4
4
x ≤ –1
-1
2
2≤x<8
8
x>5
5
Example 2: Describe the region(s) containing the indicated sign(s):
Sign
Graph
+
–
0
+
x < –4 or x > –1
-1
0 – 0 +
4
2
– (*) + +
–, 0
+
0
-4
– (*) + +
+, 0
Answer
2 < x ≤ 4 or x ≥ 5
0 – 0 +
4
2
5
5
x < 2 or 4 ≤ x ≤ 5
+ 0 – (*) + 0 –
–
-2
0
2
–2 < x < 0 or x > 2
Addition Property of Equality
For any real numbers a, b, and c,
if a = b, then a + c = b + c.
This means that you can always add the same quantity to both sides of an
equation. Because of the relation between addition and subtraction, you
also can always subtract the same quantity to both sides of an equation.
Examples: Solve the following:
x–4=9
x–4+4=9+4
x = 13
w + 2 = –3
w + 2 – 2 = –3 – 2
w = –5
Multiplication Property of Equality
For any real numbers a, b, and c, with c ≠ 0,
if a = b, then a • c = b • c.
This means that you can always multiply the same nonzero quantity to
both sides of an equation. Because of the relation between multiplication
and division, you also can always divide the same nonzero quantity to
both sides of an equation.
Examples: Solve the following:
x
3 = –5
x
3 • 3 = 3 • (–5)
x = –15
4w = 8
4w
8
=
4
4
w=2
Addition Property of Equations
You can always add or subtract the same quantity to both sides of an
equation without affecting the solution.
Example 1: Solve 3x – 4 = 2x + 3
We add the quantity (–2x + 4) to both sides of the equation :
3x – 4 = 2x + 3
3x – 4 – 2x + 4 = 2x + 3 – 2x + 4
x=
Multiplication Property
of Equations
You can always multiply or divide the same nonzero quantity to both
sides of an equation without affecting the solution.
Example 2: Solve –4x = 12
We divide both sides of the equation by –4.
–4x = 12
–4x 12
–4 = –4
x = –3
Solving First Degree Equations in
One Variable (Be GLAD)
Remember: You can always simplify any side of the equation at any
time.
1. Simplify the expressions on both sides of the equation. You must
first eliminate all of the
necessary.
Grouping symbols and
simplify if
2. If there are fractions, you may multiply both sides of the equation
by the
LCD. Simplify if necessary.
3. You want the variable terms on one side of the equation and the
constant terms on the other side. If this is not the case, you
A
should
dd or subtract to both sides of the equation a variable
term and/or a numerical value in order to get the variable terns
on one side and the constant values on the other side. Simplify
if necessary.
4. You want the variable term coefficient to be 1. If this is not the
D
case, you should
ivide both sides of the equation by the
variable term coefficient. Simplify if necessary.
5. Check your answer by substituting it into the original equation.
Note: Always make sure your final answer has the variable on the left
side of the equation!!!
Examples: Solve the following and use mathematics writing style:
Ex 1:
3x + 3 = x – 5
3x + 3 – x – 3 = x – 5 – x – 3
Ex 2:
x – 5 = 3x – 8
x – 5 – 3x + 5 = 3x – 8 – 3x + 5
–2x = –3
2x = –8
–2x –3
–2 = –2
2x –8
2 = 2
x = 3/2
x = –4
Ex 3: 3(5y – 2) – 4y = –2(y + 3)
Ex 4:
15y – 6 – 4y = –2y – 6
11y – 6 = –2y – 6
12
3z 1
z
4 +2 =3+3
3z

4

1
z

+ 2 = 12 3 + 3
11y –6 + 2y +6 = –2y –6 + 2y +6
9z + 6 = 36 + 4z
13y = 0
9z + 6 – 4z –6 = 36 + 4z – 4z –6
13y
0
13 = 13
5z = 30
y=0
5z 30
z = 5
z=6
Equations Involving Fractions
When we solve equations with fractions, we always assume that no
denominator is zero. Therefore, we can find the LCD and multiply both
sides by this nonzero factor. We must check to see that our solution does
not cause any denominator to be zero.
To solve equations with (simple) fractions:
Step 1: Identify all fractions in the equation and find the LCD.
Step 2: Multiply the both sides of the equation by the LCD.
Step 3: Solve the resulting equation.
Step 4:
Check the answer into the original problem.
(At least
check to make sure no denominator can be zero.)
WARNING:
You must know the difference between an expression and an
equation. When you solve an equation, you may multiply both sides by
the LCD and we get an equation without fractions. When you have an
expression with fractions, you must perform the indicated operation and /
or simplify. Nonsense like treating expressions like equations will not be
tolerated!!
Examples: Solve the following:
a)
2
3
=
x–6
x–8
2
3
(x – 6)(x – 8) x – 6 = (x – 6)(x – 8) x – 8
2(x – 8) = 3(x – 6)
2x – 16 = 3x – 18
–x = –2
x=2
b)
z–4
1
2
4
=
–
–
z
z2 – 2z
z2
z3 – 2z2
z–4
1
2
4
=
–
–
z(z – 2)
z
z2
z2(z – 2)
z2(z
1
z–4
2
4 

2
– 2) z(z – 2) = z (z – 2) z – z2 – z2(z – 2)
z(z – 4) = z(z – 2) – 2(z – 2) – 4
z2 – 4z = z2 – 2z – 2z + 4 – 4
z2 – 4z = z2 – 4z
0=0
all real numbers except 0, 2
c)
y–2
2
=
1
–
y–3
y2 – 9
y–2
2
y – 3 = 1 – (y + 3)(y – 3)


y–2
2

(y + 3)(y – 3) y – 3 = (y + 3)(y – 3) 1 – (y + 3)(y – 3)
(y + 3)(y – 2) = (y + 3)(y – 3) – 2
y2 + y – 6 = y2 – 9 – 2
y = –5
d)
2
1
1
x2 – 4 = x2 + x2 – 2x
2
1
1
(x + 2)(x – 2) = x2 + x(x – 2)
x2(x
1
2
1 

