aplustestbank.euaplustestbank.eu/...Intermediate-Algebra-Graphs-Models-4-E-4th-Editi… · 78 Chapter 2: Intermediate Algebra: Graphs and Models b) The range is the set of all y-values

Chapter 2

Functions, Linear Equations, and Models Exercise Set 2.1 1. correspondence 2. exactly 3. domain 4. range 5. horizontal 6. vertical 7. “f of 3,” “f at 3,” or “the value of f at 3” 8. vertical 9. The correspondence is a function because each member of the domain corresponds to just one member of the range. 10. The correspondence is not a function because a member of the domain (6) corresponds to more than one member of the range. 11. The correspondence is a function because each girl’s age corresponds to just one weight. 12. The correspondence is a function because each boy’s age corresponds to just one weight. 13. The correspondence is not a function because one member of the domain, 2008, corresponds to three musicians and another, 2009, corresponds to two musicians. 14. The correspondence is a function because each celebrity corresponds to just one birthday. 15. The correspondence is a function because each predator corresponds to just one prey. 16. The correspondence is not a function because one member of the domain, Texas, corresponds to four members of the range.

17. The correspondence is a function because each USB flash drive would have only one storage capacity. 18. The correspondence is not a function, since it is reasonable to assume that at least one member of the rock band plays more than one instrument. 19. The correspondence is a function because each team member would have only one number on his or her uniform. 20. The correspondence is a function because each triangle would have only one area. 21. a) The domain is the set of all x-values. It is { }3, 2,0,4− − .

b) The range is the set of all y-values. It is { }10,3,5,9− .

c) The correspondence is a function. 22. a) The domain is the set of all x-values. It is { }0,1,2,5 .

b) The range is the set of all y-values. It is { }1,3− .

c) The correspondence is a function. 23. a) The domain is the set of all x-values. It is { }1,2,3,4,5 .

b) The range is the set of all y-values. It is { }1 .

c) The correspondence is a function. 24. a) The domain is the set of all x-values. It is { }1 .

b) The range is the set of all y-values. It is { }1,2,3,4,5 .

c) The correspondence is not a function. 25. a) The domain is the set of all x-values. It is { }2,3,4− .

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78 Chapter 2: Intermediate Algebra: Graphs and Models

b) The range is the set of all y-values. It is { }8, 2,4,5− − .

c) The correspondence is not a function. 26. a) The domain is the set of all x-values. It is { }0,4,7,8 .

b) The range is the set of all y-values. It is { }0,4,7,8 .

c) The correspondence is a function. 27. a) Locate 1 on the horizontal axis, and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, –2. Thus, ( )1 2f = − .

b) The domain is the set of all x-values in the graph. It is { }| 2 5x x− ≤ ≤ .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. One such point exists. Its first coordinate is 4. Thus, the x-value for which ( ) 2f x = is

4. d) The range is the set of all y-values in the graph. It is { }| 3 4y y− ≤ ≤ .

28. a) Locate 1 on the horizontal axis, and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, 3. Thus, ( )1 1f = − .


c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. One such point exists. Its first coordinate is –3. Thus, the x-value for which ( ) 2f x = is –3.

d) The range is the set of all y-values in the graph. It is { }| 2 5y y− ≤ ≤ .

29. a) Locate 1 on the horizontal axis, and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, –2. Thus, ( )1 2f = − .




30. a) Locate 1 on the horizontal axis, and then find the point on the graph for which 1 is the first coordinate. From that point, look to the vertical axis to find the corresponding y- coordinate, 3. Thus, ( )1 3f = .



0. d) The range is the set of all y-values in the graph. It is { }| 5 4y y− ≤ ≤ .




Exercise Set 2.1 79






d) The range is the set of all y-values in the graph. It is { }| 0 5y y≤ ≤ .


b) The domain is the set of all x-values in the graph. It is { }| 3, 1,1,3,5x − − .


3. d) The range is the set of all y-values in the graph. It is { }| 1,0,1,2,3y − .


b) The domain is the set of all x-values in the graph. It is { }| 4, 3, 2, 1,0,1,2x − − − − .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. There are two such points. They are ( )2,2−

and ( )0,2 . Thus, the x-values for which

( ) 2f x = are –2 and 0..

d) The range is the set of all y-values in the graph. It is { }|1,2,3,4y .





( ) 2f x = are –1 and 3.








( ) 2f x = are –5 and 1.



b) The domain is the set of all x-values in the graph. It is { }| 4 5x x− < ≤ .

c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. All points in the set { }| 2 5x x< ≤ satisfy

this condition. These are the x-values for which ( ) 2f x = .

d) The range is the set of all y-values in the graph. It is { }| 1,1,2y − .



c) To determine which member(s) of the domain are paired with 2, locate 2 on the vertical axis. From there look left and right on the graph to find any points for which 2 is the second coordinate. All points in the set { }| 0 2x x< ≤ satisfy

this condition. These are the x-values for which ( ) 2f x = .

d) The range is the set of all y-values in the graph. It is { }|1,2,3,4y .

39. Domain: ; range: 40. Domain: ; range: 41. Domain: ; range: { }4

42. Domain: ; range: { }2−

43. Domain: ; range: { }| 1y y ≥ , or [ )1,∞

44. Domain: ; range: { }| 4y y ≤ , or ( ],4−∞

45. Domain: { }| is a real number 2x x and x ≠ − ;

range: { }| is a real number 4y y and y ≠ −

46. Domain: { }| is a real number 5x x and x ≠ ;

range: { }| is a real number 2y y and y ≠

47. Domain: { }| 0x x ≥ , or [ )0,∞ ;

range: { }| 0y y ≥ , or [ )0,∞

48. Domain: { }| 3x x ≤ , or ( ],3−∞ ;

range: { }| 0y y ≥ , or [ )0,∞

49. We can use the vertical line test.

Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function.


Exercise Set 2.1 81


It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function. 51. We can use the vertical line test.

Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function. 52. We can use the vertical line test.

It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.


It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function. 54. We can use the vertical line test.

Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function. 55. We can use the vertical line test.

It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.




Visualize moving the vertical line across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function. 57. ( ) 2 3g x x= +

a) ( )0 2 0 3 0 3 3g = ⋅ + = + =

b) ( ) ( )4 2 4 3 8 3 5g − = − + = − + = −

c) ( ) ( )7 2 7 3 14 3 11g − = − + = − + = −

d) ( )8 2 8 3 16 3 19g = ⋅ + = + =

e) ( ) ( )2 2 2 3

2 4 3 2 7

g a a

a a

+ = + += + + = +

f) ( ) ( )2 2 3 2 2 5g a a a+ = + + = +

58. ( ) 3 2h x x= −

a) ( )4 3 4 2 12 2 10h = ⋅ − = − =

b) ( )8 3 8 2 24 2 22h = ⋅ − = − =

c) ( ) ( )3 3 3 2 9 2 11h − = − − = − − = −

d) ( ) ( )4 3 4 2 12 2 14h − = − − = − − = −

e) ( ) ( )1 3 1 2

3 3 2 3 5

h a a

a a

− = − −= − − = −

f) ( ) ( )1 3 2 1

3 2 1 3 3

h a a

a a

− = − −= − − = −

59. ( ) 25 4f n n n= +

a) ( ) 20 5 0 4 0 0 0 0f = ⋅ + ⋅ = + =

b) ( ) ( ) ( )21 5 1 4 1 5 4 1f − = − + − = − =

c) ( ) 23 5 3 4 3 45 12 57f = ⋅ + ⋅ = + =

d) ( ) 25 4f t t t= +

e) ( ) ( )2

2 2

2 5 2 4 2

5 4 8 20 8

f a a a

a a a a

= + ⋅

= ⋅ + = +

f) ( ) ( )2 22 2 5 4 10 8f a a a a a⋅ = + = +

60. ( ) 23 2g n n n= −

a) ( ) 20 3 0 2 0 0 0 0g = ⋅ − ⋅ = − =

b) ( ) ( ) ( )21 3 1 2 1 3 2 5g − = − − − = + =

c) ( ) 23 3 3 2 3 27 6 21g = ⋅ − ⋅ = − =

d) ( ) 23 2g t t t= −

e) ( ) ( )2

2 2

2 3 2 2 2

3 4 4 12 4

g a a a

a a a a

= − ⋅

= ⋅ − = −

f) ( ) ( )2 22 2 3 2 6 4g a a a a a⋅ = − = −

61. ( ) 3

2 5

xf x

x

−=−

a) ( ) 0 3 3 3 30

2 0 5 0 5 5 5f

− − −= = = =⋅ − − −

b) ( ) 4 3 1 14

2 4 5 8 5 3f

−= = =⋅ − −

c) ( ) ( )1 3 4 4 4

12 1 5 2 5 7 7

f− − − −− = = = =− − − − −

d) ( ) 3 3 0 03 0

2 3 5 6 5 1f

−= = = =⋅ − −

e) ( ) ( )2 3

22 2 5

1 1

2 4 5 2 1

xf x

x

x x

x x

+ −+ =+ −− −= =+ − −

62. ( ) 3 4

2 5

xs x

x

−=+

a) ( ) 3 10 4 30 4 2610

2 10 5 20 5 25s

⋅ − −= = =⋅ + +

b) ( ) 3 2 4 6 4 22

2 2 5 4 5 9s

⋅ − −= = =⋅ + +

c) 31

2 2

12

3 8 52 2 2

1 3 4 4

2 2 5 1 5

5 1 5

6 6 2 6 12

s⋅ − −⎛ ⎞ = =⎜ ⎟⎝ ⎠ ⋅ + +− −= = = − ⋅ = −

d) ( ) ( )( )

3 1 4 3 4 7 71 , or

2 1 5 2 5 3 3s

− − − − −− = = = −− + − +

e) ( ) ( )( )

3 3 43

2 3 5

3 9 4 3 5

2 6 5 2 11

xs x

x

x x

x x

+ −+ =

+ ++ − += =+ + +


Exercise Set 2.1 83

63.

64.

65.

66.

67. ( ) 2 3

4A s s=

( ) 2 34 4 4 3 6.93

4A = = ≈

The area is 2 24 3 cm 6.93 cm≈ .

68. ( ) 2 3

4A s s=

( ) 2 3 36 6 36 9 3 15.59

4 4A = = = ≈

The area is 2 29 3 in 15.59 in≈ .

69. ( ) 24V r rπ=

( ) ( )23 4 3 36V π π= =

The area is 2 236 in 113.10 inπ ≈ . 70. ( ) 24V r rπ=

( ) ( )25 4 5 100V π π= =

The area is 2 2100 cm 314.16 cmπ ≈ .

71. ( ) 133

dP d = +

( ) 20 2020 1 1

33 33P = + =

The pressure at 20 ft is 20

133

atm.

( ) 30 1030 1 1

33 11P = + =


111

atm.

( ) 100 1 1100 1 1 3 4

33 33 33P = + = + =


433

atm.

72. ( ) 0.112W d d=

( ) ( )16 0.112 16 1.792W = =

The amount of water resulting from snow melting at a depth of 16 cm is 1.792 cm. ( ) ( )25 0.112 25 2.8W = =

The amount of water resulting from snow melting at a depth of 25 cm is 2.8 cm. ( ) ( )100 0.112 100 11.2W = =

The amount of water resulting from snow melting at a depth of 100 cm is 11.2 cm. 73. ( )

( ) ( )2 5

8 2 8 5

16 5

11

f x x

f

= −

= −= −=

74. ( ) 2 5

13 2 5

18 2

9

f x x

x

x

x

= −= −==



75. ( ) 2 5

5 2 5

0 2

0

f x x

x

x

x

= −− = −

==

76. ( )

( ) ( )2 5

4 2 4 5

8 5

13

f x x

f

= −

− = − −= − −= −

77. ( )

( )( )

13

1 12 3

1 12 3

81 12 2 3

7 12 3

7 32 1

212

4

4

4

f x x

x

x

x

x

x

x

= += +

− =− =− =

− =

− =

78. ( )

( )( )

13

1 13 3

1 13 3

1 12 13 3 3

13 13 3

13 33 1

4

4

4

13

f x x

x

x

x

x

x

x

= +− = +

− − =− − =

− =

− =− =

79. ( )

( ) ( )13

1 1 12 3 2

16

251 246 6 6

4

4

4

f x x

f

= +

= += += + =

80. ( )

( ) ( )13

1 1 13 3 3

361 19 9 9

359

4

4

4

f x x

f

= +

− = − +

= − + = − +=

81. ( ) 4

7 4

4 7 3

f x x

x

x

= −= −= − = −

So, ( )3 7f − = .

82. ( )

( )( )

12

12

12

1 12 5

110

5 1

5 1

1 5

5

f x x

x

x

x

x

x

= += +

− =− =

− =− =

So, ( )1 110 2f − =

83. ( ) 0.1 0.5

3 0.1 0.5

3 0.5 0.1

2.5 0.1

2.5

0.125

f x x

x

x

x

x

x

= −− = −

− + =− =− =

− =

So, ( )25 3f − = − .

