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SEVENTH EDITION and EXPANDED SEVENTH EDITION
Slide 6-1
Chapter 6
Algebra, Graphs and Functions
Copyright © 2005 Pearson Education, Inc.
6.1
Order of Operations
Copyright © 2005 Pearson Education, Inc.
Definitions



Algebra: a generalized form of arithmetic.
Variables: used to represent numbers
Algebraic expression: a collection of variables,
numbers, parentheses, and operation symbols.

Examples:
4x  2
x, x  4, 4(3 y  5),
, y 2  8y  2
3x  5
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Slide 6-4
Order of Operations




1. First, perform all operations within
parentheses or other grouping symbols
(according to the following order).
2. Next, perform all exponential operations
(that is, raising to powers or finding roots).
3. Next, perform all multiplication and
divisions from left to right.
4. Finally, perform all additions and
subtractions from left to right.
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Slide 6-5
Example: Evaluating an Expression


Evaluate the expression x2 + 4x + 5 for x = 3.
Solution:
x2 + 4x + 5
= 32 + 4(3) + 5
= 9 + 12 + 5
= 26
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Slide 6-6
Example: Substituting for Two Variables


Evaluate 4x 2  3xy  5y 2 when x = 3 and y = 4.
Solution:
2
2
4 x  3 xy  5 y
 4(3)2  3(3)(4)  5(42 )
 4(9)  36  5(16)
 36  36  80
 0  80
 80
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Slide 6-7
6.2
Linear Equations in One
Variable
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Definitions

Like terms are terms that have the same
variables with the same exponents on the
variables.
2
2
2x, 7x
 5x , 8x

Unlike terms have different variables or different
exponents on the variables.
2x, 7
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 5x , 6x
3
2
Slide 6-9
Properties of the Real Numbers
a(b + c) = ab + ac
Distributive property
a+b=b+a
Commutative property of
addition
ab = ba
Commutative property of
multiplication
(a + b) + c = a + (b + c)
Associative property of
addition
(ab)c = a(bc)
Associative property of
multiplication
Copyright © 2005 Pearson Education, Inc.
Slide 6-10
Example: Combine Like Terms


8x + 4x
= (8 + 4)x
= 12x
5y  6y
= (5  6)y
= y
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

x + 15  5x + 9
= (1 5)x + (15+9)
= 4x + 24
3x + 2 + 6y  4 + 7x
= (3 + 7)x + 6y + (2  4)
= 10x + 6y  2
Slide 6-11
Solving Equations
Addition Property of Equality
If a = b, then a + c = b + c for
all real numbers a, b, and c.
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
Find the solution to the
equation
x  9 = 24.
x  9 + 9 = 24 + 9
x = 33
Check: x  9 = 24
33  9 = 24 ?
24 = 24 true
Slide 6-12
Solving Equations continued
Subtraction Property of
Equality
If a = b, then a  c = b  c
for all real numbers a, b,
and c.
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
Find the solution to the
equation
x + 12 = 31.
x + 12  12 = 31  12
x = 19
Check: x + 12 = 31
19 + 12 = 31 ?
31 = 31 true
Slide 6-13
Solving Equations continued
Multiplication Property of
Equality
If a = b, then a • c = b • c for
all real numbers a, b, and c,
where c  0.

Find the solution to the
equation x
 9.
7
x
9
7
x
7     7(9)
7
1
7x
 63
1
7
x  63
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Slide 6-14
Solving Equations continued
Division Property of Equality
a b
If a = b, then c  c for all
real numbers a, b, and c,
c  0.
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
Find the solution to the
equation 4x = 48.
4 x  48
4 x 48

4
4
x  12
Slide 6-15
General Procedure for Solving Linear
Equations



If the equation contains fractions, multiply both sides of
the equation by the lowest common denominator (or
least common multiple). This step will eliminate all
fractions from the equation.
Use the distributive property to remove parentheses
when necessary.
Combine like terms on the same side of the equal sign
when possible.
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Slide 6-16
General Procedure for Solving Linear
Equations continued