2
+ 2)(x – 2) (x + 2)(x – 2) = x (x + 2)(x – 2) x2 + x(x – 2)
2x2 = (x + 2)(x – 2) + x(x + 2)
2x2 = x2 – 4 + x2 + 2x
2x2 = 2x2 + 2x – 4
–x = –2
x=2
no solution
Solving For a Term in a Formula
Solving for any term in a formula is similar
Remember: You can always simplify any side of the equation at any
time.
1. Simplify the expressions on both sides of the equation. You must
first eliminate all of the Grouping symbols and simplify if
necessary.
2. If there are fractions, you may multiply both sides of the equation
by the LCD. Simplify if necessary.
3. You want all of the terms with the desired variable on one side of
the equation and all other term on the other side. If this is not
the case, you should Add or subtract to both sides of the
equation various variable terms and/or a numerical value in
order to get desired variable terns on one side and everything
else on the other side. Simplify if necessary.
4. You want one variable term with a coefficient of 1. If this is not
the case, you might have to factor out the desired variable and
Divide both sides of the equation by the resulting coefficient.
Simplify if necessary.
Note: Always make sure your final answer has the desired variable on the
left side of the equation!!!
Examples: Solve the following and use mathematics writing style:
Ex 1: Solve P = 2l + 2w for w
P = 2l + 2w
P – 2l = 2w
P – 2l
2 =w
w=
P – 2l
2
h
Ex 4: Solve A = 2 (b + B) for B
h
A = 2 (b + B)
2A
h =b+B
2A
h –b=B
2A
B= h –b
Ex 2: Solve A = P(1 + rt) for r
A = P(1 + rt)
A = P + Prt
x+1
Ex 5: Solve y = x – 3 for x
A – P = Prt
x+1
y=x–3
A–P
Pt = r
y(x – 3) = x + 1
A–P
r = Pt
xy – 3y = x + 1
xy – x = 3y + 1
x(y – 1) = 3y + 1
Ex 3: Solve A = 2πrh + 2πr2 for h
A = 2πrh + 2πr2
2πrh = A – 2πr2
A – 2πr2
h = 2πr
3y + 1
x= y–1
Solving Absolute Value Equations:
Always put the absolute value on a side by itself if
possible!!!
1. If | x | = c (c > 0), then x = c or x = –c
2. If | x | = 0 ,
then x = 0
3. If | x | = –c (c > 0), then there is no solution
If you have an absolute value on both sides:
4. If | x | = | c |,
then x = c or x = –c
Note: Always make sure your final answer has the variable on the left
side of the equation!!!
Solve the following and use mathematics writing style:
Ex 1: | 2x – 5 | = –3
no solution
1
Ex 3: | 2 x – 1 | = 4
1
2 x–1=4
1
2 x – 1 = –4
Ex 2: | x + 4 | = 0
1
2 x=5
1
2 x = –3
x+4=0
x = 10
x = –6
x = –4
x = 10, –6
Ex 4: | x + 2 | = 3
Ex 5: | 2 – x | = 1 + x
x+2=3
2 – x = 1 + x 2 – x = –(1 + x)
x=1
x + 2 = –3
x = –5
x = 1, –5
–2x = –1
2 – x = –1 – x
x = 1/2
0 = –3
x = 1/2
Ex 6: | 2x + 1 | – 1 = 4
| 2x + 1 | = 5
2x + 1 = 5
2x + 1 = –5
2x = 4
2x = –6
x=2
x = –3
x = 2, –3
Ex 7: | 3x + 1 | = | x – 1 |
3x+1 = x–1
3x+1 = –(x–1)
2x = –2 3x+1 = –x+1
x = –1
4x = 0
x=0
x = –1, 0
Solving Word Problems
( Super Solvers Use C.A.P.E.S.)
1. Read the problem carefully. (Reread it several times if necessary)
2.
Categorize the problem type if possible.
(Is it a problem of
numerical expression, distance–rate–time, cost–profit, or simple
interest type?)
A
3. Decide what is asked for, and
ssign a variable to the unknown
quantity. Label the variable so you know exactly what it
represents.
4. Draw a
Picture, diagram, or chart whenever possible!!
E
5. Form an
quation (or inequality) that relates the information
provided.
6.
Solve the equation (or inequality).
7. Check your solution with the wording of the problem to be sure
it makes sense.
distance–rate–time: distance = rate • time (d = r • t)
cost–profit:
profit = revenue – cost (P = R – C)
simple interest:
interest = principal * rate (i = p • r)
Example: A grocery store bought ice cream for 59¢ a half gallon and
stored it in two freezers. During the night, one freezer "defrosted" and
ruined 14 half gallons. If the remaining ice cream is sold for 98¢ a half
gallon, how many half gallons did the store buy if it made a profit of
$42.44?
charge
=
price
•
quantity
bought
.59
x
sold
.98
x – 14
Example: Last summer, Ernie sold surfboards. One style sold for
$70 and the other sold for $50. He sold a total of 48 surfboards. How
many of each style did he sell if the receipts from each style were equal?
charge
=
price
•
quantity
one style
70
x
other style
50
48 – x
Example: Sellit Realty Company gets a 6% fee for selling improved
properties and 10% for selling unimproved land. Last week, the total sales
were $220,000 and the total fees were $14,000. What were the sales from
each of the two types of properties?
fees
=
rate
•
sales
improved
.06
x
unimproved
.10
220000 – x
Example: Maria jogs to the country at a rate of 10 mph. She returns
along the same route at 6 mph. If the total trip took 1 hour 36 minutes,
how far did she jog?
distance
=
rate
going
x
10
coming
x
6
•
time
Properties of Exponents
Property:
Example:
b0 = 1
50 = 1
b–n
1
= bn
3–2
1
1
= 32 = 9
bm • bn = bm + n
23 •25 = 28
bm
m–n
=
b
n
b
59
2 = 25
=
5
7
5
(bm)n = bm • n
(23)4 = 212 = 4096
(a • b)n = an • bn
(4x)3 = 43 • x3 = 64x3
a  n
 
b
an
= bn
3 2
 
5
32
9
= 52 = 25
Some Additional Properties
 a  –n
 
b
bn
= an
1
n
=
b
–n
b
a–n
bm
b–m = an
Examples: Simplify the following using mathematics writing style:
x–5 • x9
x3 • x
a)
x–5+9
= x3+1
x4
= x4
= 1
 2 –1 –2
w w 
 5 2
w w 
d)
=
 1 –2
w 
 7
w 
= (w–6)–2
= w12
 3 0 –5
z z 
 k 
z z
b)
=
 –2 –2
y 
 –5
y 
c)
= (y–2+5)–2
 3  –5
 z 
 k+1
z