84. ( ) 2.3 1.5

10 2.3 1.5

1.5 2.3 10

1.5 7.7

7.7 77, or 5.13

1.5 15

f x x

x

x

x

x

= −= −= −= −−= = − −

85. The graph crosses the x-axis at only one point whose coordinate is ( )2,0− , so –2 is the zero

of the function. 86. The graph crosses the x-axis at two points whose coordinates are ( )1,0− and ( )3,0 , so

the zeros of the function are –1 and 3. 87. The graph does not cross the x-axis at all so there are no zeros of this function. 88. The graph crosses the x-axis at only one point whose coordinates are ( )4,0 , so 4 is the zero

of this function. 89. The graph crosses the x-axis at two points whose coordinates are ( )2,0− and ( )2,0 , so

the zeros of this function are –2 and 2. 90. The graph does not cross the x-axis at all so there are no zeros of this function.


Exercise Set 2.1 85

91. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 5

0 5

5

f x x

x

x

= −= −=

The zero of this function is 5. 92. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 3

0 3

3

f x x

x

x

= += +

− =

The zero of this function is –3. 93. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

( )( )

12

12

12

10

0 10

10

10 2

20

f x x

x

x

x

x

= += +

− =− =

− =

The zero of this function is –20. 94. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

( )

23

23

23

32

6

0 6

6

6

9

f x x

x

x

x

x

= −= −=

==

The zero of this function is 9. 95. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( ) 2.7

0 2.7

2.7

f x x

x

x

= −= −=

The zero of this function is 2.7.


solve: ( ) 0.5

0 0.5

0.5

f x x

x

x

= −= −=

The zero of this function is 0.5. 97. We want to find any x-values for which ( ) 0f x = , so we substitute 0 for ( )f x and

solve: ( )

73

3 7

0 3 7

7 3

f x x

x

x

x

= += +

− =− =

The zero of this function is 73− .


solve: ( )

85

5 8

0 5 8

8 5

f x x

x

x

x

= −= −==

The zero of this function is 85 .

99. ( ) 5

3f x

x=

−

Since 5

3x − cannot be computed when

the denominator is 0, we find the x-value that causes 3x − to be 0: 3 0

3

x

x

− ==

Thus, 3 is not in the domain of f, while all other real numbers are. The domain of f is { }| is a real number 3x x and x ≠ .



100. ( ) 7

6f x

x=

−

Since 7

6 x− cannot be computed when

the denominator is 0, we find the x-value that causes 6 x− to be 0: 6 0

6

6

x

x

x

− =− = −

=


101. ( )2 1

xf x

x=

−

Since 2 1

x

x − cannot be computed when

the denominator is 0, we find the x-value that causes 2 1x − to be 0: 2 1 0

2 1

1

2

x

x

x

− ==

=

Thus, 1

2 is not in the domain of f, while

all other real numbers are. The domain

of f is 1

| is a real number 2

x x and x⎧ ⎫≠⎨ ⎬⎩ ⎭

.

102. ( ) 2

4 3

xf x

x=

+

Since 2

4 3

x

x + cannot be computed when

the denominator is 0, we find the x-value that causes 4 3x + to be 0: 4 3 0

4 3

3

4

x

x

x

+ == −

= −

Thus, 3

4

x = − is not in the domain of f,

while all other real numbers are. The domain

of f is 3


x x and x⎧ ⎫≠ −⎨ ⎬⎩ ⎭

.

103. ( ) 2 1f x x= +

Since we can compute 2 1x + for any real

number x, the domain is the set of all real numbers. 104. ( ) 2 3f x x= +

Since we can compute 2 3x + for any real number x, the domain is the set of all real numbers. 105. ( ) 5g x x= −

Since we can compute 5 x− for any real

number x, the domain is the set of all real numbers. 106. ( ) 3 4f x x= −

Since we can compute 3 4x − for any

real number x, the domain is the set of all real numbers.

107. ( ) 5

9f x

x=

−

Since 5

9x − cannot be computed when

the denominator is 0, we find the x-value that causes 9x − to be 0: 9 0

9

x

x

− ==


108. ( ) 3

1f x

x=

+

Since 3

1x + cannot be computed when

the denominator is 0, we find the x-value that causes 1x + to be 0: 1 0

1

x

x

+ == −

Thus, 1 x = − is not in the domain of f, while all other real numbers are. The domain of f is { }| is a real number 1x x and x ≠ − .

109. ( ) 2 9f x x= −

Since we can compute 2 9x − for any real number x, the domain is the set of all real numbers.


Exercise Set 2.1 87

110. ( ) 2 2 1f x x x= − +

Since we can compute 2 2 1x x− + for any real number x, the domain is the set of all real numbers.

111. ( ) 2 7

5

xf x

−=

Since we can compute 2 7

5

x − for any


112. ( ) 5

8

xf x

+=

Since we can compute 5

8

x + for any


113. ( ) , if 0,

2 1, if 0

x xf x

x x

<⎧= ⎨ + ≥⎩

a) ( )5f −

Since 5 0− < , ( )f x x= . Thus

( )5 5f − = − .

b) ( )0f

Since 0 0≥ , ( ) 2 1f x x= + . Thus,

( )0 2 0 1 1f = ⋅ + = .

c) ( )10f

Since 10 0≥ , ( ) 2 1f x x= + . Thus,

( )10 2 10 1 20 1 21f = ⋅ + = + = .

114. ( ) 5, if 5,

3 , if 5

x xg x

x x

− ≤⎧= ⎨ >⎩

a) ( )0g

Since 0 5≤ , ( ) 5g x x= − . Thus,

( )0 0 5 5g = − = − .

b) ( )5g

Since 5 5≤ , ( ) 5g x x= − . Thus,

( )5 5 5 0g = − = .

c) ( )6g

Since 6 5> , ( ) 3g x x= . Thus,

( )6 3 6 18g = ⋅ = .

115. ( )5, if 1,

, if 1 2

2, if 2

x x

G x x x

x x

− < −⎧⎪= − ≤ ≤⎨⎪ + >⎩

a) ( )0G

Since 1 0 2− ≤ ≤ , ( )G x x= . Thus,

( )0 0G = .

b) ( )2G

Since 1 2 2− ≤ ≤ , ( )G x x= . Thus,

( )2 2G = .

c) ( )5G

Since 5 2> , ( ) 2G x x= + . Thus,

( )5 5 2 7G = + =

116. ( )2 , if 0,

, if 0 3

5 , if 3

x x

F x x x

x x

≤⎧⎪= < ≤⎨⎪− >⎩

a) ( )1F −

Since 1 0− ≤ , ( ) 2F x x= . Thus,

( ) ( )1 2 1 2F − = − = − .

b) ( )3F

Since 0 3 3< ≤ , ( )F x x= . Thus,

( )3 3F = .

c) ( )10F

Since 10 3> , ( ) 5F x x= − . Thus,

( ) ( )10 5 10 50F = − = − .

117. ( )2

2

2

10 if 10

, if 10 10

10, if 10

x x

f x x x

x x

⎧ − < −⎪= − ≤ ≤⎨⎪ + >⎩

a) ( )10f −

Since 10 10 10− ≤ − ≤ , ( )f x x2= .

Thus, ( ) ( )210 10 100f − = − = .



b) ( )10f

Since 10 10 10− ≤ ≤ , ( )f x x2= . Thus,

( ) 210 10 100f = = .

c) ( )11f

Since 11 10> , ( ) 2 10f x x= + . Thus,

( ) 211 11 10 121 10 131f = + = + = .

118. ( )2

2

2 3, if 2

, if 2 4

5 7, if 4

x x

f x x x

x x

⎧ − ≤⎪= < <⎨⎪ − ≥⎩

a) ( )0f

Since 0 2≤ , ( ) 22 3f x x= − . Thus,

( ) ( )20 2 0 3 0 3 3f = = − = − = − .

b) ( )3f

Since 2 3 4< < , ( ) 2f x x= . Thus,

( ) 23 3 9f = = .

c) ( )6f

Since 6 4≥ , ( ) 5 7f x x= − . Thus,

( )6 5 6 7 30 7 23f = ⋅ − = − = .

119. Thinking and Writing Exercise. The

expression 3

2

x + is defined for all real

numbers, but the expression 2

3x + is not

defined for 3x = − . 120. Thinking and Writing Exercise. The notation ( )n z implies that n is a function of z, or that

the value of n depends on the variable z. Thus, z is the independent variable.

121. 6 3 3 1

2 7 9 3

− = − = −− −

122. ( )2 4 2

5 8 3

− − −= −

−

123. ( )

( )5 5 0

03 10 13

− − −= =

− −

124. ( )2 3 5

13 2 5

− −= − = −

− −

125. 2 8

2 8

2 8

x y

y x

y x

− =− = − +

= −

126. 5 5 10

5 5 10

2

x y

y x

y x

+ == − += − +

127. 2 3 6

3 2 6

22

3

x y

y x

y x

+ == − +

= − +

128. 5 4 8

4 5 8

52

4

x y

y x

y x

− =− = − +

= −

129. Thinking and Writing Exercise. Jaylan should choose the number of fish as the independent variable, since the amount of food depends on the number of fish. Therefore, the amount of food is the dependent variable, and the number of fish is the independent variable. 130. Thinking and Writing Exercise ( )0f refers

to the output from the function f when the input is 0. That is, we substitute the value 0 for the independent variable and determine the value of the function. When we find the zeros of a function, we are trying to determine those input values which result in an output value of 0. That is, we are trying to find any x-values for which ( ) 0f x = , so we substitute

0 for ( )f x and solve.

131. To find ( )( )4f g − , we first find ( )4g − :

( ) ( )4 2 4 5 8 5 3g − = − + = − + = − .

Then ( )( ) ( ) ( )24 3 3 3 1

3 9 1 27 1 26.

f g f− = − = − −

= ⋅ − = − =

To find ( )( )4g f − , we first find ( )4f − :

( ) ( )24 3 4 1 3 16 1 48 1 47f − = − − = ⋅ − = − = .


Exercise Set 2.2 89

Then ( )( ) ( )4 47 2 47 5

94 5 99.

g f g− = = ⋅ +

= + =

132. To find ( )( )1f g − , we first find ( )1g − .

( ) ( )1 2 1 5 2 5 3g − = − + = − + = .

Then ( )( ) ( ) 21 3 3 3 1

27 1 26

f g f− = = ⋅ −

= − =

To find ( )( )1g f − , we first find ( )1f − .

( ) ( )21 3 1 1 3 1 1 2f − = − − = ⋅ − =

Then ( )( ) ( )1 2 2 2 5 9g f g− = = ⋅ + = .

133. To find ( )( )( )( )f f f f tiger , we start with

the innermost function and work our way out. Since ( )f tiger dog= , we have

( )( )( )( ) ( )( )( )f f f f tiger f f f dog= .

Since ( )f dog cat= , we have

( )( )f f cat .

Since ( )f cat fish= , we have

( )f fish .

Finally, ( )f fish worm= . So,

( )( )( )( )f f f f tiger worm= .

134. To find the strength of the largest contraction, locate the highest point on the graph, and find the corresponding pressure on the y-axis, which is the second coordinate of the point. The largest contraction was approximately 22 mm of mercury. 135. To find the time during the test when the largest contraction occurred, locate the highest point on the graph, and find the corresponding time on the x-axis, which is the first coordinate of the point. The time of the largest contraction was approximately 2 min 50 sec into the test. 136. Writing Exercise. About 12 mm; we would expect the contraction at 7 min to be about the same size as the contraction at 4 min since the largest contractions occurred about 3 min apart.

137. The two largest contractions occurred at about 2 minutes, 50 seconds and 5 minute, 40 seconds. The difference in these times is 2 minutes 50 seconds, so the frequency is about 1 every 3 minutes. 138. 139. We know that ( )1, 7− − and ( )3,8 are both

solutions of ( )g x mx b= + . Substituting, we

have

( )7 1m b− = − + , or 7 m b− = − + ,

and ( )8 3m b= + , or 8 3m b= + .

Solve the first equation for b and substitute that expression into the second equation.

( )

7 First equation

7 Solving for

8 3 Second equation

8 3 7 Substituting

8 3 7

8 4 7

15 4

15

4

m b

m b b

m b

m m

m m

m

m

m

− = − +− == += + −= + −= −=

=

We know that 7m b− = , so 15

74

b− = , or

13

4b− = . We have

15

4m = and

13

4b = − , so

( ) 15 13

4 4g x x= − .

Exercise Set 2.2 1. f 2. c 3. e 4. d



5. a 6. b 7. Graph ( ) 2 1f x x= − .

We make a table of values. Then we plot the corresponding points and connect them.

( )1 1

2 3

3 5

5 9

x f x

8. Graph ( ) 3 4g x x= +


( )0 4

1 1

2 2

3 5

x g x

−− −− −

9. Graph ( ) 1

3 2g x x= − + .


( )0 2

3 1

6 0

9 1

x g x

−

10. Graph ( ) 1

2 5f x x= − − .


( )10 0

8 1

4 3

0 5

x f x

−− −− −

−

11. Graph ( ) 25 4h x x= − .


( )0 4

5 2

10 0

x h x

−−

12. Graph ( ) 4

5 2h x x= + .


( )0 2

5 6

10 10

x h x

13. 5 3y x= +

The y-intercept is ( )0,3 , or simply 3.

14. 2 11y x= −

The y-intercept is ( )0, 11− , or simply –11.

15. ( ) 1g x x= − −

The y-intercept is ( )0, 1− , or simply –1.

16. ( ) 4 5g x x= − +


17. 3

8 4.5y x= − −

The y-intercept is ( )0, 4.5− , or simply –4.5.

18. 15

7 2.2y x= +

The y-intercept is ( )0,2.2 , or simply 2.2.