Use the addition or subtraction property to
collect all terms with a variable on one side of
the equal sign and all constants on the other
side of the equal sign. It may be necessary to
use the addition or subtraction property more
than once. This process will eventually result in
an equation of the form ax = b, where a and b
are real numbers.
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Slide 6-17
General Procedure for Solving Linear
Equations continued

Solve for the variable using the division or
multiplication property. This will result in an
answer in the form x = c, where c is a real
number.
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Slide 6-18
Example: Solving Equations

Solve 3x  4 = 17.
3 x  4  17
3 x  4  4  17  4
3 x  21
3 x 21

3
3
x 7
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Slide 6-19
Solve 21 = 6 + 3(x + 2)

21  6  3( x  2)
21  6  3 x  6
21  3 x  12
21  12  3 x  12  12
9  3x
9 3x

3
3
3x
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Slide 6-20
Solve 8x + 3 = 6x + 21

8 x  3  6 x  21
8 x  3  3  6 x  21  3
8 x  6 x  18
8 x  6 x  6 x  6 x  18
2 x  18
2 x 18

2
2
x 9
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Slide 6-21
Solve 6(x  2) + 2x + 3 = 4(2x  3) + 2


6( x  2)  2 x  3  4(2 x  3)  2
6 x  12  2 x  3  8 x  12  2
8 x  9  8 x  10
8 x  8 x  9  8 x  8 x  10
9  10
False, the equation has no solution. The
equation is inconsistent.
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Slide 6-22
Solve 4(x + 1)  6(x + 2) = 2(x + 4)


4( x  1)  6( x  2)  2( x  4)
4 x  4  6 x  12  2 x  8
2 x  8  2 x  8
2 x  2 x  8  2 x  2 x  8
8  8
8  8  8  8
00
True, 0 = 0 the solution is all real numbers.
Copyright © 2005 Pearson Education, Inc.
Slide 6-23
Proportions

A proportion is a statement of equality between
two ratios.

Cross Multiplication

a c
If  , then ad = bc, b  0, d  0.
b d
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Slide 6-24
To Solve Application Problems Using
Proportions


Represent the unknown quantity by a variable.
Set up the proportion by listing the given ratio on the
left-hand side of the equal sign and the unknown and
other given quantity on the right-hand side of the equal
sign. When setting up the right-hand side of the
proportion, the same respective quantities should
occupy the same respective positions on the left and
right.
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Slide 6-25
To Solve Application Problems Using
Proportions continued

For example, an acceptable proportion might be
miles miles

hour
hour


Once the proportion is properly written, drop the
units and use cross multiplication to solve the
equation.
Answer the question or questions asked.
Copyright © 2005 Pearson Education, Inc.
Slide 6-26
Example

A 50 pound bag of
fertilizer will cover an
area of 15,000 ft2. How
many pounds are needed
to cover an area of
226,000 ft2?
Copyright © 2005 Pearson Education, Inc.

50 pounds
x

2
15,000 ft
226,000 ft 2
(50)(226,000)  15,000 x
11,300,000  15,000 x
11,300,000 15,000 x

15,000
15,000
753.33  x

754 pounds of fertilizer would
be needed.
Slide 6-27
6.3
Formulas
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Definitions

A formula is an equation that typically has a
real-life application.
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Slide 6-29
Perimeter


The formula for the perimeter of a rectangle is
Perimeter = 2 • length + 2 • width or P = 2l + 2w.
Use the formula to find the perimeter of a yard when
l = 150 feet and w = 100 feet.

P = 2l + 2w
P = 2(150) + 2(100)
P = 300 + 200
P = 500 feet
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Slide 6-30
Example

The formula for the volume of a cylinder is V = r2h. Use the
formula to find the height of a cylinder with a radius of 6 inches and
a volume of 565.49 in3.