= (y3)–2
= (z2–k)–5
= y3•(–2)
= z–10+5k
= y–6
= z5k–10
1
= y6
e)
(x–2y–3)4 (x4y–2)–2
(x3y–1)–3 (x–4y)4
x–8y–12 x–8y4
= x–9y3 x–16y4
x–16y–8
= x–25y7
= x9y–15
x9
= y15
Multiplying Polynomials
The product of two polynomials can be found by using some form of
the distributive property
a(b + c) = ab + ac.
Example: Multiply the given polynomials and simplify.
a)
(x + 4)(2x – 3)
= x(2x – 3) + 4(2x – 3)
= 2x2 – 3x + 8x – 12
= 2x2 + 5x – 12
b)
(3x – 5)(x – 2)
= 3x(x – 2) – 5(x – 2)
= 3x2 – 6x – 5x + 10
= 3x2 – 11x + 10
c)
(2x – 3y)(x + 2y)
= 2x(x + 2y) – 3y(x + 2y)
= 2x2 + 4xy – 3xy – 6y2
= 2x2 + xy – 6y2
d)
(z – 1)(z – 2)
= z(z – 2) – 1(z – 2)
= z2 – 2z – z + 2
= z2 – 3z + 2
e)
=
=
=
(x + 2y – 3)(2x – y – 2)
x(2x – y – 2) + 2y(2x – y – 2) – 3(2x – y – 2)
2x2 – xy – 2x + 4xy – 2y2 – 4y – 6x + 3y + 6
2x2 – 2y2 + 3xy – 8x – y + 6
=
=
=
(x + 2)(x2 – 2x + 4)
x(x2 – 2x + 4) + 2(x2 – 2x + 4)
x3 – 2x2 + 4x + 2x2 – 4x + 8
x3 + 8
f)
Special Products of Polynomials
Take time and learn these formulas. You will need them later and you
will be tested on them!
(F + S)(F – S)
= F2 – S2
Difference of squares
(F + S)2
= F2 + 2FS + S2
Perfect square trinomial
(F – S)2
= F2 – 2FS + S2
Perfect square trinomial
(F + S)(F2 – FS + S2) = F3 + S3
Sum of cubes
(F – S)(F2 + FS + S2) = F3 – S3
Difference of cubes
WARNING:
Unless x or y is zero, (x + y)2 does not equal x2 + y2!! For example
(3 + 4)2 = 72 = 49, but 32 + 42 = 9 + 16 = 25. (Note that 25 is less than 49
by 2(3)(4) = 24, the missing middle term.)
x
y
x
x
2
xy
y
xy
y
2
Table to demonstrate that (x + y)2 = x2 + 2xy + y2
Nonsense like thinking that (x + y)2 equals x2 + y2 will not be
tolerated!!
We use the table below for reference to help with multiplication.
(F + S)(F – S)
(F + S)2
(F – S)2
= F2 – S2
= F2 + 2FS + S2
= F2 – 2FS + S2
Difference of squares
Perfect square trinomial
Perfect square trinomial
Example: Perform the indicated products and simplify.
F)
c)
e)
g)
(x + 3y)(x – 3y)
S)
(2xy – 5z)(2xy + 5z)
= (F + S)(F – S)
= (F – S)(F + S)
= F2 – S2
= F2 – S2
= x2 – 9y2
= 4x2y2 – 25z2
(6w2 + 7v5)(6w2 – 7v5)
d)
(x – 4)2
= (F + S)(F – S)
= (F – S)2
= F2 – S2
= F2 – 2FS + S2
= 36w4 – 49v10
= x2 – 8x + 16
(4x – 3y)2
f)
(7w + 5)2
= (F – S)2
= (F + S)2
= F2 – 2FS + S2
= F2 + 2FS + S2
= 16x2 – 24xy + 9y2
= 49w2 + 70w + 25
(x3 + 8y)2
h)
(2x2z – 3y4)2
= (F + S)2
= (F – S)2
= F2 + 2FS + S2
= F2 – 2FS + S2
= x6 + 16x3y + 64y2
= 4x4z2 – 12x2y4z + 9y8
Special Products:
Perfect Square Polynomials
Complete the table:
(F + S)2
= F2 + 2FS + S2
3
(x + 3)2
= _______________
z
4
(z + 4)2
= _______________
w
5
(w + 5)2
= _______________
x
–2
(x – 2)2
= _______________
y
–1
(y – 1)2
= _______________
z
–6
(z – 6)2
= _______________
2x
3y
(2x + 3y)2 = _______________
3x
–5y
(3x – 5y)2 = _______________
4x
–y
(4x – y)2
2x2
5yz3
F
S
x
F2
2FS
S2
= _______________
(2x2 + 5yz3)2 = _______________
Special Products:
Difference of Squares
Complete the table:
F2
S2
(F + S)(F – S)
=
F2 – S2
F
S
x
3
(x + 3)(x – 3)
= _______________
z
4
(z + 4)(z – 4)
= _______________
w
5
(w + 5)(w – 5)
= _______________
x
2
(x – 2)(x + 2)
= _______________
y
1
(y – 1)(y + 1)
= _______________
z
6
(z – 6)(z + 6)
= _______________
2x
3y
(2x + 3y)(2x – 3y) = _______________
3x
5y
(3x – 5y)(3x + 5y) = _______________
4x
y2
(4x – y2)(4x + y2)
2x2
5yz3
= _______________
(2x2+5yz3)(2x2–5yz3) = _______________
FACTORING
Factoring is the inverse operation to multiplying polynomials.
Remember: In order to be good at factoring polynomials, you must be
good at multiplying polynomials. Make sure you are very good at
multiplying polynomials and using the special products.
Special Polynomial Factors
Common factor
ax + ay
Sum/Diff. of squares
 2
x
 x2
+ y2
– y2
Sum/Diff. of cubes
x3

 x3
+ y3
– y3
 2
x
 x2
+ 2xy + y2
– 2xy + y2
Perfect square trinomial
=
a(x + y)
=
does