19. ( ) 11.3

4f x x= −

The y-intercept is 1

0,4

⎛ ⎞−⎜ ⎟⎝ ⎠, or simply

1

4− .


Exercise Set 2.2 91

20. ( ) 11.2

5f x x= − +


0,5

⎛ ⎞⎜ ⎟⎝ ⎠

, or simply 1

5.

21. 17 138y x= +


22. 52 260y x= − −

The y-intercept is ( )0, 260− , or simply 260− .

23. Slope = Change in 3 11 8

4Change in 8 10 2

y

x

− −= = =− −

24. Slope = Change in 4 9 5 1

Change in 12 2 10 2

y

x

− −= = = −−

25. Slope = Change in 7 4 11 1

Change in 20 13 33 3

y

x

− − −= = =− − −

26. Slope = ( )( )

21 11Change in 10 10

Change in 8 5 3 3

y

x

− − − −= = =− − − −

27. Slope = ( )

( )( )

1 2 1 46 3 6 6

31 1 16 2 6 6

56 5 6

6 226

30 512 2

Change in

Change in

y

x

− − += =− −

= = −−

= − = −

28. Slope = ( )

( )( )

1 24 5

313 4

5 8 320 20 20 3 912

20 5 259 5412 12 12

Change in

Change in

y

x

− − −= =

−− + = = − = −

− −

29. Slope = ( )Change in 43.6 43.6

Change in 4.5 9.7

0 00

4.5 9.7 14.2

y

x

−= =− −

= =+

30. Slope = ( )( )

2.6 3.1Change in

Change in 1.8 2.8

2.6 3.1 0.5 1

1.8 2.8 1 2

y

x

− − −=− − −− += = =− +

31. a) The graph of 3 5y x= − has a positive

slope, 3, and the y-intercept is ( )0, 5− .

Thus, graph II matches this equation. b) The graph of 0.7 1y x= + has a positive

slope, 0.7, and the y-intercept is ( )0,1 .

Thus graph IV matches this equation. c) The graph of 0.25 3y x= − − has a

negative slope, –0.25, and the y-intercept is ( )0, 3− . Thus graph III matches this

equation. d) The graph of 4 2y x= − + has a negative

slope, –4, and the y-intercept is ( )0,2 .

Thus graph I matches this equation. 32. a) The graph of 1

2 5y x= − has a positive

slope, 12 , and the y-intercept is ( )0, 5− .

Thus graph II matches this equation. b) The graph of 2 3y x= + has a positive

slope, 2, and the y-intercept is ( )0,3 .

Thus graph IV matches this equation. c) The graph of 3 1y x= − + has negative

slope, –3, and the y-intercept is ( )0,1 .

Thus graph I matches this equation. d) The graph of 3

4 2y x= − − has a negative

slope, 34− , and the y-intercept is ( )0, 2− .

Thus graph III matches this equation.

33. 5

32

y x= −

Slope is 5

2; y-intercept is (0, –3).

From the y-intercept, we go up 5 units and to the right 2 units. This gives us the point (2, 2). We can now draw the graph.

As a check, we can rename the slope and find another point.

5 5 1 5

2 2 1 2

− −= ⋅ =− −

From the y-intercept, we go down 5 units and to the left 2 units. This gives us the point (–2, –8). Since (–2, –8) is on the line, we have a check.



34. Slope is 2

;5

y-intercept is (0, –4).

35. 5( ) 22

f x x= − +

Slope is 52

− , or 52− ; y-intercept is (0, 2).

From the y-intercept, we go down 5 units and to the right 2 units. This gives us the point (2, –3). We can now draw the graph.


5 52 2− = −

From the y-intercept, we go up 5 units and to the left 2 units. This gives us the point (–2, 7). Since (–2, 7) is on the line, we have a check.

36. Slope is 2

;5

− y-intercept is (0, 3).

37. ( ) 2 1F x x= +

Slope is 2

2, or 1

; y-intercept is (0, 1).

From the y-intercept, we go up 2 units and to the right 1 unit. This gives us the point (1, 3). We can now draw the graph.


2 3 6

21 3 3

= ⋅ =

From the y-intercept, we go up 6 units and to the right 3 units. This gives us the point (3, 7). Since (3, 7) is on the line, we have a check.

38. Slope is 3; y-intercept is (0, –2).

39. Convert to a slope-intercept equation.

4 3

4 3

x y

y x

+ == − +

Slope is 4

4, or 1

−− ; y-intercept is (0, 3).

From the y-intercept, we go down 4 units and to the right 1 unit. This gives us the point (1, –1). We can now draw the graph.


4 4 1 4

1 1 1 1

− − −= ⋅ =− −

From the y-intercept, we go up 4 units and to the left 1 unit. This gives us the point (–1, 7). Since (–1, 7) is on the line, we have a check.

40. 4 1, or 4 1x y y x− = = −

Slope is 4; y-intercept is (0, –1).


6 6

6 6

11

6

y x

y x

y x

+ == − +

= − +

Slope is 1 1

, or 6 6



Exercise Set 2.2 93

From the y-intercept, we go down 1 unit and to the right 6 units. This gives us the point (6, 0). We can now draw the graph.

As a check, we choose some other value for x, say –6, and determine y:

( )16 1 1 1 2

6y = − − + = + =

We plot the point (–6, 2) and see that it is on the line.

42. 14 20 or 54

y x y x+ = , = −

Slope is 14

; y -intercept is (0, –5).

43. ( ) 0 25g x x= − .

Slope is –0.25, or 14− ; y-intercept is (0, 0).

From the y-intercept, we go down 1 unit and to the right 4 units. This gives us the point (4, –1). We can now draw the graph.


1 1 1 14 4 1 4− − −= ⋅ =− −

From the y -intercept, we go up 1 unit and to

the left 4 units. This gives us the point (–4, 1). Since (–4, 1) is on the line, we have a check.

44. Slope is 1.5; y-intercept is (0, 0).


4 5 10

5 4 10

4 25

x y

y x

y x

− =− = − +

= −

Slope is 45

; y-intercept is (0, –2).

From the y-intercept, we go up 4 units and to the right 5 units. This gives us the point (5, 2). We can now draw the graph.


4 ( 5) 2 4 2 65

y = − − = − − = −

We plot the point (–5, –6) and see that it is on the line.

46. 55 4 4 or 14

x y y x+ = , = − +

Slope is 54

− ; y-intercept is (0, 1).


2 3 6

3 2 6

22

3

x y

y x

y x

+ == − +

= − +

Slope is 2

3− ; y-intercept is (0, 2).

From the y-intercept, we go down 2 units and to the right 3 units. This gives us the point (3, 0). We can now draw the graph.




( )23 2 2 2 4

3y = − − + = + =


48. 33 2 8 or 42

x y y x− = , = −

Slope is 32

; y-intercept is (0, –4).


5 3

3 5

3 5

y x

y x

y x

− =− = −

= − +

Slope is 3

3, or 1


From the y-intercept, we go down 3 units and to the right 1 unit. This gives us the point (1, 2). We can now draw the graph.

As a check, we choose some other value for x, say –1, and determine y: ( )3 1 5 3 5 8y = − − + = + =


50. 3 2 or 2 3y x y x+ = , = −

Slope is 2; y-intercept is (0, –3).

51. ( ) 4 5 0 4 5g x x= . = + .

Slope is 0; y-intercept is (0 4 5), . .

From the y-intercept, we go up or down 0 units and any number of nonzero units to the left or right. Any point on the graph will lie on a

horizontal line 4.5 units above the x-axis. We draw the graph.

52. Slope is 0; y-intercept is 1

0, .2

⎛ ⎞⎜ ⎟⎝ ⎠

53. Use the slope-intercept equation,

( ) ,f x mx b= +

with m = 2 and b = 5. ( )

( ) 2 5

f x mx b

f x x

= += +


( ) ,f x mx b= +

with m = –4 and b = 1. ( ) 4 1f x x= − +


( ) ,f x mx b= +

with 2

3m = − and b = –2.

( )

( ) 22

3

f x mx b

f x x

= +

= − −

56. Use the slope-intercept equation, ( ) ,f x mx b= +

with 3

4m = − and b = –5.

( ) 35

4f x x= − −


( ) ,f x mx b= +

with 7m = − and 1

.3

b =

( )

( ) 17

3

f x mx b

f x x

= +

= − +


Exercise Set 2.2 95

58. Use the slope-intercept equation, ( ) ,f x mx b= +

with m = 8 and b = 1

4− .

( ) 18

4f x x= −

59. We can use the coordinates of any two points

on the line. Let’s use (0, 5) and (4, 6).

change in 6 5 1Rate of changechange in 4 0 4

yx

−= = =−

The distance from home is increasing at a rate

of 14

km per minute.


on the line. We’ll use (0, 50) and (2, 200).

change in 200 50Rate of change 75change in 2 0

yx

−= = =−

The number of pages read is increasing at a rate of 75 pages per day.



change in

Rate of changechange in

40 100 60 209 0 9 3

yx

=

− −= = = −−

,

or 263

−

The distance from the finish line is decreasing at a rate of 2

36 m per second.



change in 14 8Rate of change 6change in 1 2

yx

−= = = −−

The value is decreasing at a rate of $600 per year.


on the line. We’ll use (3, 2.5) and (6, 4.5).

change in


2 22.5 4.53 6 3 3

yx

=

−−= = =− −

The number of bookcases stained is increasing

at a rate of 2

3 bookcase per quart of stain used.

64. We can use the coordinates of any two points on the line. We’ll use (0, 1) and (1, 4).

change in 34 1Rate of change or 3change in 1 0 1

yx

−= = = ,−

The distance is increasing at a rate of 3 miles per hour.


on the line. We’ll use (35, 490) and (45, 500). change in


500 490 10 or 145 35 10

yx

=

−= = ,−

The average SAT math score is increasing at a rate of 1 point per thousand dollars of family income.


on the line. We’ll use (25, 465) and (35, 480). change in


3480 465 15 or 1.535 25 10 2

yx

=

−= = = ,−

The average SAT verbal score is increasing at a rate of 1.5 points per thousand dollars of family income.

67. a) Graph II indicated that 200 ml of fluid

was dripped in the first 3 hr, a rate of 2003

ml/hr. It also indicates that 400 ml of fluid was dripped in the next 3 hr, a rate of 400

3

ml/hr, and that this rate continues until the end of the time period shown. Since the rate of 400

3 ml/hr is double the rate of 2003

ml/hr, this graph is appropriate for the given situation.

b) Graph IV is the only graph that shows a slope of 0 from 7 PM to 10 PM. Thus, it is the appropriate graph for the given situation.

c) Graph I is the only graph that shows a constant rate for 5 hours, in this case from 3 PM to 8 PM. Thus, it is appropriate for the given situation.



d) Graph III indicates that 100 ml of fluid was dripped in the first 4 hr, a rate of 100/4, or 25 ml/hr. In the next 3 hr, 200 ml was dripped. This is a rate of 200/3, or

2366 ml/hr. Then 100 ml was dripped in

the next hour, a rate of 100 ml/hr. In the last hour 200 ml was dripped, a rate of 200 ml/hr. Since the rate at which the fluid was given gradually increased, this graph is appropriate for the given situation.

68. The marathoner’s speed is given by

change in distancechange in time

. Note that the runner

reaches the 22-mi point 56 min after the 15-mi point was reached or after 2 hr, 56 min. We will express time in hours: 2 hr, 56 min = 14

152

hr. Then change in distance 22 15 7

change in time 14 142 215 1515 157 , or 7.5 mph14 2

−= =−

= ⋅ =

The marathoner’s speed is 7.5 mph. 69. The skier’s speed is given by

change in distancechange in time

. Note that the skier

reaches the 12-km mark 45 min after the 3-km mark was reached or after 15 45+ , or 60 min. We will express time in hours: 15 min 0 25 hr= . and 60 min 1 hr= . Then

change in distance 12 3 9 12

change in time 1 0 25 0 75−= = =− . . .

The speed is 12 km/h. 70. The rate at which the number of recycling

groups increased is given by change in number of groups

.change in time

change in number of groups 4224 2936

change in time 28 months

128846

28

−=

= =

The number of recycling groups is increasing at a rate of 46 groups per month.

71. The work rate is given by change in portion of house painted

change in time⋅

change in portion of house paintedchange in time

2 1 53 4 5 512 18 0 8 12 8 96

−= = = ⋅ =−

The painter’s work rate is 596

of the house per

hour. 72. The average rate of descent is given by

change in altitudechange in time

. We will express time in

minutes:

60 min311 hr hr 90 min

2 2 1 hr= ⋅ =

2 hr 10 min, = 2 hr + 10 min

60 min2 hr 10 min 120 min 10 min1 hr

130 min

= ⋅ + = +

=

Then

change in altitude 0 12 000

change in time 130 90

12 000 300.40

− ,= −− ,= = −

The average rate of descent is 300 ft/min. 73. The rate at which the number of hits is

increasing is given by change in number of hits

change in time.

change in number of hits 430 000 80 000change in time 2009 2007

350 000175 000

2

, − ,= −,= = ,

The number of hits is increasing at a rate of 175,000 hits/yr.

74. a) Graph III is appropriate, because it shows

that the rate before January 1 is $3000/month while it is $2000/month after January 1.

b) Graph IV is appropriate, because it shows that the rate before January 1 is $3000/month while it is $4000− /month after January 1.