V   r 2h
565.49   (62 )h
565.49  36 h
565.49 36 h

36
36
5.000  h

The height of the cylinder is 5 inches.
Copyright © 2005 Pearson Education, Inc.
Slide 6-31
Exponential Equations: Carbon Dating


Carbon dating is used by scientists to find the
age of fossils, bones, and other items. The
 t / 5600
formula used in carbon dating is P  P0 2
.
If 15 mg of C14 is present in an animal bone
recently excavated, how many milligrams will be
present in 4000 years?
Copyright © 2005 Pearson Education, Inc.
Slide 6-32
Exponential Equations: Carbon Dating
continued

P  P0 2
 t / 5600
P  15(2)4000 / 5600
P  15(2).71
P  15(0.61)
P  9.2 mg

In 4000 years, approximately 9.2 mg of the original 15
mg of C14 will remain.
Copyright © 2005 Pearson Education, Inc.
Slide 6-33
Solving for a Variable in a Formula or
Equation

Solve the equation 3x + 8y  9 = 0 for y.
3 x  8y  9  0
3 x  8y  9  9  0  9
3 x  8y
3 x  3 x  8y
8y
8y
8
9
 9  3x
 9  3x
9  3x

8
9  3x
y
8
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Slide 6-34
Solve

h
A  (b1  b2 )
2
for b2.
h
A  (b1  b2 )
2
h

2   A   2   (b1  b2 ) 
2

2 A  h(b1  b2 )
2 A h(b1  b2 )

h
h
2A
 b1  b2
h
2A
 b1  b2
h
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Slide 6-35
6.4
Applications of Linear
Equations in One Variable
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Translating Words to Expressions
Phrase
Mathematical Expression
Ten more than a number
x + 10
Five less than a number
x5
Twice a number
2x
Eight more than twice a number
2x + 8
The sum of three times a
number decreased by 9.
3x  9
Copyright © 2005 Pearson Education, Inc.
Slide 6-37
To Solve a Word Problem







Read the problem carefully at least twice to be sure that
you understand it.
If possible, draw a sketch to help visualize the problem.
Determine which quantity you are being asked to find.
Choose a letter to represent this unknown quantity.
Write down exactly what this letter represents.
Write the word problem as an equation.
Solve the equation for the unknown quantity.
Answer the question or questions asked.
Check the solution.
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Slide 6-38
Example
The bill (parts and labor) for the repairs of a car
was $496.50. The cost of the parts was $339.
The cost of the labor was $45 per hour. How
many hours were billed?


Let h = the number of hours billed
Cost of parts + labor = total amount
339 + 45h = 496.50
Copyright © 2005 Pearson Education, Inc.
Slide 6-39
Example continued


339  45h  496.50
339  339  45h  496.50  339
45h  157.50
45h 157.50

45
45
h  3.5
The car was worked on for 3.5 hours.
Copyright © 2005 Pearson Education, Inc.
Slide 6-40
Example

Sandra Cone wants to fence in a rectangular
region in her backyard for her lambs. She only
has 184 feet of fencing to use for the perimeter
of the region. What should the dimensions of
the region be if she wants the length to be 8 feet
greater than the width?
Copyright © 2005 Pearson Education, Inc.
Slide 6-41
continued, 184 feet of fencing, length 8
feet longer than width



Let x = width of region
Let x + 8 = length
P = 2l + 2w
x
x+8
184  2( x )  2( x  8)
184  2 x  2 x  16
184  4 x  16
168  4 x
42  x
The width of the region is 42 feet
and the length is 50 feet.
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Slide 6-42
6.5
Variation
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Direct Variation



Variation is an equation that relates one variable to one
or more other variables.
In direct variation, the values of the two related variables
increase or decrease together.
If a variable y varies directly with a variable x, then
y = kx where k is the constant of proportionality
(or the variation constant).
Copyright © 2005 Pearson Education, Inc.
Slide 6-44
Example


The amount of interest earned on an investment, I,
varies directly as the interest rate, r. If the interest
earned is $50 when the interest rate is 5%, find the
amount of interest earned when the interest rate is 7%.
I = rx
$50 = 0.05x
1000 = x
Copyright © 2005 Pearson Education, Inc.
Slide 6-45
Example continued


x = 1000, r = 7%
I = rx
I = 0.07(1000)
I = $70
The amount of interest earned is $70.
Copyright © 2005 Pearson Education, Inc.
Slide 6-46
Inverse Variation

When two quantities vary inversely, as one
quantity increases, the other quantity
decreases, and vice versa.