 (x –
not factor
y)(x + y)
=
(x

(x
+ y)(x2 – xy + y2)
– y)(x2 + xy + y2)
=

(x
 (x
+ y)2
– y)2
Common Factors & Negative Exponents
Anytime you factor, remember to take out the common factor
first!! Also, the biggest factor uses the smallest exponent.
Examples: Factor the following and use mathematics writing style:
a)
x2 + 5x
b)
= 4z4 (2z3 + 3)
= x (x + 5)
c)
12x3y2 – 32x2y3 + 20xy4
d)
= 4xy2 (3x2 – 8xy + 5y2)
e)
x–5 + 5x–7
x–3y2 – x–4y6
= x–4y2 (x – y4)
9a9b4 – 12a8b6 – 6a7b8
= 3a7b4 (3a2 – 4ab2 + 2b4)
f)
= x–7 (x2 + 5)
g)
8z7 + 12z4
8x–9 – 6x–8
= 2x–9 (4 – 3x)
h)
12a–6b–8 – 16a–8b–6
= 4a–8b–8 (3a2 – 4b2)
Sum / Difference of Squares & Cubes
You are required to determine at a quick glance any special product
polynomial.
F2 + S2
F2 – S2
does not factor,
= (F – S)(F + S)
F3 + S3 = (F + S)(F2 – FS + S2)
F3 – S3 = (F – S)(F2 + FS + S2)
Examples: Is the polynomial is the sum of squares, difference of squares,
difference of cubes, or the sum of cubes? If so, then factor it, if possible.
a) A2 + 64
s.s.
does not factor
b) w2 – 25
d.s.
(w – 5)(w + 5)
c) z2 – 8
no
d) z3 – 8
d.c
(z – 2)(z2 + 2x + 4)
e) 4x2 – 9y4
d.s
(2x – 3y2)(2x + 3y2)
f) z3 + 125
s.c.
(z + 5)(z2 – 5z + 25)
g) B2 + 9
s.s.
does not factor
h) c2 – 16
d.s.
(c – 4)(c + 4)
i) c3 – 16
no
j) x2y6 – 81z10
d.s.
(xy3 – 9z5)(xy3 + 9z5)
k) 8y3z6 + 27w3
s.c.
(2yz2 + 3w)(4y2z4 – 6yz2w + 9w2)
Perfect Square Trinomials
You are required to determine at a quick glance any special product
polynomial.
F2 + 2FS + S2
F2 – 2FS + S2
=
=
(F + S)2
(F – S)2
Examples: Is the polynomials a perfect square trinomial? If so, then
factor it.
x2 – x + 1
no
x2 – 2x + 1
yes
(x – 1)2
x2 + 6x + 9
yes
(x + 3)2
x2 – 4x – 4
no
x2 – 4x + 4
yes
(x – 2)2
a2 – 10a + 25
yes
(a – 5)2
9B2 – 12BC + 4C2
yes
(3B – 2C)2
z2 + 13z + 36
no
25z2 + 60z + 36
yes
(5z + 6)2
Factoring By Grouping
Examples: Factor the following and use mathematics writing style:
a)
c)
e)
ax + ay + bx + by
b)
ax + ay – x – y
= a(x + y) + b(x + y)
= a(x + y) – 1(x + y)
= (a + b) (x + y)
= (a – 1) (x + y)
x2 + 4xy – 2x – 8y
d)
xy – 3x – 4y + 12
= x(x + 4y) – 2(x + 4y)
= x(y – 3) – 4(y – 3)
= (x – 2)(x + 4y)
= (x – 4)(y – 3)
x3 – 4x2 + 3x – 12
f)
2x2 – 7xy + 6y2
= x2(x – 4) + 3(x – 4)
= 2x2 – 4xy – 3xy + 6y2
= (x2 + 3)(x – 4)
= 2x(x – 2y) – 3y(x – 2y)
= (2x – 3y)(x – 2y)
Factoring Polynomials
Using The ac–Method
ax2 + bx + c
We assume that a, b, and c are integers.
Step 1. Multiply a • c
Step 2. If you can, find two integers such that:
a. their product is ac
b. their sum is b
Step 3. If two integers don't exist, STOP because the problem cannot
be factored. Otherwise, move on to Step 4.
Step 4. Rewrite the middle term (bx) using the two integers from
Step 2 as coefficients.
Step 5. Factor by grouping (like when there are 4 terms).
Examples: Factor the following using the ac–method:
a)
4x2 – x – 18
= 4x2 – 9x + 8x– 18
= x(4x – 9) + 2(4x – 9)
= (x + 2)(4x – 9)
b)
12x2 – 23x – 24
= 12x2 – 32x + 9x– 24
= 4x(3x – 8) + 3(3x – 8)
= (4x + 3)(3x – 8)
Factoring With Substitution
Examples: Factor the following and use mathematics writing style:
a)
x2y6 – 81z10
b)
x3 + 8
= F2 – S2
= F3 + S3
= (F – S)(F + S)
= (F + S)(F2 – FS + S2)
= (xy3 – 9z5)(xy3 + 9z5)
= (x + 2)(x2 – 2x + 4)
c) x2 – (y + 2)2
d) (x + 2y)2 – 25
= F2 – S2
= F2 – 25
= (F – S)(F + S)
= (F – 5)(F + 5)
= [x – (y + 2)][x + (y + 2)]
= [(x + 2y) – 5][(x + 2y) + 5]
= (x – y – 2)(x + y + 2)
= (x + 2y – 5)(x + 2y + 5)
e) (2x – 5)2 – 2(2x – 5) – 8
f)
(x – 3)2 – 9(y + 2)2
= F2 – 2F – 8
= F2 – 9S2
= (F – 4)(F + 2)
= (F – 3S)(F + 3S)
= [(2x – 5) – 4][(2x – 5) + 2]
= [(x–3) –3(y+2)][(x–3) +3(y+2)]
= [2x – 5 – 4][2x – 5 + 2]
= [x –3 – 3y –6)][x –3 + 3y + 6]
= (2x – 9)(2x – 3)
= (x – 3y – 9)(x + 3y + 3)
Factoring Completely
You are always required to continue to factor until every expression can
be factored no further.
Steps For Factoring Polynomials
1. Always take out the greatest common factor first!!!
(This includes the case with negative exponents.)
2. See if you can use one of the special factors. Check how many
terms are present.
A. If there are 2 terms:
a. See if it is a difference of squares
b. See if it is a sum or difference of cubes
B. If there are 3 terms:
See if it is a perfect square trinomial
C. If there are 4 terms:
See if you can factor by grouping.
3. Try factoring by any other method you can.
(Trial and error, ac method, etc.)
Hint: Whenever you have to factor, DO NOT MULTIPLY out the terms
unless you absolutely have to !! Consider using a substitution.