Exercise Set 2.2 97

c) Graph I is appropriate, because it shows that the rate is $1000/month before January 1 and $2000/month after January 1.

d) Graph II is appropriate, because it shows that the rate is $4000/month before January 1 and $2000− /month after January 1.

75. ( ) 0.75 30C d d= +

0.75 signifies the cost per mile is $0.75; 30 signifies that the minimum cost to rent a truck is $30.

76. ( ) 0 05 200P x x= . +

0.05 signifies that a salesperson earns a 5% commission on sales. 200 indicates that a salesperson’s base salary is $200 per week.

77. 1( ) 52

L t t= +

12

signifies that Lauren’s hair grows 12

in. per

month. 5 signifies that her hair is 5 in long immediately after she gets it cut.

78. ( ) 1913439

5D t t= +

191

5 signifies that the demand increases

191

5

billion kWh per year, for years after 2000. 3439 signifies that the demand was 3439 billion kWh in 2000.

79. ( ) 175.5

8A t t= +

1

8 signifies that the life expectancy of

American females increases 1

8 of a year, per

year, for years after 1970. 75.5 signifies that the life expectancy for a female born in 1970 was 75.5 years.

80. ( ) 12

8G t t= +

1

8 signifies that the grass grows

1

8in. per day.

2 signifies that the grass is 2 in. long immediately after is it cut.

81. ( ) 0.89 16.63P t t= + 0.89 signifies that the average price of a ticket increases by $0.89 per year, for years after 2000. 16.63 signifies that the cost of a ticket is $16.63 in 2000.

82. ( ) 2 2.5C d d= +

2 signifies that the cost per mile of a taxi ride is $2. 2.5 signifies that the minimum cost of a taxi ride is $2.50.

83. ( ) 849 5960C t t= +

849 signifies that the number of acres of organic cotton increases by 849 acres per year, for years after 2006. 5960 signifies that 5960 acres were planted with organic cotton in 2006.

84. ( ) 25 75C x x= +

25 signifies that the cost per person is $25. 75 signifies that the setup cost for the party is $75.

85. ( ) 5000 90 000F t t= − + ,

a) –5000 signifies that the truck’s value depreciates $5000 per year; 90,000 signifies that the original value of the truck was $90,000.

b) We find the value of t for which ( ) 0F t = .

0 5000 90 000

5000 90 000

18

t

t

t

= − + ,= ,=

It will take 18 yr for the truck to depreciate completely.

c) The truck’s value goes from $90,000 when 0t = to $0 when 18t = , so the

domain of F is { }0 18x t| ≤ ≤ .

86. ( ) 2000 15 000V t t= − + ,

a) –2000 signifies that the color separator’s value depreciates $2000 per year; 15,000 signifies that the original value of the separator was $15,000.

b) 0 2000 15 000

2000 15 000

7 5

t

t

t

= − + ,= ,= .

It will take 7.5 yr for the machine to depreciate completely.



c) The machine’s value goes from $15,000 when 0t = to $0 when 7 5t = . , so the

domain of V is { }0 7 5t t| ≤ ≤ . .

87. ( ) 200 1800v n n= − + a) –200 signifies that the depreciation is

$200 per year; 1800 signifies that the original value of the bike was $1800.

b) We find the value of n for which ( ) 600v n = .

600 200 1800

1200 200

6

n

n

n

= − +− = −

=

The trade-in value is $600 after 6 yrs of use.

c) First we find the value of n for which ( ) 0v n = .

0 200 1800

1800 200

9

n

n

n

= − +− = −

=

The value of the bike goes from $1800 when n = 0, to $0 when n = 9, so the domain of v is { }0 9n n| ≤ ≤ .

88. ( ) 300 2400T x x= − +

a) –300 signifies that the mower’s value depreciates $300 per summer of use; 2400 signifies that the original value of the mower was $2400.

b) 1200 300 2400

1200 300

4

x

x

x

= − +− = −

=

The mower’s value will be $1200 after 4 summers of use.

c) 0 300 2400

2400 300

8

x

x

x

= − +− = −

=

The domain of T is { }0 8x x| ≤ ≤ .

89. Thinking and Writing Exercise. Find the

slope-intercept form of the equation. 4 5 12

5 4 12

4 12

5 5

x y

y x

y x

+ == − +

= − +

This form of the equation indicates that the line has a negative slope and thus should slant

down from left to right. The student

apparently graphed 4 12

5 5y x= + .

90. Thinking and Writing Exercise. Using algebra,

we find that the slope-intercept form of the

equation is 5 3

2 2y x= − . This indicates that

the y-intercept is 3

0,2

⎛ ⎞−⎜ ⎟⎝ ⎠, so a mistake has

been made. It appears that the student graphed 5 3

2 2h x= + .

91. ( )( )

8 8 8 8 00

6 6 6 6 12

− − − − += = =− − +

92. ( )2 2 2 2 4

undefined3 3 3 3 0

− − − − −= = =− − − − +

93. 3 0 2 9

0 2 9

2 9

9

2

y

y

y

y

⋅ − =− =− =

= −

94. 4 7 0 3

4 0 3

4 3

3

4

x

x

x

x

− ⋅ =− =

=

=

95. ( )

( ) ( )2 7

0 2 0 7 0 7 7

f x x

f

= −= − = − = −

96. ( ) 2 7

0 2 7

7 2

7

2

f x x

x

x

x

= −= −=

=

97. Thinking and Writing Exercise. Yes, the

population can be modeled as a linear function with –10%, or –0.10 as the slope and the current population for the y-intercept.


Exercise Set 2.2 99

98. Thinking and Writing Exercise. a) Answers will vary. b) The profit increases each year but not as

much as in the previous year. 99. a) Graph III indicates that the first 2 mi and

the last 3 mi were traveled in approximately the same length of time and at a fairly rapid rate. The mile following the first two miles was traveled at a much slower rate. This could indicate that the first two miles were driven, the next mile was swum and the last three miles were driven, so this graph is most appropriate for the given situation.

b) The slope in Graph IV decreases at 2 mi and again at 3 mi. This could indicate that the first two miles were traveled by bicycle, the next mile was run, and the last 3 miles were walked, so this graph is most appropriate for the given situation.

c) The slope in Graph I decreases at 2 mi and then increases at 3 mi. This could indicate that the first two miles were traveled by bicycle, the next mile was hiked, and the last three miles were traveled by bus, so this graph is most appropriate for the given situation.

d) The slope in Graph II increases at 2 mi and again at 3 mi. This could indicate that the first two miles were hiked, the next mile was run, and the last three miles were traveled by bus, so this graph is most appropriate for the given situation.

100. Look for the section of the graph that has the

greatest positive slope. Ponte sul Pesa to Panzano is the steepest part of the trip.

101. The longest uphill climb is the widest rising

line. It is the trip from Sienna to Castellina in Chianti.

102. From Le Bolle to Passo dei Pecorai and from

Passo dei Pecorai to Strada in Chianti the road’s grade is about the same. The grade of the road from Strada in Chianti to Poggio Ugolino is about 1

10 of the previous two

sections. Brittany begin her ride in Le Bolle.

103. Reading from the graph the trip from Castellina in Chianti to Ponte sul Pesa is downhill, then to Panzano is uphill and then to Creve in Chianti is downhill. All sections are about the same grade. So the trip began at Castellina in Chianti.

104. From Ponte sul Pesa to Panzano the elevation

is about 4% as shown.

change in elevationgrade

change in horizontal distance

500 300 m 200 m35 30 km 5 km200 m 2 0.04 4%

5000 m 50

=

−= =−

= = = =

105.

( )

rx py s ry

ry py rx s

y r p rx s

sry xr p r p

+ = −+ = − ++ = − +

= − ++ +

The slope is rr p

− + , and the y-intercept

is ( )0 .sr p

, +

106. We first solve for y .

rx py s

py rx s

sry xp p

+ == − +

= − +

The slope is rp

− , and the y-intercept is

( )0 .sp

,

107. Since 1 1( , )x y and 2 2( , )x y are two points on

the graph of y mx b= + , then 1 1y mx b= +

and 2 2y mx b= + . Using the definition of

slope, we have:

( ) ( )

2 1

2 1

2 1

2 1

2 1

2 1

Slope

( )

.

y y

x x

mx b mx b

x x

m x x

x x

m

−=−+ − +

=−

−=−

=



108. Let 2c = and 3d = . Then ( ) (2 3) (6) 6 6f cd f f m b m b= ⋅ = = ⋅ + = + ,

but

2 2

( ) ( ) (2) (3)

( 2 )( 3 )

6 5 .

f c f d f f

m b m b

m mb b

== ⋅ + ⋅ += + +

Thus, the given statement is false. 109. Let 1c = and 2d = . Then

( ) (1 2) (3) 3f c d f f m b+ = + = = + , but

( ) ( ) ( ) (2 ) 3 2f c f d m b m b m b+ = + + + = + .

The given statement is false. 110. Let 5c = and 2d = . Then

( ) (5 2) (3) 3 3f c d f f m b m b− = − = = ⋅ + = +, but ( ) ( ) (5) (2)

( 5 ) ( 2 )

5 2 3

f c f d f f

m b m b

m b m b m

− = −= ⋅ + − ⋅ += + − − =

.

Thus, the given statement is false. 111. Let 2k = . Then ( ) (2 ) 2f kx f x mx b= = + ,

but ( ) 2( ) 2 2kf x mx b mx b= + = + . The given

statement is false. 112. Observe that parallel lines rise or fall at the

same rate. Thus, their slopes are the same. For the line containing ( 3 )k− , and (4 8), ,

8 8slope3 4 7

k k− −= =− − − .

For the line containing (5 3), and (1 6),− ,

6 3 9slope1 5 4− −= =− .

Then we have

8 97 4

4 32 63

4 31

314

k

k

k

k

− =−− = −

= −

= −

113. a) ( )6 5 5

5 4 4

c c c c

b b b b

− − − = = −− −

b) ( )

0

d e d e

b b

+ − =−

Since we cannot divide

by 0, the slope is undefined.

c) ( ) ( )( ) ( )

( )

2 2

2

2

2

a d a d a d a d

c f c f c f c f

a d

f

a d

f

a d

f

− − − + − − − −=− − + − − −

− −=−

− +=−+=

114.

115.

Exercise Set 2.3 1. horizontal 2. undefined 3. vertical 4. y-axis 5. 0; x 6. 0; y 7. parallel 8. standard 9. linear 10. –1


Exercise Set 2.3 101

11. 9 3

12

y

y

− ==

The graph of 12y = is a horizontal line.

Since 9 3y − = is equivalent to 12y = , the

slope of the line 9 3y − = is 0.

12. 1 7

6

x

x

+ ==

The graph of 6x = is a vertical line. Since 1 7x + = is equivalent to 6x = , the slope of the line 1 7x + = is undefined. 13. 8 6

6

83

4

x

x

x

=

=

=

The graph of 3

4x = is a vertical line. Since

8 6x = is equivalent to 3

4x = , the slope of

the line 8 6x = is undefined. 14. 3 5

8

y

y

− ==

The graph of 8y = is a horizontal line.

Since 3 5y − = is equivalent to 8y = , the

slope of the line 3 5y − = is 0.

15. 3 28

28

3

y

y

=

=

The graph of 28

3y = is a horizontal line.

Since 3 28y = is equivalent to 28

3y = , the

slope of the line 3 28y = is 0.

16. 19 6

19

6

y

y

= −

− =

The graph of 19

6y = − is a horizontal line.

Since 19 6y= − is equivalent to 19

6y = − ,

the slope of the line 19 6y= − is 0.

17. 9 12

3

x

x

+ ==

The graph 3x = is a vertical line. Since 9 12x+ = is equivalent to 3x = , the slope of the line 9 12x+ = is undefined. 18. 2 18

9

x

x

==

The graph of 9= is a vertical line. Since 2 18x = is equivalent to 9x = , the slope of the line 2 18x = is undefined. 19. 2 4 3

2 7

7

2

x

x

x

− ==

=

The graph of 7

2x = is a vertical line. Since

2 4 3x − = is equivalent to 7

2x = , the slope

of the line 2 4 3x − = is undefined. 20. 5 1 16

5 17

17

5

y

y

y

− ==

=

The graph of 17


Since 5 1 16y − = is equivalent to 17

5y = , the

slope of the line 5 1 16y − = is 0.

21. 5 4 35

5 39

39

5

y

y

y

− ==

=

The graph of 39


Since 5 4 35y − = is equivalent to 39

5y = ,

the slope of the line 5 4 35y − = is 0.



22. 2 17 3

2 20

10

x

x

x

− ===

The graph of 10x = is a vertical line. Since 2 17 3x − = is equivalent to 10x = , the slope of the line 2 17 3x − = is undefined. 23. 4 3 9 3

4 9

9

4

y x x

y

y

− = −=

=

The graph of 9


Since 4 3 9 3y x x− = − is equivalent to

9

4y = , the slope of the line is zero.

24. 4 12 4

12

x y y

x

− = −=

The graph of 12x = is a vertical line. Since 4 12 4x y y− = − is equivalent to 12x = , the

slope of the line 4 12 4x y y− = − is

undefined. 25. 5 2 2 7

5 2 5

5

3

x x

x x

x

− = −= −

= −

The graph of 5

3x = − is a vertical line. Since

5 2 2 7x x− = − is equivalent to 5

3x = − , the

slope of the line 5 2 2 7x x− = − is undefined. 26. 5 3 9

4 6

3

2

y y

y

y

+ = +=

=

The graph of 3


Since 5 3 9y y+ = + is equivalent to 3

2y = ,

the slope of the line 5 3 9y y+ = + is 0.