If a variable y varies inversely with a variable, x,
then y = k/x where k is the constant of
proportionality.
Copyright © 2005 Pearson Education, Inc.
Slide 6-47
Example

Suppose y varies
inversely as x. If y = 12
when x = 18, find y when
x = 21.
k
y
x
k
12 
18
216  k
Copyright © 2005 Pearson Education, Inc.

Now substitute 216 for k
in y = k/x and find y when
x = 21.
k
x
216
y
21
y  10.3
y
Slide 6-48
Joint Variation

One quantity may vary directly as the product of
two or more other quantities.

The general form of a joint variation, where y,
varies directly as x and z, is
y = kxz where k is the constant of
proportionality.
Copyright © 2005 Pearson Education, Inc.
Slide 6-49
Example


The area, A, of a triangle varies jointly as its
base, b, and height, h. If the area of a triangle is
48 in2 when its base is 12 in. and its height is 8
in., find the area of a triangle whose base is 15
in. and whose height is 20 in.
A = kbh
Copyright © 2005 Pearson Education, Inc.
Slide 6-50
Example continued

A  kbh
48  k (12)(8)
48  k (96)
48
k
96
1
k
2
Copyright © 2005 Pearson Education, Inc.

A  kbh
1
A  (15)(20)
2
A  150 in.2
Slide 6-51
Combined Variation


A varies jointly as B and C and inversely as the
square of D. If A = 1 when B = 9, C = 4, and
D = 6, find A when B = 8, C = 12, and D = 5.
Write the equation.
kBC
A 2
D
Copyright © 2005 Pearson Education, Inc.
Slide 6-52
Combined Variation continued

Find the constant of
proportionality.
kBC
A 2
D
k (9)(4)
1
62
36k
1
36
1 k
Copyright © 2005 Pearson Education, Inc.

Now find A.
kBC
A 2
D
(1)(8)(12)
A
62
96
A
25
A  3.84
Slide 6-53
6.6
Linear Inequalities
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Symbols of Inequality




a < b means that a is less than b.
a  b means that a is less than or equal to b.
a > b means that a is greater than b.
a  b means that a is greater than or equal to b.
Copyright © 2005 Pearson Education, Inc.
Slide 6-55
Example: Graphing


Graph the solution set of x  4, where x is a real
number, on the number line.
The numbers less than or equal to 4 are all the
points on the number line to the left of 4 and 4
itself. The closed circle at 4 shows that 4 is
included in the solution set.
Copyright © 2005 Pearson Education, Inc.
Slide 6-56
Example: Graphing


Graph the solution set of x > 3, where x is a real
number, on the number line.
The numbers greater than 3 are all the points
on the number line to the right of 3. The open
circle at 3 is used to indicate that 3 is not
included in the solution set.
Copyright © 2005 Pearson Education, Inc.
Slide 6-57
Solve 3x  8 < 10 and graph the
solution set.
3 x  8  10
3 x  8  8  10  8
3 x  18
3 x 18

3
3
x6

The solution set is all real numbers less than 6.
Copyright © 2005 Pearson Education, Inc.
Slide 6-58
Compound Inequality



Graph the solution set of the inequality
4 < x  3
a) where x is an integer.
The solution set is the integers between 4 and
3, including 3.
Copyright © 2005 Pearson Education, Inc.
Slide 6-59
Compound Inequality continued


b) where x is a real number
The solution set consists of all real numbers
between 4 and 3, including the 3 but not the
4.
Copyright © 2005 Pearson Education, Inc.
Slide 6-60
Example

A student must have an average (the mean) on
five tests that is greater than or equal to 85%
but less than 92% to receive a final grade of B.
Jamal’s grade on the first four tests were 98%,
89%, 88%, and 93%. What range of grades on
the fifth test will give him a B in the course?
Copyright © 2005 Pearson Education, Inc.
Slide 6-61
Example continued

98  89  88  93  x
85 
 92
5
368  x
85 
 92
5
5(85)  368  x  92(5)
425  368  x  460
425  368  368  368  x  460  368
57  x  92
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Slide 6-62
6.7
Graphing Linear Equations
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Rectangular Coordinate System



The horizontal line is
called the x-axis.
The vertical line is called
the y-axis.
The point of intersection
is the origin.
y-axis
Quadrant II
Quadrant I
x-axis
origin
Quadrant III
Copyright © 2005 Pearson Education, Inc.
Quadrant IV
Slide 6-64
Plotting Points


Each point in the
xy-plane corresponds to
a unique ordered pair
(a, b).
Plot the point (2, 4).