Examples: Factor completely the following and use mathematics writing
style:
a)
c)
w–7 – 4w–9
4 + 4z–1 + z–2
= w–9(w2 – 4)
= z–2(4z2 + 4z + 1)
= w–9(w – 2)(w + 2)
= z–2(2z + 1)2
(x2 – x)2 – 18(x2 – x) + 72
= F2 – 18F + 72
e)
b)
d) x8 – y8
= (F – 6)(F – 12)
= (x4 – y4)(x4 + y4)
= (x2 – x – 6)(x2 – x – 12)
= (x2 – y2)(x2 + y2)(x4 + y4)
= (x – 3)(x + 2)(x – 4)(x + 3)
= (x–y)(x+y)(x2+y2)(x4+y4)
(x2 + 3x – 10)2 – (x – 2)2
= F2 – S2
= (F + S)(F – S)
= [(x2 + 3x – 10) + (x – 2)][(x2 + 3x – 10) – (x – 2)]
= [x2 + 3x – 10 + x – 2][x2 + 3x – 10 – x + 2]
= (x2 + 4x – 12)(x2 + 2x – 8)
= (x + 6)(x – 2)(x + 4)(x – 2)
= (x + 6)(x + 4)(x – 2)2
Basic Properties
of Rational Expressions
A rational expression is any expression of the form
P
Q
where P and Q are polynomials and Q ≠ 0. In the following properties, no
denominator is allowed to be zero.
Basic Property:
Example:
P R
2
Q = S if and only if P • S = Q • R 3
6
= 9
P
P•R
Q = Q•R
3
3•4
12
5 = 5 • 4 = 20
P
–P
P
–P
– Q = Q = –Q = – –Q
7
–7
7
–7
– 4 = 4 = –4 = – –4
P
–P
P
–P
–
Q = – Q = –Q = –Q
5
–5
5
–5
–
8 = – 8 = –8 = –8
Notice that
–P
P = –1.
Since it is true that
x–a
=
– (–x + a)
=
– (a – x),
It must also be true that
x–a
a – x = –1.
WARNING:
You must reduce only factors!! If the terms are not factors, they
cannot be factored out.
Nonsense like "canceling out" nonfactors will not be tolerated!!
2x + 8
2
=
x+4
x2 – 9
x–3
=
x+3
Multiplication and Division of
Rational Expressions
To multiply two rational expression, multiply the numerators together
and multiply the denominators together.
P
R
•
Q
S
=
P•R
Q•S .
To divide two rational expressions, invert the one immediately after
the division sign, and do a rational expression multiplication.
P
R
Q ÷ S =
P
S
Q • R
=
P•S
Q•R
Example: Perform the indicated operation and simplify:
x2 + 2x – 3
4x + 2
2x2 – 9x – 5
x2 – 3x – 10 ÷ x2 – x • x2 – 2x + 1
=
x2 + 2x – 3
x2 – x
2x2 – 9x – 5
x2 – 3x – 10 • 4x + 2 • x2 – 2x + 1
=
(x + 3)(x – 1)
x(x – 1)
(2x + 1)(x – 5)
•
•
(x – 5)(x + 2)
2(2x + 1)
(x – 1)2
=
x(x + 3)
2(x + 2)
Rule for Adding or Subtracting
Fractions with Equal Denominators
(RASFED)
To add or subtract two rational expressions whose denominators are
equal, simply add or subtract the numerators. Make sure you use
parentheses when appropriate!
P
R
Q + Q
=
P+R
Q
P
R
Q – Q =
P–R
Q
Examples: Perform the indicated operation and simplify:
a)
x
2
+
x2 – 4
x2 – 4
b)
x2
4x – x2
x2 – 4 – x2 – 4
x+2
= x2 – 4
x2 – (4x – x2)
=
x2 – 4
x+2
= (x + 2)(x – 2)
x2 – 4x + x2
=
x2 – 4
1
= x–2
2x2 – 4x
= x2 – 4
2x(x – 2)
= (x + 2)(x – 2)
2x
= (x + 2)
Finding the LCD
Step 1. Factor each denominator completely, including the prime
factors of any constant factor.
Step 2.
Form the product of all the factors that appears in the
complete factorizations.
Step 3. The number of times any factors appears in the LCD is the
most number of times it appears in any one factorization.
Examples: Find the LCD for the given denominators:
a) Denominators are 24, 30, and 36
24 =
8 • 3=
23 • 3
30 =
6 • 5=
2•3•5
36 =
4 • 9=
22 • 32
LCD
=
23 • 32 • 5 =
360
b) Denominators are x3 – x2 and x3 – x
x3 – x2
x3 – x
=
x(x2 – 1)
LCD
=
x2(x – 1)
=
x(x + 1)(x – 1)
=
x2(x + 1)(x – 1)
Rule for Adding or Subtracting
Fractions with Unequal Denominators
(FLEAS)
1.
Factor the rational expression.
2. Find the
Least Common Denominator (LCD).
E
3.
qualize each denominators by replacing each fraction with an
equivalent one whose denominator is the LCD.
4.
Add or Subtract using RASFED.
Example:
1
2
3
2 – 3 + 4
=
6
8
9
12 – 12 + 12
=
6–8+9
12
=
7
12
Examples: Perform the indicated operation and simplify.
2
1
–
x2 – 1
x2 + 2x + 1
a)
b)
x+2
x2 + 2x
x – 2 – x2 – 4
2
1
= (x + 1)(x – 1) – (x + 1)2
x+2
x2 + 2x
= x – 2 – (x + 2)(x – 2)
2(x + 1)
x–1
= (x + 1)2(x – 1) – (x + 1)2(x – 1)
(x + 2)2
x2 + 2x
= (x + 2)(x – 2) – (x + 2)(x – 2)
=
2(x + 1) – (x – 1)
(x + 1)2(x – 1)
(x2 + 4x + 4) – (x2 + 2x)
=
(x + 2)(x – 2)
2x + 2 – x + 1
= (x + 1)2(x – 1)
x2 + 4x + 4 – x2 – 2x
=
(x + 2)(x – 2)
x+3
= (x + 1)2(x – 1)
2x + 4
= (x + 2)(x – 2)
2(x + 2)
= (x + 2)(x – 2)
2
= (x – 2)
Complex Fractions
A simple fraction is any rational expression whose numerator and
denominator contain no rational expression.
A complex fraction is any rational expression whose numerator or
denominator contains a rational expression.
To simplify complex fractions:
Step 1: Identify all fractions in the numerator and denominator and
find the LCD.
Step 2: Multiply the numerator and denominator by the LCD.
Examples:
1 1
2–3
3 1
4–6
a)
12
=
12
1