27. 2

53

y x= − +

The equation is written in slope-intercept

form. We see that the slope is 2

3− .

28. 3

42

y x= − +

The equation is written in slope-intercept

form. We see that the slope is 3

2−

29. Graph 5y = .

This is a horizontal line that crosses the y-axis at ( )0,5 . If we find some ordered pairs, note

that, for any x-value chosen, y must be 5.

2 5

0 5

3 5

x y

−

30. Graph 1x = − . This is a vertical line that crosses the x-axis at ( )1,0− . If we find some ordered pairs,

note that, for any y-value chosen, x must be –1.

1 5

1 0

1 3

x y

−−− −

31. Graph 3x = . This is a vertical line that crosses the x-axis at ( )3,0 . If we find some ordered pairs,

note that, for any y-value chosen, x must be 3.

3 5

3 0

3 3

x y

−



32. Graph 2y = .

This is a horizontal line that crosses the y-axis at ( )0,2 . If we find some ordered pairs, note

that, for any x-value chosen, y must be 2.

2 2

0 2

3 2

x y

−

33. Graph ( ) 2f x = − .

This is a horizontal line that crosses the y-axis at ( )0, 2− . If we find some ordered pairs,

for any x-value chosen, y must be 2− .

3 2

0 2

2 2

x y

− −−−

34. Graph ( ) 3g x = − .

This is a horizontal line that crosses the y-axis at ( )0, 3− . If we find some ordered pairs,

for any x-value chosen, y must be 3− .

2 3

0 3

2 3

x y

− −−−

35. Graph 3 15x = − . Since y does not appear, we solve for x. 3 15

5

x

x

= −= −

This is a vertical line that crosses the x-axis at ( )5,0− .

36. Graph 2 10x = . Since y does not appear, we solve for x. 2 10

5

x

x

==

This is a vertical line that crosses the x-axis at ( )5,0 .

37. Graph ( )3 15g x⋅ = .

Solve for ( )g x .

( )( )

3 15

5

g x

g x

⋅ =

=

This is a horizontal line that crosses the y-axis at ( )0,5 .

38. Graph ( )3 2f x− = .

First solve for ( )f x .

( )( )

3 2

1

f x

f x

− =

=

This is a horizontal line that crosses the vertical axis at 0,1 .



39. Graph 4x y+ = .

To find the y-intercept, let 0x = and solve for y. 4

0 4

4

x y

y

y

+ =+ =

=

The y-intercept is ( )0,4 .

To find the x-intercept, let 0y = and solve

for x. 4

0 4

4

x y

x

x

+ =+ =

=

The x-intercept is ( )4,0 .

Plot these points and draw a line. A third point could be used as a check. 40. Graph 5x y+ = .

To find the y-intercept, let 0x = and solve for y. 5

0 5

5

x y

y

y

+ =+ =

=



for x. 5

0 5

5

x y

x

x

+ =+ =

=


Plot these points and draw a line. A third point could be used as a check.

41. Graph ( ) 2 1f x x= − .

To find the y-intercept, let 0x = and solve for y. ( )

( ) ( )( )

2 1

2 0 1

1

f x x

f x

f x

= −

= −

= −

The y-intercept is ( )0, 1− .

To find the x-intercept, let ( ) 0f x = and

solve for x. ( )

12

2 1

0 2 1

2 1

f x x

x

x

x

= −= −==

The x-intercept is ( )12 ,0 .

Plot these points and draw a line. A third point could be used as a check. 42. Graph ( ) 3 12f x x= + .

To find the y-intercept, let 0x = and solve for y. ( )

( ) ( )( )

3 12

3 0 12

12

f x x

f x

f x

= +

= +

=



solve for x. ( ) 3 12

0 3 12

3 12

4

f x x

x

x

x

= += += −= −

The x-intercept is ( )4,0− .



Plot these points and draw a line. A third point could be used as a check. 43. Graph 3 5 15x y+ = − .

To find the y-intercept, let 0x = and solve for y. 3 5 15

3 0 5 15

5 15

3

x y

y

y

y

+ = −⋅ + = −

= −= −



for x. 3 5 15

3 5 0 15

3 15

5

x y

x

x

x

+ = −+ ⋅ = −

= −= −



44. Graph 5 4 20x y− = .


5 0 4 20

4 20

5

x y

y

y

y

− =⋅ − =

− == −



for x.

5 4 20

5 4 0 20

5 20

4

x y

x

x

x

− =− ⋅ =

==



45. Graph 2 3 18x y− = .


2 0 3 18

3 18

6

x y

y

y

y

− =⋅ − =

− == −



for x. 2 3 18

2 3 0 18

2 18

9

x y

x

x

x

− =− ⋅ =

==





46. Graph 3 2 18x y+ = − .


3 0 2 18

2 18

9

x y

y

y

y

+ = −⋅ + = −

= −= −



for x. 3 2 18

3 2 0 18

3 18

6

x y

x

x

x

+ = −+ ⋅ = −

= −= −



47. Graph 3 12y x= − .

To find the y-intercept, let 0x = and solve for y.

( )3 12

3 12 0

3 0

0

y x

y

y

y

= −= −==



for x.

( )3 12

3 0 12

0 12

0

y x

x

x

x

= −= −= −=

The x-intercept is ( )0,0 . Since the x- and y-

intercepts are the same, we obtain one additional point. Letting 2x = ,

( )3 12

3 12 2

3 24

8

y x

y

y

y

= −= −= −= − ,

We obtain the point ( )2, 8− .

Plot these points and draw a line. A third point could be used as a check. 48. Graph 5 15y x= .

To find the y-intercept, let 0x = and solve for y. 5 15

5 15 0

5 0

0

y x

y

y

y

== ⋅==



for x. 5 15

5 0 15

0 15

0

y x

x

x

x

=⋅ ===

The x-intercept is ( )0,0 . Since the x- and y-

intercepts are the same, we obtain one additional point. Letting 2x = ,

5 15

5 15 2

5 30

6

y x

y

y

y

== ⋅== ,

we obtain the point ( )2,6 .

Plot these points and draw a line. A third point to be used as a check.



49. Graph ( ) 3 7f x x= − .

To find the y-intercept, let 0x = and solve

for ( )f x .

( )( )( )( )

3 7

3 0 7

0 7

7

f x x

f x

f x

f x

= −

= ⋅ −

= −

= −




0 3 7

7 3

7

3

f x x

x

x

x

= −= −=

=

The x-intercept is 7

,03

⎛ ⎞⎜ ⎟⎝ ⎠

.

Plot these points and draw a line. A third point to be used as a check.

50. Graph ( ) 2 9g x x= − .


for ( )g x .

( )( )( )( )

2 9

2 0 9

0 9

9

g x x

g x

g x

g x

= −

= ⋅ −

= −

= −


To find the x-intercept, let ( ) 0g x = and


0 2 9

9 2

9

2

g x x

x

x

x

= −= −=

=


,02

⎛ ⎞⎜ ⎟⎝ ⎠

.


51. Graph 5 5y x− = .


5 0 5

5 5

1

y x

y

y

y

− =− =

==


To find the x-intercept, let 0y = and

solve for x. 5 5

5 0 5

5

5

y x

x

x

x

− =⋅ − =

− == −

The x-intercept is ( 5,0)− .


52. Graph 3 3y x− = .


3 0 3

3

y x

y

y

− =− ⋅ =

=

The y-intercept is (0,3) .


solve for x.



3 3

0 3 3

3 3

1

y x

x

x

x

− =− =− =

= −



53. Graph 0.2 1.1 6.6y x− = .

To find the y-intercept, let 0x = and solve for y. 0.2 1.1 6.6

0.2 1.1 0 6.6

0.2 6.6

33

y x

y

y

y

− =− ⋅ =

==



solve for x. 0.2 1.1 6.6

0.2 0 1.1 6.6

1.1 6.6

6

y x

x

x

x

− =⋅ − =

− == −



54. Graph 1 1

13 2

x y+ = .

To find the y-intercept, let 0x = and solve for y.

1 1

13 2

1 10 1

3 21

12

2

x y

y

y

y

+ =

⋅ + =

=

=



solve for x.

1 1

13 2

1 10 1

3 21

13

3

x y

x

x

x

+ =

+ ⋅ =

=

=

The x-intercept is (3,0) .


55. ( ) 20 4f x x= −


for ( )f x .

( )( )( )( )

20 4

20 4 0

20 0

20

f x x

f x

f x

f x

= −

= − ⋅

= −

=



solve for x.

( ) 20 4

0 20 4

4 20

5

f x x

x

x

x

= −= −==




Choice c) with window [ ]10,10, 10,30− −

will display both intercepts. 56. ( ) 3 7g x x= +


for ( )g x .

( )( )( )( )

3 7

3 0 7

0 7

7

g x x

g x

g x

g x

= +

= ⋅ +

= +

=


To find the x-intercept, let ( ) 0g x = and


0 3 7

7 3

7

3

g x x

x

x

x

= += +

− =

− =


,03

⎛ ⎞−⎜ ⎟⎝ ⎠.

Choice a) with window [ ]10,10, 10,10− −

will display both intercepts. 57. ( ) 35 7000p x x= − +


for ( )p x .

( )( )( )( )

35 7000

35 0 7000

0 7000

7000

p x x

p x

p x

p x

= − +

= − ⋅ +

= +

=


To find the x-intercept, let ( ) 0p x = and

solve for x. ( ) 35 7000

0 35 7000

35 7000

200

p x x

x

x

x

= − += − +==


Choice d) with window [ ]0,500,0, 10,000

will display both intercepts.

58. ( ) 0.2 0.01r x x= −


for ( )r x .

( )( )( )( )

0.2 0.01

0.2 0.01 0

0.2 0

0.2

r x x

r x

r x

r x

= −

= − ⋅

= −

=

The y-intercept is ( )0,0.2 .

To find the x-intercept, let ( ) 0r x = and

solve for x. ( ) 0.2 0.01

0 0.2 0.01

0.01 0.2

0.220

0.01

r x x

x

x

x

= −= −=

= =


Choice b) with window [ ]5,30, 1,1− − will

display both intercepts. 59. We first solve for y and determine the slope of each line. 8

8

x y

y x

+ == +

The slope of 8y x= + is 1.

5

5

y x

y x

− = −= −

The slope of 5y x= − is 1.

The slopes are the same; the lines are parallel. 60. We first solve for y and determine the slope of each line. 2 3

2 3

x y

y x

− == −

The slope of 2 3x y− = is 2.

2 9

2 9

y x

y x

− == +

The slope of 2 9y x− = is 2.

The slopes are the same; the lines are parallel.



61. We first solve for y and determine the slope of each line. 9 3

3 9

y x

y x

+ == −

The slope of 9 3y x+ = is 3.

3 2

3 2

3 2

x y

x y

y x

− = −+ =

= +

The slope of 3 2x y− = − is 3.

The slopes are the same; the lines are parallel. 62. We first solve for y and determine the slope of each line. 8 6

6 8

y x

y x

+ = −= − −

The slope of 8 6y x+ = − is –6.

2 5

2 5

x y

y x

− + == +

The slope of 2 5x y− + = is 2.

The slopes are not the same; the lines are not parallel. 63. We determine the slope of each line. ( ) 3 9f x x= +

The slope of ( ) 3 9f x x= + is 3.

2 6 2

3 1

y x

y x

= − −= − −

The slope of 2 6 2y x= − − is –3.

The slopes are not the same; the lines are not parallel. 64. We determine the slope of each line. ( ) 7 9f x x= − −

The slope of ( ) 7 9f x x= − − is –7.

3 21 7

77

3

y x

y x

− = +

= − −

The slope of 7

73

y x= − − is –7.

The slopes are the same; the lines are parallel. 65. We determine the slope of each line. The slope of ( ) 4 3f x x= − is 4.

4 7

4 7

1 7

4 4

y x

y x

y x

= −= − +

= − +

The slope of 4 7y x= − is 1

4− .

The product of their slopes is 1

44

⎛ ⎞−⎜ ⎟⎝ ⎠, or –1;

the lines are perpendicular. 66. We determine the slope of each line. 2 5 3

2 3 5

2 3

5 52 3

5 5

x y

x y

x y

y x

− = −+ =

+ =

= +

The slope of 2 5 3x y− = − is 2

5.

2 5 4

5 2 4

2 4

5 5

x y

y x

y x

+ == − +

= − +

The slope of 2 5 4x y+ = is 2

5− .

The product of their slopes is

2 2 4

15 5 25⎛ ⎞− = − ≠ −⎜ ⎟⎝ ⎠

, so the lines are not

perpendicular. 67. We determine the slope of each line. 2 7

2 7

1 7

2 2

x y

y x

y x

+ == − +

= − +

The slope of 2 7x y+ = is 1

2− .

2 4 4

4 2 4

2 4

4 41

12

x y

y x

y x

y x

+ == − +

= − +

= − +

The slope of 2 4 4x y+ = is 1

2− .



The product of their slopes is 1 1

2 2⎛ ⎞− −⎜ ⎟⎝ ⎠

, or

1

4, so the lines are not perpendicular.