4 units
2 units
Move 2 units right
Move 4 units up
Copyright © 2005 Pearson Education, Inc.
Slide 6-65
Graphing Linear Equations

Graph the equation
y = 5x + 2
x
y
0
2
2/5
0
1
3
Copyright © 2005 Pearson Education, Inc.
Slide 6-66
To Graph Equations by Plotting Points




Solve the equation for y.
Select at least three values for x and find their
corresponding values of y.
Plot the points.
The points should be in a straight line. Draw a
line through the set of points and place arrow
tips at both ends of the line.
Copyright © 2005 Pearson Education, Inc.
Slide 6-67
Graphing Using Intercepts


The x-intercept is found by letting y = 0 and solving
for x.
 Example:
y = 3x + 6
0 = 3x + 6
6 = 3x
2= x
The y-intercept is found by letting x = 0 and solving
for y.
 Example: y = 3x + 6
y = 3(0) + 6
y=6
Copyright © 2005 Pearson Education, Inc.
Slide 6-68
Graph 3x + 2y = 6


Find the x-intercept.

3x + 2y = 6
 3x + 2(0) = 6

3x = 6

x=2
Find the y-intercept.

3x + 2y = 6
 3(0) + 2y = 6

2y = 6

y=3
Copyright © 2005 Pearson Education, Inc.
Slide 6-69
Slope

The ratio of the vertical change to the horizontal
change for any two points on the line.
vertical change
Slope =
horizontal change
y 2  y1
m
x2  x1
Copyright © 2005 Pearson Education, Inc.
Slide 6-70
Types of Slope

The slope of a vertical
line is undefined.
zero

The slope of a
horizontal line is zero.
positive
negative
undefined
Copyright © 2005 Pearson Education, Inc.
Slide 6-71
Example: Finding Slope

Find the slope of the line through the points
(5, 3) and (2, 3)
y 2  y1
m
x2  x1
3  ( 3)
m
2  5
3  3
m
7
0
m
0
7
Copyright © 2005 Pearson Education, Inc.
Slide 6-72
Graphing Equations by Using the
Slope and y-Intercept

Slope-Intercept Form of the Equation of the Line
y = mx + b where m is the slope of the line and
(0, b) is the y-intercept of the line.
Copyright © 2005 Pearson Education, Inc.
Slide 6-73
Steps





Solve the equation for y to place the equation in
slope-intercept form.
Determine the slope and y-intercept from the
equation.
Plot the y-intercept.
Obtain a second point using the slope.
Draw a straight line through the points.
Copyright © 2005 Pearson Education, Inc.
Slide 6-74
Example


Graph 2x  3y = 9
Write in slope-intercept
form.
2x  3y  9
3 y  2 x  9
3 y 2 x 9


3
3 3
2
y  x 3
3
Copyright © 2005 Pearson Education, Inc.
The y-intercept is (0,3)
and the slope is 2/3.
Slide 6-75
Example continued

Plot a point at (0,3) on
the y-axis, then move up
2 units and to the right 3
units.
Copyright © 2005 Pearson Education, Inc.
Slide 6-76
Horizontal Lines



Graph y = 3
y is always equal to 3,
the value of y can never
be 0.
The graph is parallel to
the x-axis.
Copyright © 2005 Pearson Education, Inc.
Slide 6-77
Vertical Lines

Graph x = 3
x always equals 3, the
value of x can never be
0.
The graph is parallel to
the y-axis.
Copyright © 2005 Pearson Education, Inc.
Slide 6-78
6.8
Linear Inequalities in
Two Variables
Copyright © 2005 Pearson Education, Inc.
To Graph Inequalities in Two Variables



Mentally substitute the equal sign for the inequality sign
and plot points as if you were graphing the equation.
If the inequality is < or >, draw a dashed line through the
points. If the inequality is  or , draw a solid line
through the points.
Select a test point not on the line and substitute the x
and y-coordinates into the inequality. If the substitution
results in a true statement, shade the area on the same
side of the line as the test point. If the test point results
in a false statement, shade the area on the opposite
side of the line as the test point.
Copyright © 2005 Pearson Education, Inc.
Slide 6-80
Graph 3x + 4y > 12


Draw a dashed line.
Select a test point.