2
3

4
=
6–4
9–2
=
2
7
x+y
x–1 + y–1
b)
1
– 3
1
– 6
=
x+y
1 1
x+y
=
xy (x + y)
1
1

xy x + y
=
xy (x + y)
y+x
=
xy
Example: Simplify the following:
1
9 – y2
a)
1
3–y
=

1 

y2 9 – y2

1

y2 3 – y
b)
1
1
–
x+h x
h
1
1
x(x + h) + h – x
=
x(x + h) h


x
x – (x + h)
= x(x + h) h
9y2 – 1
= 3y2 – y
x–x–h
= x(x + h) h
(3y + 1)(3y – 1)
=
y(3y – 1)
–h
= x(x + h) h
3y + 1
=
y
–1
= x(x + h)
Rational Exponents
If a is a real number and n is a positive integer, we sometimes speak
of the nth root of a. If it exists, it should be some number b whose nth
power is equal to a. That is, we should have bn = a, and then it seems
reasonable to write b = a1/n.
If a is a positive real number and n is a positive integer, the principal
nth root of a is the positive real number whose nth power is equal to a. We
1
denote the principal nth root of a by using n as an exponent and we write
a1/n .
To avoid any ambiguity, use the principal nth root of a to calculate a1/n.
If a is any real number and n is a positive integer:
a>0
a1/n is a positive
real number
a=0
a<0
n odd
a1/n is a negative
real number
n even
a1/n = 0
a1/n is not a
real number
More Rational Exponents
If a is any real number, m any nonzero integer, and n a positive integer
such that a1/n is defined. Then we have
(a1/n)m
am/n =
(am)1/n
=
To calculate am/n, you should take the root first!!
(a1/n)m .
am/n =
Radicals
If a is a real number, and n is a positive integer such that a1/n is
defined, then we have
n
a=
a1/n.
Thus we also have
am/n
=

n

m

a
n
=
am
Again, to calculate am/n, you should take the root first!!
am/n
=

n

m

a
.
Examples: Simplify the following expressions.
a)
161/2
b)
= 4
d)
16 1/2
 
25
j)
84/3
e)
27 1/3
 
8
h)
8–2/3
= (81/3)–2
= 24
= 2–2
= 16
1
= 4
k)
–25–3/2
811/4
= 3
f)
3
= 2
= (81/3)4
25–3/2
c)
= 5
4
= 5
g)
1251/3
 1  1/5