68. We determine the slope of each line. The slope of 7y x= − + is 1− .

The slope of ( ) 3f x x= + is 1.

The product of their slopes is ( )1 1 1− = − , so

the lines are perpendicular. 69. 7

8 3y x= −

a) The slope of this line is 7

8, so the slope

of a parallel line is also 7

8.

b) The reciprocal of this slope is 8

7, and the

opposite of this number is 8

7− , so the

slope of a perpendicular line is 8

7− .

70. 9

10 4y x= − +


10− , so the

slope of a parallel line is also 9

10− .


9− , and

the opposite of this number is 10

9, so the


9.

71. 51

4 8y x= − −


4− , so the slope


4− .

b) The reciprocal of this slope is −4, and the opposite of this number is 4, so the slope of a perpendicular line is 4. 72. 31

6 11y x= −


6, so the slope


6.

b) The reciprocal of this slope is 6, and the opposite of this number is −6, so the slope of a perpendicular line is −6. 73. 20 12x y− =

We rewrite the second equation in slope- intercept form: 20 12

20 12

x y

y x

− == −

a) The slope of this line is 20, so the slope of a parallel line is also 20.


20, and


20− , so

the slope of a perpendicular line is 1

20− .

74. 15 30y x+ =

We rewrite the second equation in slope- intercept form: 15 30

15 30

y x

y x

+ == − +

a) The slope of this line is −15, so the slope of a parallel line is also −15.


15− , and


15, so the


15.

75. 4x y+ =

We rewrite the second equation in slope- intercept form: 4

4

x y

y x

+ == − +

a) The slope of this line is −1, so the slope of a parallel line is also −1.


11

− = − ,

and the opposite of this number is 1, so the slope of a perpendicular line is 1.



76. 19x y− =

We rewrite the second equation in slope- intercept form: 19

19

x y

y x

− == −

a) The slope of this line is 1, so the slope of a parallel line is also 1.


11= , and

the opposite of this number is −1, so the slope of a perpendicular line is −1. 77. The slope of the given line is 3. Therefore, the slope of a line parallel to it is also 3. The y-intercept is ( )0,9 , so the equation of the

desired function is ( ) 3 9f x x= + .

78. The slope of the given line is –5. Therefore, the slope of a line parallel to it is also –5. The y-intercept is ( )0, 2− , so the equation of the

desired function is ( ) 5 2f x x= − − .

79. First we find the slope of the given line. 2 3

2 3

x y

y x

+ == − +

The slope of the given line is –2. Therefore, the slope of a line parallel to it is also –2. The y-intercept is ( )0, 5− , so the equation of the

desired function is ( ) 2 5f x x= − − .


3 10

3 10

x y

x y

y x

= +− =

= −

The slope of the given line is 3. Therefore, the slope of a line parallel to it is also 3. The y-intercept is ( )0,1 , so the equation of the

desired function is ( ) 3 1f x x= + .

81. First we find the slope of the given line. 2 5 8

5 2 8

2 8

5 5

x y

y x

y x

+ == − +

= − +

The slope of the given line is 2

5− . Therefore,

the slope of a line parallel to it is also 2

5− .


0,3

⎛ ⎞−⎜ ⎟⎝ ⎠, so the equation of

the desired function is ( ) 2 1

5 3f x x= − − .

82. First we find the slope of the given line. 3 6 4

3 4 6

6 3 4

1 2

2 3

x y

x y

y x

y x

− =− =

= −

= −


2. Therefore,

the slope of a line parallel to it is also 1

2. The

y-intercept is 4

0,5

⎛ ⎞⎜ ⎟⎝ ⎠

, so the equation of the

desired function is ( ) 1 4

2 5f x x= + .


0 4

y

y x

== +

The slope of the given line is 0. Therefore, the slope of a line parallel to it is also 0. The y-intercept is ( )0, 5− , so the equation of the

desired function is ( ) 5f x = − .

84. First we find the slope of the given line.

5 10

10 5

10

2

y

y

y x

==

= +

The slope of the given line is 0. Therefore, the slope of a line parallel to it is also 0. The y-intercept is ( )0,12 , so the equation of the

desired function is ( ) 12f x = .

85. The slope of the given line is 1. The slope of a line perpendicular to it is the opposite of the reciprocal of 1, or –1. The y-intercept is ( )0,4 , so we have 4y x= − + .



86. The slope of the given line is 2. The slope of a line perpendicular to it is the opposite of the

reciprocal of 2, or 1

2− . The y-intercept is

( )0, 3− , so we have 1

32

y x= − − .

87. First find the slope of the given line.

2 3 6

3 2 6

22

3

x y

y x

y x

+ == − +

= − +


3− . The slope

of a line perpendicular to it is the opposite of

the reciprocal of 2

3− , or

3

2. The y-intercept

is ( )0, 4− , so we have 3

42

y x= − .

88. First find the slope of the given line. 4 2 8

2 4 8

2 4

x y

y x

y x

+ == − += − +

The slope of the given line is 2− . The slope of a line perpendicular to it is the opposite of

the reciprocal of 2− , or 1

2. The y-intercept

is ( )0,8 , so we have 1

82

y x= + .

89. First find the slope of the given line. 5 13

5 13

5 13

x y

x y

y x

− =− =

= −

The slope of the given line is 5. The slope of a line perpendicular to it is the opposite of

the reciprocal of 5, or 1

5− . The y-intercept

is 1

0,5

⎛ ⎞⎜ ⎟⎝ ⎠

, so we have 1 1

5 5y x= − + .

90. First find the slope of the given line. 2 5 7

2 7 5

5 2 7

2 7

5 5

x y

x y

y x

y x

− =− =

= −

= −


5. The slope

of a line perpendicular to it is the opposite of

the reciprocal of 2

5, or

5

2− . The y-intercept

is 1

0,8

⎛ ⎞−⎜ ⎟⎝ ⎠, so we have

5 1

2 8y x= − − .

91. This equation is in the standard form for a linear equation, Ax By C+ = , with 5A = ,

3B = − , and 15C = . Thus, it is a linear equation. Solve for y to find the slope. 5 3 15

3 5 15

55

3

x y

y x

y x

− =− = − +

= −

The slope is 5

3.

92. We write 3 5 15 0x y+ + = in standard form

for a linear equation, Ax By C+ = , as

3 5 15x y+ = − with 3A = , 5B = , and

15C = − . Thus, it is a linear equation. Solve for y to find the slope. 3 5 15

5 3 15

33

5

x y

y x

y x

+ = −= − −

= − −

The slope is 3

5− .

93. We write 16 4 10y+ = in standard form

for a linear equation, Ax By C+ = , as

0 4 6x y+ = − with 0A = , 4B = , and

6C = − . Thus, it is a linear equation. Solve for y to find the slope.

4 6

30

2

y

y x

= −

= + −

The slope is 0 .



94. We write 3 12 0x − = in standard form for a linear equation, Ax By C+ = , as

3 0 12x y+ = with 3A = , 0B = , and

12C = . Thus, it is a linear equation and the line is vertical. 95. 10xy =

The equation cannot be written in standard form for a linear equation, Ax By C+ = ,

since there is an xy term. Thus, the equation is not linear.

96. 10

yx

=

10xy =

When trying to convert the equation into standard linear form, multiply both sides by x. The equation cannot be written in standard form for a linear equation, since there is an xy term. The equation is not linear. 97. ( )3 7 2 4y x= −

The equation can be written in standard form for a linear equation, Ax By C+ =

( )3 7 2 4

3 14 28

14 3 28

y x

y x

x y

= −= −

− + = −

with 14A = − , 3B = , and 28C = − . Thus, it is a linear equation. Solve for y to find the slope. 14 3 28

3 14 28

14 28

3 3

x y

y x

y x

− + = −= −

= −

The slope is 14

3.

98. ( )2 5 3 5

10 6 5

10 5 6

6 5 10

x y

x y

y x

x y

− =− =

= ++ =

This is a linear equation since it is in the form Ax By C+ = , with 6A = , 5B = , and

10C = . Solve for y to find the slope. 6 5 10

5 6 10

62

5

x y

y x

y x

+ == − +

= − +

The slope is 6

5− .

99. ( ) 1g x

x=

Replacing ( )g x with y and attempt to write

the equation in standard form.

1

1

yx

xy

=

=

The equation is not linear because it has an xy-term. 100. ( ) 3f x x=

Replace ( )f x with y and attempt to write the

equation in standard form. 3

3 0

y x

x y

=

− + =

The equation is not linear because it has an 3x -term.

101. ( ) 2

5

f xx=

Replace ( )f x with y and attempt to write the

equation in standard form.

2

2

2

5

5

5 0

yx

y x

x y

=

=

− + =

The equation is not linear because it has an 2x -term.

102. ( )

32

g xx= +

Replace ( )g x with y and attempt to write the

equation in standard form.

32

6 2

2 6

yx

y x

x y

= +

= +− + =

The equation is linear. From the next-to-last step above, we see that the slope is 2.



103. Thinking and Writing Exercise. For each 10 km/hr increase, the corresponding force of resistance does not remain constant but increases. The slope is not constant, so a linear function does not give an approximate fit.

104. Thinking and Writing Exercise. For each 5°

temperature decrease, the corresponding decrease in wind chill temperature is 6° or 7°. Thus, the slope is nearly constant, so a linear function will give an approximate fit.

105. 3 10 30

110 3 30

⎛ ⎞− = − = −⎜ ⎟⎝ ⎠

106. 1 2 1 2

2 12 1 2 2

⎛ ⎞ ⎛ ⎞− = − = − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

107. ( ) ( )3 1 3 1 3 3x x x− ⎡ − − ⎤ = − + = − −⎣ ⎦

108. ( ) [ ]10 7 10 7 10 70x x x− ⎡ − − ⎤ = − + = −⎣ ⎦

109. 2 1 2 1

1 13 2 3 2

2 2 2 21

3 6 3 3

x x

x x

⎡ ⎤⎛ ⎞ ⎡ ⎤− − − = + −⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦⎣ ⎦

= + − = −

110. 3 2 3 6

3 32 5 2 10

3 3 3 123

2 5 2 5

x x

x x

⎛ ⎞− − − = − + −⎜ ⎟⎝ ⎠

= − + − = − −

111. Thinking and Writing Exercise. Consider the

equation Ax By C+ = . Writing this equation

in slope-intercept form we have A C

y xB B

= − + . If we choose a value for x for

which is not a multiple of B, the corresponding y-value will be a fraction. Similarly, if Ax− is a multiple of B but C is

not, then the corresponding y-value will be a fraction. Under these conditions, Jim will avoid fractions if he graphs using intercepts.

112. Thinking and Writing Exercise. A line’s

x- and y-intercepts coincide only when the line passes through the origin. The equation for such a line is of the form y mx= .

113. The line contains the points ( )5,0 and

( )0, 4− . We use the points to find the slope.

4 0 4 4

Slope = 0 5 5 5

− − −= =− −

Then the slope-intercept equation is

4

45

y x= − . We rewrite this equation in

standard form.

4

45

5 4 20

4 5 20

y x

y x

x y

= −

= −− + = −

This equation can also be written as 4 5 20x y− = .

114. ( )0y mx b m= + ≠

To find the x-intercept, let 0y = .

0 mx b

b mx

bx

m

= +− =

− =

Thus, the x-intercept is ,0b

m⎛ ⎞−⎜ ⎟⎝ ⎠

.

115. 23rx y p s+ = −

The equation is in standard form with A r= , 3B = , and 2C p s= − . It is linear.

116. 2 9py sx r y= − −

Try to put the equation in standard form.

( )

2

2

2

9

9

9

py sx r y

sx py r y

sx p r y

= − −− + + = −

− + + = −

The equation is in standard form with A s= − , 2B p r= + , and 9C = − . It is

linear.



117. 2 5r x py= +

Try to put the equation in standard form.

2

2

5

5

r x py

r x py

= +− =

The equation is in standard form with 2A r= , B p= − , and 5C = . It is linear.

118. 17x

pyr− =

The equation is in standard form with 1

Ar

= ,

B p= − , and 17C = . It is linear.

119. Let equation A have intercepts ( ),0a and

( )0,b . Then equation B has intercepts

( )2 ,0a and ( )0,b .

Slope of A = 0

0

b b

a a

− = −−

Slope of B = 0 1

0 2 2 2

b b b

a a a

− ⎛ ⎞= − = −⎜ ⎟⎝ ⎠−

The slope of equation B is 1

2 the slope of

equation A. 120.

( ) ( )3 5 8

5 3 8

ax y x by

a x b y

+ = − +− + + =

If the graph is a horizontal line, then the coefficient of x is 0.

5 0

5

a

a

− ==

Then we have ( )3 8b y+ = .

If the graph passes through ( )0,4 , we have

( )3 4 8

3 2

1

b

b

b

+ =+ == −

121. First write the equation in standard form.

( ) ( )

3 5 8

5 3 8

5 3 8

ax y x by

ax x y by

a x b y

+ = − +− + + =

− + + =

If the graph is a vertical line, then the coefficient of y is 0.