Try (0, 0)
3x + 4y > 12
3(0) + 4(0) > 12
0 + 0 > 12
0 > 12 false
Shade the opposite half
plane.
Copyright © 2005 Pearson Education, Inc.
Slide 6-81
Graph 3x + y  6


Draw a solid line.
Select a test point.






Try (0, 0)
3x + y  6
3(0) + (0)  6
0 + 0  6
0  6 false
Shade the opposite
half plane.
Copyright © 2005 Pearson Education, Inc.
Slide 6-82
Graph y > 2


Draw a dashed line.
Select a test point.



Try (0, 0)
0 > 2 true
Shade the half plane
containing (0, 0).
Copyright © 2005 Pearson Education, Inc.
Slide 6-83
6.9
Solving Quadratic Equations by
Using Factoring and by
Using the Quadratic Formula
Copyright © 2005 Pearson Education, Inc.
FOIL






A binomial is an expression that contains two
terms.
To multiply two binomials, we use the FOIL
method.
F = First
L
F
O = Outer
I = Inner
(a + b)(c + d) = ac + ad + bc + bd
L = Last
I
O
Copyright © 2005 Pearson Education, Inc.
Slide 6-85
Example

Multiply: (2x + 4)(x + 6)
F
O
I
L
(2 x  4)( x  6)  2 x  x  2 x  6  4  x  4  6
 2 x 2  12 x  4 x  24
 2 x 2  16 x  24
Copyright © 2005 Pearson Education, Inc.
Slide 6-86
To Factor Trinomial Expressions of the
Form x2 + bx + c


Find two numbers whose product is c and
whose sum is b.
Write the factors in the form

(x +
) (x +
One number
from step 1

)
Other number
from step 1
Check your answer by multiplying the factors
using the FOIL method.
Copyright © 2005 Pearson Education, Inc.
Slide 6-87
Factoring Example



Factor x2  7x + 12.
We need to find two numbers whose product is 12 and
whose sum is 7.
Factors of 12
Sum of Factors
Factors of 12
Sum of Factors
1(12)
1 + 12 = 13
1(12)
1 + 12 = 13
(2)(6)
2+6=8
(2)(6)
2 + 6 = 8
(3)(4)
3+4=7
(3)(4)
3 + 4 = 7
(x  3)(x  4)
Copyright © 2005 Pearson Education, Inc.
Slide 6-88
Factoring Trinomials of the Form
ax2 + bc + c, a  1.




Write all pairs of factors of the coefficient of the squared
term, a.
Write all pairs of the factors of the constant, c.
Try various combinations of these factors until the sum
of the products of the outer and inner terms is bx.
Check your answer by multiplying the factors using the
FOIL method.
Copyright © 2005 Pearson Education, Inc.
Slide 6-89
Example: Factoring



Factor 3x2 + 14x + 8.
(3x + )(x + )
Possible Factors
Sum of Outer and Inner
Terms
(3x + 1)(x + 8)
25x
(3x + 8)(x + 1)
11x
(3x + 4)(x + 2)
10x
(3x + 2)(x + 4)
14x
Correct middle term
Thus, 3x2 + 14x + 8 = (3x + 2)(x + 4).
Copyright © 2005 Pearson Education, Inc.
Slide 6-90
Solving Quadratic Equations by
Factoring

Standard Form of a Quadratic Equation


ax2 + bx + c = 0, a  0
Zero-Factor Property

If a • b = 0, then a = 0 or b = 0.
Copyright © 2005 Pearson Education, Inc.
Slide 6-91
To Solve a Quadratic Equation by
Factoring



Use the addition or subtraction property to make
one side of the equation equal to 0.
Factor the side of the equation not equal to 0.
Use the zero-factor property to solve the
equation.
Copyright © 2005 Pearson Education, Inc.
Slide 6-92
Example: Solve by Factoring