–32
1
= –2
i)
(–4)–1/2
not real
l)
3 • 16–5/4
= (251/2)–3
= –(251/2)–3
= 3 • (161/4)–5
= 5–3
= –5–3
= 3 • 2–5
1
= 125
1
= –125
1
= 3 • 32
3
= 32
Examples: Simplify the following expressions. Assume that all variables
are positive.
a)
d)
(2x1/3)3
b)
y2/7 • y3/7
4z1/2 • 3z1/3
c)
= 23 (x1/3)3
= y(2/7 + 3/7)
= 12z(1/2 + 1/3)
= 8x
= y5/7
= 12z5/6
x2/3
x1/4
= x(2/3 – 1/4)
= x5/12
e)
(x1/4y1/2)3
x1/2y–1/4
x3/4y3/2
= x1/2y–1/4
= x(3/4 – 1/2)y(3/2 + 1/4)
= x1/4y7/4
Properties of Radicals
Remember that if no index is shown, then it is by default 2. Thus
2
x means x . The properties of radicals come directly from the
definition and the properties of exponents. In particular, we have
Property
n
n
m
a•b
n
n
a
b
n
=
=
a
=
n
Example
n
a • b
3
8x12 =
4
16
81 =
a
b
mn
5
a
x =
3
4
4
10
3
8 • x12 = 2x4
16
81
2
= 3
x
Absolute Value Again
The absolute value can be defined in terms of the square root:
|x|
=
x2
Simplifying Radicals
A radical expression is considered to be simplified when:
1. The radicand has no factors with an exponent
greater than or equal to the index of the radical,
3
x6y11 =
3
x6y9y2 =
3
3
3
x6y9 y2 = x2y3 y2
2. No radical contains a fraction,
3
3
2x4
3
=
12x2
8x6 =
3
12x2
2x2
3. No denominator contains a radical, and
2
1–3 x
2(1 + 3 x)
2(1 + 3 x)
= (1 – 3 x)(1 + 3 x) = 1 – 9x
4. The index and the power of the radicand have no
common factor except 1.
9
x6 =
3
x2
Examples: Simplify the following expressions. Assume that all variables
are positive.
a)
98
=
3
b)
49 • 2
=
49 2
=
=
= 7 2
3
3
56
8•7
=
3
8 7
=
3
e)
4
=
=
2
4
16x
9y
16x•9y3
81y4
3
=
3
3
3
32x
3
=
12
x
2x2
64x3
3
x
=
5x
= 2x
3 2x2
= 4x
4
3x5
8y7
4
3x5 • 2y
8y7 • 2y
4
6x5y
16y8
h)
3
•
3
5x
4x2
4
81 3
32x
3
9xy3
4
81 • 3
3
f)
=
g)
4
5
4x
d)
= 3 3
3y
3
4
243
4
= 2 7
4
4
c)
2x2
=
2x2
=
4
x 6xy
= 2y2
Examples: Simplify the following expressions. Assume that all variables
are positive.
a)
2 10 – 6 10
b)
32x + 18x
=
= –4 10
c)
12 + 18 + 27
16•2x + 9•2x
= 4•3 + 9•2
+ 9•3
= 4 2x + 3 2x
= 7 2x
= 2 3 +3 2 +3 3
= 5 3 +3 2
d)
5 + 45 – 15
e)
8x – 3 2x + 18x
=
5 + 9•5 – 15
=
=
5 + 3 5 – 15
= 2 2x – 3 2x + 3 2x
= 4 5 – 15
3
f)
=
=
3
3
3
5x – 40x
3
5x – 8•5x
3
5x – 2 5x
3
= – 5x
4•2x – 3 2x + 9•2x
= 2 2x
g)
=
4
x7 + x 81x3
4
4
x4•x3 +x 81•x3
4
= xy2 z + 4xy2 z
4
4
= 5xy2 z
= x x3 + 3x x3
4
= 4x x3
h)
5
5
x5y10z + 4xy2 z
5
5
5
i) ( 7 – 2)( 7 + 3)
j) ( 6 – 2 5 )( 6 + 5 )
= ( 7 )2 + 3 7 – 2 7 – 6
= 6 + 30 – 2 30 – 2•5
= 7+ 7 –6
= 6 – 30 – 10
= 1+ 7
= –4 – 30
k) ( 2 + 5)( 2 – 5)
l) ( 3 + 2 )2
= ( 2 )2 – (5)2
= 3+2 3 2 +2
= 2 – 25
= 3+2 6 +2
= –23
= 5+2 6
m) (3 + 2 )2
n) ( 6 + 2 )2
= 32 + 2•3 2 + 2
= ( 6 )2 + 2 6 2 + ( 2 )2
= 9+6 2 +2
= 6 + 2 12 + 2
= 11 + 6 2
= 6 + 2•2 3 + 2
= 8+4 3
2
4–3 2
o)
31
6+ 5
p)
=
2(4 + 3 2)
(4 – 3 2)(4 + 3 2)
=
31(6 – 5)
(6 + 5)(6 – 5)
=
2(4 + 3 2)
16 – 9•2
=
31(6 – 5)
36 – 5
=
2(4 + 3 2)
16 – 18
=
31(6 – 5)
31
=
2(4 + 3 2)
–2
=
6– 5
=
–(4 + 3 2 )
1
q) 2 7 – 3 2
2+ 5
3– 5
r)
1(2 7 + 3 2)
= (2 7 – 3 2)(2 7 + 3 2)
(2 + 5)(3 + 5)
= (3 – 5)(3 + 5)
2 7+3 2
= 4•7 – 9•2
=
2 7+3 2
= 28 – 18
11 + 5 5
=
4
=
2 7+3 2
10
6+5 5+5
9–5
4 2
2 2+3
s)
=
4 2(2 2 – 3)
(2 2 + 3)(2 2 – 3)
=
8•2 – 12 2
4•2 – 9
=
16 – 12 2
8–9
=
16 – 12 2
–1
=
–16 + 12 2
x–5
x– 5
t)
=
(x – 5)( x + 5)
( x – 5)( x + 5)
=
(x – 5)( x + 5)
x–5
=
x + 5
Graphing Equations
Caution:
We use parentheses to represent both an ordered pair
and an open interval. The context of the discussion should clarify whether
points in the plane or intervals is needed. You must know the difference
between the two concepts!
y
5.00
-5.00
5.00 x
-5.00
!x= 1.00
!y= 1.00
Notice that:
1. Every pair of real number corresponds to a unique point in the
plane and every point in the plane corresponds to a unique pair of
real numbers.
2. The plane is divided into four regions called quadrants.
y
5.00
-5.00
5.00 x
-5.00
!x= 1.00
!y= 1.00
Graphing Linear Equations In Two Variables
1. Calculate any three points that satisfy the equation.
2. Plot these points in the graph.
3. If a straight line passes through these points, draw the line. If not,
go back to step 1.
Example: Graph the lines
a) x + y = 2
c) y = 4
b) 2x – 3y = 6
d) x = –3
y
5.00
-5.00
5.00 x
-5.00
!x= 1.00
!y= 1.00
Example: Graph the following (nonlinear) equations:
a) y = x2
b) y = x2 – 4
y
5.00
-5.00
y
5.00
5.00 x
-5.00
-5.00
5.00 x
-5.00
!x= 1.00
!y= 1.00
c) y = | x |
d) y = | x | + 1
y
5.00
-5.00
!x= 1.00
!y= 1.00
y
5.00
5.00 x
-5.00
-5.00
5.00 x
-5.00
!x= 1.00
!y= 1.00
!x= 1.00
!y= 1.00
Introduction to Functions
A relation is a correspondence between two sets that assigns to
each element of the first set an element of the second set.
For us, we will consider a relation to be a set of ordered
pairs.
The domain, D, of a relation is the set of all first components in
the relation.
The range, R, of a relation is the set of all second components in
the relation.
Example: Find the domain and range for each of the following relations.
1. r = {(2, 3), (4, –3), (–1, 1), (4, 0), (–3, 9)}
Solution: D = {–3, –1, 2, 4}
(all of the x–coordinates)
R = {–3, 0, 1, 3, 9} (all of the y–coordinates)
2. r = {(1, 2), (3, 1), (2, –2), (–5, 1)}
Solution: D = {–5, 1, 2, 3}
R = {–2, 1, 2}
(all of the x–coordinates)
(all of the y–coordinates)
A function is a relation in which each domain element has only
one corresponding range element.
Example: Determine whether or not each of the following relations is a
function.
1. r = {(2, 3), (4, –3), (–1, 1), (4, 0), (–3, 9)}
2. r = {(1, 2), (3, 1), (2, –2), (–5, 1)}
3. r = {(2, 3), (1, 6), (2, 5), (0, –1)}
4. r = {(–1, 1), (1, 5), (0, 3)}
Example: Determine whether or not each of the following relations is a
function.