3 0

3

b

b

+ == −

Then we have ( )5 8a x− =

If the line passes through ( )4,0 , we have:

( )5 4 8 Substituting 4 for

5 2

7

a x

a

a

− =− =

=

122. a) 3.6x =

b) 1.52x = − c) 3 5 7 2

7 4

7

4

x x

x

x

− = +− =

− =

d) ( )2 5 10

2 10 10

20

x x

x x

x

− = +− = +

=



123. a) Solve each equation for y, enter each on the equation-editor screen, and then examine a table of values for the two functions. Since the difference between the y-values is the same for all x-values, the lines are parallel. b) Solve each equation for y, enter each on the equation-editor screen, and then examine a table of values for the two functions. Since the difference between the y-values is not the same for all x- values, the lines are not parallel. Exercise Set 2.4 1. True 2. False 3. False 4. True 5. True 6. True 7. True 8. True 9. False 10. True 11. The point-slope form of a line is:

( )1 1 .y y m x x− = −

Substitute 1 13, 5, and 2m x y= = = to get:

( )2 3 5 .y x− = −

To graph the line, plot the point (5 2)., Then

use change in 3 3change in 1 1

ym

x−= = =−

to generate a

second point ( ) ( )( ) ( )5 1 2 3 4, 1 .+ − , + − = −

Plot this second point and draw the line

passing through both points.

12. The point-slope form of a line is:

( )1 1 .y y m x x− = −

Substitute 1 12, 3, and 4m x y= = = to get:

( )4 2 3 .y x− = −

To graph the line, plot the point (3,4). Then


ym

x−= = =−

to generate a

second point ( ) ( )( ) ( )3 1 4 2 2,2 .+ − , + − = Plot

this second point and draw the line passing through both points.


( )1 1 .y y m x x− = −

Substitute 1 14, 1, and 2m x y= − = = to get:

( )2 4 1 .y x− = − −

To graph the line, plot the point (1 2)., Then


ym

x−= = − = to generate a

second point ( )( ) ( )1 1 2 4 2, 2 .+ , + − = − Plot





( )1 1 .y y m x x− = −


( )4 5 1 .y x− = − −

To graph the line, plot the point (1,4). Then


ym


second point ( )( ) ( )1 1 4 5 2, 1 .+ , + − = − Plot



( )1 1 .y y m x x− = −

Substitute 1 11 , 2, and 42

m x y= = − = − to get:

( ) ( )( ) ( )1 14 2 , or 4 2 .

2 2y x y x− − = − − + = +

To graph the line, plot the point ( )2, 4 .− −

Then use change in 1change in 2

ym

x= = to generate a

second point ( ) ( )2 2 4 1 0, 3 .− + , − + = − Plot



( )1 1 .y y m x x− = −

Substitute 1 11, 5, and 7m x y= = − = − to get:

( ) ( )( )7 1 5 , or 7 5.y x y x− − = − − + = +

To graph the line, plot the point ( )5, 7 .− −

Then use change in 1change in 1

ym

x= = to generate a

second point ( ) ( )5 1 7 1 4, 6 .− + ,− + = − − Plot



( )1 1 .y y m x x− = −


( ) ( )0 1 8 , or 8 .y x y x− = − − = − −

To graph the line, plot the point ( )8,0 . Then


ym

x= = − =

−to generate a

second point ( )( ) ( )8 1 0 1 7,1 .+ − , + = Plot this

second point and draw the line passing through both points.


( )1 1 .y y m x x− = −

Substitute 1 13, 2, and 0m x y= − = − = to get:

( )( ) ( )0 3 2 , or 3 2 .y x y x− = − − − = − +

To graph the line, plot the point ( )2,0 .− Then


ym


second point ( )( ) ( )2 1 0 3 1, 3 .− + , + − = − −

Plot this second point and draw the line




19. ( )

( )1 1

13 5

4y x

y y m x x

− = −

− = −

1 1

1, 5, 3,

4m x y= = = so the slope is

1

4 and

the point ( )5,3 is on the graph.

20. ( )

( )1 1

5 6 1y x

y y m x x

− = −− = −

1 16, 1, 5,m x y= = = so the slope is 6 and the

point ( )1,5 is on the graph.

21. ( )

( ) ( )

( )1 1

1 7 2

1 7 2

y x

y x

y y m x x

+ = − −− − = − −− = −

1 17, 2, 1,m x y= − = = − so the slope is 7− and

the point ( )2, 1− is on the graph.

22. ( )

( )( )

( )1 1

24 8

32

4 83

y x

y x

y y m x x

− = − +

− = − − −

− = −

1 1

2, 8, 4,

3m x y= − = − = so the slope is

2

3−

and the point ( )8,4− is on the graph.

23. ( )

( )( )

( )1 1

106 4

310

6 43

y x

y x

y y m x x

− = − +

− = − − −

− = −

1 1

10, 4, 6,

3m x y= − = − = so the slope is

10

3−

and the point ( )4,6− is on the graph.

24. ( )( ) ( )

( )1 1

1 9 7

1 9 7

y x

y x

y y m x x

+ = − −− − = − −− = −

1 19, 7, 1,m x y= − = = − so the slope is 9− and

the point ( )7, 1− is on the graph.

25.

( )

( )1 1

5

0 5 0

y x

y x

y y m x x

=− = −− = −

1 15, 0, 0,m x y= = = so the slope is 5 and the

point ( )0,0 is on the graph.

26.

( )

( )1 1

4

54

0 05

y x

y x

y y m x x

=

− = −

− = −

1 1

4, 0, 0,

5m x y= = = so the slope is

4

5and

the point ( )0,0 is on the graph.

27. ( )

( ) ( )1 1

1 1

Point-slope equation

4 2 1 Substituting 2 for ,

and (1, 4) for ( , )

4 2 2 Simplifying

2 6 Subtracting 4 from

y y m x x

y x m

x y

y x

y x

− = −

− − = −−

+ = −= −

both sides

( ) 2 6 Using function notationf x x= −

To graph the line, plot the point ( )1, 4− and the

y-intercept ( )0, 6 ,− and then draw the line




28. ( )( )( )

( ) ( )

( )

1 1

1 1


Substituting 4 for 5 4 1

and 1,5 for ,

5 4 4 Simplifying

4 1 Adding 5 to both sides

Function notation4 1

y y m x x

my x

x y

y x

y x

f x x

− = −−− = − − −

−− = − −

= − += − +

To graph the line, plot the point ( )1,5− and the

y-intercept ( )0,1 , and then draw the line passing

through both points.

29. ( )

( )( )( ) ( )

( )

1 1

3355

1 1

3 125 5

3 285 5

3 285 5


Substituting for 8 4

and 4,8 for ,

Simplifying8

Adding 8 to both sideds

Function notation

y y m x x

my x

x y

y x

y x

f x x

− = −−− = − − −

−− = − −

= − += − +


y-intercept 28 3

0, 0,5 ,5 5

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ and then draw the

line passing through both points.

30. ( )

( )( )( ) ( )

( )

1 1

1155

1 1

1 25 5

315 5

315 5



and 2,1 for ,

Simplifying1

Adding 1 to both sides

Function notation

y y m x x

my x

x y

y x

y x

f x x

− = −−− = − − −

−− = − −

= − += − +


y-intercept 3

0, ,5

⎛ ⎞⎜ ⎟⎝ ⎠ and then draw the line


31. ( )

( ) ( )( )

( ) ( )

( )

1 1

1 1


Substituting 0.6 4 0.6 3for and

3, 4 for ,

4 0.6 1.8 Simplifying

0.6 5.8 Subtracting 4 from

both sides

Function notation0.6 5.8

y y m x x

y xm

x y

y x

y x

f x x

− = −−− − = − − −

− −+ = − −

= − −

= − −

To graph the line, plot the point ( )3, 4− − and the

y-intercept ( )0, 5.8 ,− and then draw the line


32. ( )

( ) ( )( ) ( )

( )

1 1

1 1


Substituting 2.3 for 5 2.3 4

and 4, 5 for ,

5 2.3 9.2 Simplifying

2.3 14.2 Subtracting 5 from

both sides

Function notation2.3 14.2

y y m x x

my x

x y

y x

y x

f x x

− = −

− − = −−

+ = −= −

= −


y-intercept ( )0, 14.2 ,− and then draw the line




33. ( )

( ) ( )( ) ( )

( )

1 1

2277

1 1

27

27

27



and 0, 6 for ,

Simplifying6 0

Subtracting 6 from 6

both sides

Function notation6

y y m x x

my x

x y

y x

y x

f x x

− = −

− − = −−

+ = −= −

= −

To graph the line, plot the point ( )0, 6 .− Then,

since the given point is the y-intercept, use change in 2change in 7

ym

x= = to generate a second point

( ) ( )0 7, 6 2 7, 4 .+ − + = − Plot this second point

and draw the line passing through both points.

34. ( )

( )( ) ( )

( )

1 1

1144

1 1

14

14

14



and 0,3 for ,

Simplifying3

Adding 3 to both sides3

Function notation3

y y m x x

my x

x y

y x

y x

f x x

− = −

− = −

− == += +

To graph the line, plot the point ( )0,3 . Then,

since the given point is the y-intercept, use change in 1change in 4

ym

x= = to generate a second point

( ) ( )0 4,3 1 4,4 .+ + = Plot this second point and

draw the line passing through both points.

35. ( )

( )( )

( )

1 1

3355

1

1

3 125 5

3 425 5

3 425 5



, 4 for , and 6

for

Simplifying6

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −

− = − −−

− = += += +


y-intercept 42 2

0, 0,8 ,5 5

⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ and then draw the

line passing through both points.

36. ( )

( ) ( )

( )

1 1

2277

1

1

2 127 7

2327 7

2327 7



, 6 for , and 5

for

Simplifying5

Simplifying

Function notation

y y m x x

y x

m x

y

y x

y x

f x x

− = −−− − = − −

−

+ = − += − −= − −


y-intercept 23 2

0, 0, 3 ,7 7

⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ and then draw

the line passing through both points.



37. First find the slope of the line.

7 3 4

43 2 1

m−= = =−

Use the point-slope equation with 4m = and

( ) ( )1 12,3 ,x y= (or ( ) ( )1 13,7 ,x y= ).

( )

( )

3 4 2

3 4 8

4 5

4 5 Function notation

y x

y x

y x

f x x

− = −− = −

= −= −


4 8 4

21 3 2

m− −= = =− −


( ) ( )1 13,8 ,x y= (or ( ) ( )1 11,4 ,x y= ).

( )

( )

8 2 3

8 2 6

2 2


y x

y x

y x

f x x

− = −− = −

= += +


( )5 4 5 4 9

4.53.2 1.2 3.2 1.2 2

m− − += = = =− −

Use the point-slope equation with 4.5m =

and ( ) ( )1 11.2, 4 ,x y− = (or ( ) ( )1 13.2, 5 ,x y= ).

( ) ( )

( )

4 4.5 1.2

4 4.5 5.4

4.5 9.4

4.5 9.4 Function notation

y x

y x

y x

f x x

− − = −+ = −

= −= −


( )( )

8.5 2.5 8.5 2.5 112.2

4 1 4 1 5m

− − += = = =− − +

Use the point-slope equation with 2.2m =

and ( ) ( )1 11, 2.5 ,x y− − = (or ( ) ( )1 14,8.5 ,x y= ).

( ) ( )( )( )

( )

2.5 2.2 1

2.5 2.2 1

2.5 2.2 2.2

2.2 0.3

2.2 0.3 Function notation

y x

y x

y x

y x

f x x

− − = − −

+ = ++ = +

= −= −


( )1 5 1 5 4

20 2 0 2 2

m− − − − += = = = −

− − −

Observe that the y-intercept is ( )0, 1− . Using

2m = − and the slope-intercept equation immediately gives:

2 1

( ) 2 1 Function notation

y x

f x x

= − −= − −

.

One could also use the point-slope equation with 2m = − and either point as ( )1 1, .x y


( )7 0 7 0 7

0 2 0 2 2m

− − − −= = = −− − +

Observe that the y-intercept is ( )0, 1− . Using

7

2m = − and the slope-intercept equation

immediately gives: 7

727

( ) 7 Function notation2

y x

f x x

= − −

= − −.

One could also use the point-slope equation

with 7

2m = − and either point as ( )1 1,x y .

43. First find the slope of the line

( )( )

5 10 5 10 5

3 6 3 6 3m

− − − − += = =− − − − +

Use the point-slope equation with5

3m = and

( ) ( )1 16, 10 ,x y− − = (or ( ) ( )1 13, 5 ,x y− − = ).

( ) ( )( )

( )

( )

510 6

35

10 635

10 1035

35

Function notation3

y x

y x

y x

y x

f x x

− − = − −

+ = +

+ = +

=

=




( )( )

9 3 9 3 62

4 1 4 1 3m

− − − − + −= = = =− − − − + −


( ) ( )1 11, 3 ,x y− − = (or ( ) ( )1 14, 9 ,x y− − = ).

( ) ( )( )( )

( )

3 2 1

3 2 1

3 2 2

2 1


y x

y x

y x

y x

f x x

− − = − −

+ = ++ = +

= −= −

45. Plot the given data values using the

incandescent wattage as the first coordinate and the corresponding CFL wattage as the second coordinate.

To predict the CFL wattage that creates light equivalent to a 75-watt incandescent bulb, locate the point on the line directly above 75. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 19 watts. To predict the CFL wattage that creates light equivalent to a 120-watt incandescent bulb, locate the point on the line directly above 120. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 30 watts.

46. The data values were plotted as shown.

To predict the CFL wattage that creates light

equivalent to a 40-watt incandescent bulb, locate the point on the line directly above 40. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage

is about 9 watts. To predict the CFL wattage that creates light equivalent to a 150-watt incandescent bulb, locate the point on the line directly above 150. Then move horizontally to the vertical axis and read the CFL wattage value. The wattage is about 38 watts.