Solve 4x2 + 17x  15 = 0.
4 x 2  17 x  15  0
(4 x  3)( x  5)  0
4x  3  0
4x  3
or x  5  0
or
x  5
3
x
4

The solutions are 5 and ¾.
Copyright © 2005 Pearson Education, Inc.
Slide 6-93
Quadratic Formula

For a quadratic equation in standard form,
ax2 + bx + c = 0, a  0, the quadratic formula is
b  b2  4ac
x
2a
Copyright © 2005 Pearson Education, Inc.
Slide 6-94
Example: Using the Quadratic Formula


Solve the equation
3x2 + 2x  7 = 0.
a = 3, b = 2 and
c = 7
b  b2  4ac
x
2a
Copyright © 2005 Pearson Education, Inc.

2  22  4(3)( 7)
2(3)

2  4  84
6

2  88
6
2  2 22  1 2  1 2 22


3
6
6
1  22
1  22
or
3
3
Slide 6-95
6.10
Functions and Their Graphs
Copyright © 2005 Pearson Education, Inc.
Function



A function is a special type of relation where
each value of the independent variable
corresponds to a unique value of the dependent
variable.
Domain the set of values used for the
independent variable.
Range The resulting set of values obtained for
the dependent variable.
Copyright © 2005 Pearson Education, Inc.
Slide 6-97
Vertical Line Test

If a vertical line can be
drawn so that it intersects
the graph at more than
one point, then each x
does not have a unique
y.
Copyright © 2005 Pearson Education, Inc.
Slide 6-98
Types of Functions

Linear: y = ax + b

Quadratic: y = ax2 + bx + c
Copyright © 2005 Pearson Education, Inc.
Slide 6-99
Graphs of Quadratic Functions

y  ax  bx  c
a0
2

y  ax  bx  c
a0
2
axis of symmetry
axis of symmetry
Copyright © 2005 Pearson Education, Inc.
Slide 6-100
Graphs of Quadratic Functions continued

Axis of Symmetry of a Parabola
b
x
2a
Copyright © 2005 Pearson Education, Inc.
Slide 6-101
General Procedure to Sketch the
Graph of a Quadratic Equation






Determine whether the parabola opens upward or
downward.
Determine the equation of the axis of symmetry.
Determine the vertex of the parabola.
Determine the y-intercept by substituting x = 0 into the
equation.
Determine the x-intercept (if they exist) by substituting
y = 0 into the equation and solving for x.
Draw the graph, making use of the information gained
in steps 1 through 5. Remember the parabola will be
symmetric with respect to the axis of symmetry.
Copyright © 2005 Pearson Education, Inc.
Slide 6-102
Graph y = x2 + 2x 3.


Since a = 1, the parabola opens up
Axis: x  b  2  2  1
2a

2(1)
2
y-coordinate of vertex (1, 4)
y  x 2  2x  3
y  (1)2  2(1)  3  1  2  3  4

y-intercept: (0, 3)
y  x 2  2x  3
y  (0)2  2(0)  3  3
Copyright © 2005 Pearson Education, Inc.
Slide 6-103
Graph y = x2 + 2x 3 continued

x-intercepts:
x 2  2x  3  0
( x  1)( x  3)  0
x  1 or x  3

Plot the points and
sketch.
Copyright © 2005 Pearson Education, Inc.
Slide 6-104
Graphs of Exponential Functions

Graph y = 3x.


x
y = 3x
(x, y)
0
1
(0, 1)
1
3
(1, 3)
2
9
(2, 9)
3
27
(3, 27)
1
1/3
(1, 1/3)
2
1/9
(2, 1/9)
3
1/27
(3, 1/27)
Copyright © 2005 Pearson Education, Inc.
Domain: all real numbers
Range is y > 0
Slide 6-105
Graph
x
 1
y   .
3


x
 1
y  
3
Domain: all real numbers
Range is y > 0
x
(x, y)
0
1
(0, 1)
1
3
(1, 3)
2
9
(2, 9)
3
27
(3, 27)
1
1/3
(1, 1/3)
2
1/9
(2, 1/9)
3
1/27
(3, 1/27)
Copyright © 2005 Pearson Education, Inc.
Slide 6-106