1. y = 2x + 1
2. x = y2
x
3. y = x + 3
A function is a relation in which each domain element has only
one corresponding range element.
It then follows that a function is a relation in which no two ordered pairs
have the same first element. Graphically, this means that if two or more
points in a relation have a common vertical line passing through them,
then the relation is not a function.
Vertical Line Test: If any vertical line intersects a graph of a
relation in more than one point, then the relation is not a function.
Example: Which of the following graphs is the graph of a function?
y
5.00
-5.00
y
5.00
5.00 x
a)
-5.00
5.00 x
b)
-5.00
!x= 1.00
!y= 1.00
y
5.00
-5.00
y
5.00
5.00 x
c)
-5.00
!x= 1.00
!y= 1.00
-5.00
5.00 x
d)
-5.00
!x= 1.00
!y= 1.00
-5.00
!x= 1.00
!y= 1.00
Function Notation
Sometimes we describe functions in terms of a formula. To do this,
we need three things:
1) The name of the function
f
2) The “dummy” or “place holder” variable
x
3) The formula that describes the function
2x + 1
For example, we could have the function defined by f(x) = 2x + 1. Then
we can calculate other values as follows.
f(x)
f(3)
f(–7)
=
=
=
2x + 1
2(3) + 1
2(–7) + 1
=
=
7
–13
Please don’t confuse the notation f(x) with multiplication!
1
Example: Let f(x) = x2 – 2x and g(x) = x . Calculate the following:
a)
f(4)
b) g(6)
c)
f(–3)
d) g(0.25)
e)
f(2)
f) g(0)
Slopes
The slope m of the line passes through the points (x1,y1) and (x2,y2) is given
y2 – y1
by m = x – x .
2
1
Example: Find the slope of the line passing though the given points:
a) (–1, 2) and (3, 5)
m =
=
b) (0, 1) and (2, 6)
5–2
3 – (–1)
3
4
m =
=
c) (5, 1) and (1, 3)
m =
=
=
m =
=
=
d) (2, –3) and (–1, –9)
3–1
1–5
2
–4
1
–2
e) (1, 3) and (4, 6)
6–3
4–1
3
3
1
6–1
2–0
5
2
m =
=
=
f) (3, 6) and (1, 6)
m =
=
=
6–6
1–3
0
–2
0
–9 – (–3)
–1 – 2
–6
–3
2
g) (–3, 2) and (–3, 5)
5–2
–3 – (–3)
3
= 0
slope undefined
m =
Various Linear Equations
Standard Form
An equation that can be written in the form
Ax + By = C,
(A, B not both zero)
is a linear equation in the variables x and y. This form is called the standard form.
Intercepts: The x–intercept is the point (if any) where the line crosses the x–
axis; and the y–intercept is the point (if any) where the line crosses the y–axis.
To find the x–intercept, let y = 0 and solve the equation for x.
To find the y–intercept, let x = 0 and solve the equation for y.
Point–Slope Form
The equation of the line passing through the point (x1, y1) with slope m is
given by
y – y1 = m(x – x1),
and we call this the point–slope form of the equation of the line. Since we need
point information and slope information, this is the form used most often to find the
equation.
Slope–Intercept Form
The equation of the line with slope m and y–intercept b is given by
y = mx + b,
and we call this the slope–intercept form of the equation of the line.
Given the equation of a line, if we can write the equation of the line into
this form (by solving for y), we can determine the slope and the y–
intercept of the line.
Horizontal Line Form
The equation of the horizontal line that passes through the point (a, b)
is given by
y = b.
Vertical Line Form
The equation of the vertical line that passes through the point (a, b) is
given by
x = a.
Examples: Find the slope and the y–intercept of each of the following
lines:
a) –2x + 3y = 6
3y = 2x + 6
2
y=3 x+2
b) y = –3x + 2
slope –3
y–intercept 2
slope 2/3
y–intercept 2
c) 5x – 4y = 20
–4y = –5x + 20
5
y=4 x–5
slope 5/4
y–intercept –5
e) y = 6
slope 0
y–intercept 6
d) 3y + 6 = 0
3y = –6
y = –2
slope 0
y–intercept –2
f) x = 3
slope undefined
no y–intercept
Finding Equations Of Lines
To find the equation of any line, you always need two types of information:
slope information and point information.
Point information includes:
any point the line passes through
y–intercept of the line
x–intercept of the line
Slope information includes:
slope of line
two points the line passes through
knowing line is horizontal or vertical
any parallel or perpendicular line
The basic equations of the line we use are listed:
1. Point–Slope
y – y1 = m(x – x1)
2. Slope–Intercept
y = mx + b
3. Horizontal line
y=b
4. Vertical line
x=a
Helpful information to remember:
y2 – y1
a. The slope of the line passing through two points is m = x – x .
2
1
b. Slopes of parallel lines are equal.
c. Slopes of perpendicular lines are negative reciprocals.
Examples: Find the equation of the line: (Put answers in slope–intercept form if
possible)
a) through (2, 3) with slope –1/2
b) through (–4, –1) with slope 2
1
y – (3) = – 2 (x – 2)
1
y–3=–2 x+1
1
y=–2 x+4
c) through (–1, 2) and (4, –2)
m =
–2 – 2
4 – (–1)
4
=
–5
4
y – 2 = – 5 (x – (–1))
4
y – 2 = – 5 (x + 1)
4
4
y–2=–5 x–5
y – (–1) = 2(x – (–4))
y + 1 = 2(x + 4)
y + 1 = 2x + 8
y = 2x + 7
d) through (6, –2) and (2, 0)
m =
0 – (–2)
2–6
1
=
–2
1
y – 0 = – 2 (x – 2)
1
y=–2 x+1
4
6
y=–5x+5
e) through (2, 3) and horizontal
y=3
f) through (–2, 6) and vertical
x = –2
g) with slope 3 and y–intercept 5
y = 3x + 5
h) with slope 2 and x–intercept 7
y – 0 = 2(x – 7)
y = 2x – 14
i) through (2, 3) parallel to
5x + 3y = 1
find slope: 5x + 3y = 1
y = –5/3 x + 1/3
m =
–5/3
5
y – 3 = – 3 (x – 2)
5
10
y–3=–3 x+ 3
5
19
y=–3 x+ 3
k) through (–1, –2) parallel to
x+y=6
find slope: x + y = 6
y = –x + 6
m = –1
y – (–2) = –1(x – (–1))
y + 2 = –1(x + 1)
y + 2 = –x – 1
y = –x – 3
j) through (0, 0) perpendicular to
2x + y = 1
find slope:
2x + y = 1
y = –2x + 1
1
m= 2
1
y – 0 = 2 (x – 0)
1
y= 2 x
l) through (3, 0) perpendicular to
x + 2y = 3
find slope:
x + 2y = 3
y = –1/2 x + 3/2
m=2
y – 0 = 2(x – 3)
y = 2x – 6