47. Plot the given data values using body weight

as the first coordinate and the corresponding number of drinks as the second coordinate.

To estimate the number of drinks that a 140-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 140. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 3.5. To estimate the number of drinks that a 230-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 230. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 5.75.

48. The data values were plotted as shown.

To estimate the number of drinks that a 120-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 120. Then move horizontally to the vertical axis and read the corresponding number of drinks. The estimated number of drinks is 3. To estimate the number of drinks that a 250-lb person would have to drink to be considered intoxicated, locate the point on the line directly above 250. Then move horizontally to the vertical axis and read the



corresponding number of drinks. The estimated number of drinks is 6.25.

49. a) The problem says to let t = the number of

years after 2000, and a(t) = the world production capacity, in millions of vehicles, for the year t. Then the production capacity of 84 million vehicles in 2008, and of 97 million in 2015 correspond to data points: (2008 2000,84) (8,84)− = and

(2015 2000,97) (15,97).− =

Find the slope of the function that fits the data:

(15) (8)

15 897 84 13

7 7

a am

−=−

−= =

Use the value of m, and either data point, and substitute them into the point-slope equation for a line.

( )13( ) 84 8

713 104

( ) 847 7

13 104 588( )

7 7 713 484

( ) , 7 7

13 484or ( )

7

a t t

a t t

a t t

a t t

ta t

− = −

= − +

= − +

= +

+=

b) In 2013, t = 2013 – 2000 = 13.

( ) ( )13 13 484 169 48413 =

7 7653 2

93 93.37 7

a+ +=

= = ≈

The model predicts the production capacity in 2013 will be about 93.3 million vehicles.

c) Set ( ) 100a t = and solve for t. 13 484

1007

13 484 700

13 700 484 216

21616.6

132000 16.6 2016.6

t

t

t

t

+ =

+ == − =

= ≈

+ =

The model predicts that the production

capacity will reach 100 million vehicles in 2016.


years after 2000, and v(t) = the number of Las Vegas convention attendees, in millions, for the year t. Then the 4.6 million attendees in 2002, and 6.1 million in 2006 correspond to data points: (2002 2000,4.6) (2,4.6)− = and

(2006 2000,6.1) (6,6.1).− =


(6) (2)

6 26.1 4.6 1.5

0.3754 4

v vm

−=−−= = =


( )( ) 4.6 0.375 2

( ) 0.375 0.75 4.6

( ) 0.375 3.85

v t t

v t t

v t t

− = −= − += +

b) In 2011, t = 2011 – 2000 = 11. ( ) ( )11 0.375 11 3.85

4.125 3.85

7.975

v = += +=

The model predicts there will be 7.975 million attendees in 2011.

c) Set ( ) 8v t = and solve for t. 0.375 3.85 8

0.375 8 3.85 4.15

4.1511.1

0.3752000 11.1 2011.1

t

t

t

+ == − =

= ≈

+ =

The model predicts that the number of attendees will reach 8 million in 2011.


years after 1990, and E(t) = the life expectancy, in years, for females born in the year t. Then the expected value of 79.0 years in 1994, and 80.2 years in 2006 correspond to data points: (1994 1990,79.0) (4,79.0)− = and

(2006 1990,80.2) (16,80.2).− =




(16) (4)

16 480.2 79.0 1.2

0.112 12

E Em

−=−

−= = =


( )( ) 79 0.1 4

( ) 0.1 0.4 79

( ) 0.1 78.6

E t t

E t t

E t t

− = −= − += +

b) In 2012, t = 2012 – 1990 = 22. ( )(22) 0.1 22 78.6

2.2 78.6 80.8

E = += + =

The model predicts that the life expectancy for a female born in 2012 will be 80.8 years.


years after 1990, and E(t) = the life expectancy, in years, for males born in the year t. Then the expected value of 72.4 years in 1994, and 75.1 years in 2006 correspond to data points: (1994 1990,72.4) (4,72.4)− = and

(2006 1990,75.1) (16,75.1).− =


(16) (4)

16 475.1 72.4 2.7

0.22512 12

E Em

−=−

−= = =


( )( ) 72.4 0.225 4

( ) 0.225 0.9 72.4

( ) 0.225 71.5

E t t

E t t

E t t

− = −= − += +

b) In 2012, t = 2012 – 1990 = 22. ( )(22) 0.225 22 71.5

4.95 71.5 76.45

E = += + =

The model predicts that the life expectancy for a male born in 2012 will be 76.45 years.


years after 2000, and N(t) = the amount of solid waste recycled, in millions of tons, in the year t. Then the 53 million tons recycled in 2000, and the 61 million

tons recycled in 2008 correspond to data points: (2000 2000,53) (0,53)− = and

(2008 2000,61) (8,61).− =


(8) (0)

8 061 53 8

18 8

N Nm

−=−

−= = =


( )( ) 53 1 0

( ) 53

N t t

N t t

− = −= +

b) In 2012, t = 2012 – 2000 = 12. (12) 12 53

65

N = +=

The model predicts that 65 million tons of solid waste will be recycled in 2012.


years after 2000, and A(t) = the contributions by PACs, in millions of dollars, to federal candidates in the year t. Then the $282 million contributed in 2002, and the $412.8 million contributed in 2008 correspond to data points: (2002 2000,282) (2,282)− = and

(2008 2000,412.8) (8,412.8).− =


( ) ( )8 2

8 2412.8 282 130.8

21.86 6

A Am

−=

−−= = =


( )( ) 282 21.8 2

( ) 21.8 43.6 282

( ) 21.8 238.4

A t t

A t t

A t t

− = −= − += +

b) In 2012, t = 2012 – 2000 = 12. ( )(12) 21.8 12 238.4

261.6 238.4 500

A = += + =

The model predicts that PACs will donate $500 million to federal candidates in 2012.



55. a) The problem says to let t = the number of years after 2006, and C(t) = the percentage of Americans familiar with the term "carbon footprint" in the year t. Then the 38% of Americans familiar with the term in 2007, and the 57% familiar with the term in 2009 correspond to data points: (2007 2006,38) (1,38)− = and

(2009 2006,57) (3,57).− =


( ) ( )3 1

3 157 38 19

9.52 2

C Cm

−=

−−= = =


( )( ) 38 9.5 1

( ) 9.5 9.5 38

( ) 9.5 28.5

C t t

C t t

C t t

− = −= − += +

b) In 2012, t = 2012 – 2006 = 6. ( )(6) 9.5 6 28.5

57 28.5 85.5

C = += + =

The model predicts that 85.5% of Americans will be familiar with the term "carbon footprint" in 2012.

c) Set ( ) 100C t = and solve for t.

9.5 28.5 100

9.5 100 28.5

9.5 71.5

71.57.5

9.52006 7.5 2013.5

t

t

t

t

+ == −=

= ≈

+ =

The model predicts all Americans will be familiar with the term “carbon footprint” in 2013.


years after 2000, and M(t) = the Medicaid long-term care expenses, in billions of dollars, in the year t. Then the $92 billion in expenses in 2002, and the $109 billion in expenses in 2006 correspond to data points: (2002 2000,92) (2,92)− = and

(2006 2000,109) (6,109).− =

Find the slope of the function that fits the

data: ( ) ( )6 2

6 2109 92 17

4.254 4

M Mm

−=

−−= = =


( )( ) 92 4.25 2

( ) 4.25 8.5 92

( ) 4.25 83.5

M t t

M t t

M t t

− = −= − += +

b) In 2010, t = 2010 – 2000 = 10. ( )(10) 4.25 10 83.5

42.5 83.5 126

M = += + =

The model predicts that Medicaid's long-term medical expenses will be $126 billion in 2010.

c) Set ( ) 150M t = and solve for t.

4.25 83.5 150

4.25 150 83.5 66.5

66.515.6

4.252000 15.6 2015.6

t

t

t

+ == − =

= ≈

+ =

The model predicts Medicaid long-term care expenses will reach $150 billion in 2015.


years after 2000, and N(t) = the number of American households, in millions, that conducted some online banking in the year t. Then the 54 million households using online banking in 2009, and the 66 million households using online banking in 2014 correspond to data points: (2009 2000,54) (9,54)− = and

(2014 2000,66) (14,66).− =


(14) (9)

14 966 54 12

2.45 5

N Nm

−=−

−= = =

Use the value of m, and either data point, and substitute them into the point-slope



equation for a line. ( )

( )

( ) 54 2.4 9

( ) 2.4 21.6 54

2.4 32.4

N t t

N t t

N t t

− = −= − += +

b) In 2019, t = 2019 – 2000 = 19. ( ) ( )19 2.4 19 32.4

45.6 32.4 78

N = += + =

The model predicts that 78 million American households will use online banking in 2019.

c) Set ( ) 100N t = and solve for t. 2.4 32.4 100

2.4 100 32.4 67.6

67.628.2

2.42000 28.2 2028.2

t

t

t

+ == − =

= ≈

+ =

The model predicts that the number of American households using online banking will reach 100 million in 2028.


years after 1990, and A(t) = the amount of land, in millions of acres, in the National Park System. Then the 74.9 million acres in 1994, and the 84 million acres in 2010 correspond to data points : (1994 1990,74.9) (4,74.9)− = and

(2010 1990,84) (20,84).− =


( ) ( )20 4

20 484 74.9 9.1

0.5687516 16

A Am

−=

−−= = =

Use the value of m, and either data point, and substitute them into the point-slope equation for a line. ( ) ( )

( )

74.9 0.56875 4

74.9 0.56875 2.275

0.56875 2.275 74.9

0.56875 72.625

A t t

A t

A t t

t

− = −− = −

= − += +

b) In 2014, t = 2014 – 1990 = 24 ( ) ( )24 0.56875 24 72.625

13.65 72.625 86.275

A = += + =

The model predicts that in 2014, there will be 86.275 million acres of land in the National Park system.

59. a) The problem says to let t = the number of years after 1999, and R(t) = the record time, in seconds, in the 100-m run in the year t. Then the record of 9.79 seconds in 1999, and 9.58 seconds in 2009, correspond to data points: (1999 1999,9.79) (0,9.79)− = and

(2009 1999,9.58) (10,9.58).− =


( ) ( )10 0

10 09.58 9.79 0.21

0.02110 10

R Rm

−=

−−= = − = −


( )( ) 9.79 0.021 0

( ) 0.021 9.79

R t t

R t t

− = − −= − +

b) In 2015, t = 2015 – 1999 = 16. ( )(16) 0.021 16 9.79

0.336 9.79 9.454

R = − += − + =

The model predicts that the record in the 100-m run will be 9.454 seconds in 2015. In 2030, t = 2030 – 1999 = 31.

( )(31) 0.021 31 9.79

0.651 9.79 9.139

R = − += − + =

The model predicts that the record in the 100-m run will be 9.139 seconds in 2030.

c) Set ( ) 9.5R t = and solve for t. 0.021 9.79 9.5

0.021 9.5 9.79 0.29

0.2913.8

0.0211999 13.8 2012.8

t

t

t

− + =− = − = −

−= ≈−

+ =

The model predicts that the record in the 100-m run will be 9.5 seconds in 2012.

60. a) The problem says to let d = the depth, in

feet, below the ocean's surface, and P(d) = the pressure, in atms, at a depth d. Then the pressure of 4 atm at 100 feet, and 7 atm at 200 feet, correspond to data points: (4,100) and (7,200).




( ) ( )200 100

200 1007 4 3

0.03100 100

P Pm

−=

−−= = =


( )( ) 4 0.03 100

( ) 0.03 3 4

( ) 0.03 1

P d d

P d d

P d d

− = −= − += +

b) Find P(690). ( )(690) 0.03 690 1

20.7 1 21.7

P = += + =

The model predicts that the pressure at 690 feet below the ocean's surface is 21.7 atmospheres.

61. The points lie approximately in a straight line,

so the graph of this data is linear. 62. The points lie approximately in a straight line,

so the graph of this data is linear. 63. The points do not lie on a straight line, so the

data are not linear. 64. The points do not lie on a straight line, so the

data are not linear. 65. The points lie approximately in a straight line,

so the graph of this data is linear. 66. The points do not lie on a straight line, so the

data are not linear. (They do appear to lie on the absolute value of a linear function.)

67. a) The problem says to let x = the number

of years since 1900, and W = the life expectancy, in years, of a female born in the year x. Enter the data with the number of years since 1900 in 1L and life

expectancy in 2L .

Use the LinReg feature of the graphing calculator to find the equation of the line.

The second screen indicates that the equation is 0.1611 63.6983.W x= +

b) In 2012, 2012 1900 112.x = − = To predict the life expectancy of a female born in 2012, find W(112).

( )(112) 0.1611 112 63.6983

18.0432 63.6983

81.7415 81.7

W = += += ≈

This estimate is 81.7 – 80.8 = 0.9 of a year higher than the previous estimate.


of years since 1900, and M = the life expectancy, in years, of a male born in the year x. Enter the data with the number of years since 1900 in 1L and life

expectancy in 2L .

Use the LinReg feature of the graphing calculator to find the equation of the line:

The second screen indicates that the equation is 0.2008 53.8983.M x= +

b) In 2012, 2012 1900 112.x = − = To predict the life expectancy of a male born in 2012, find M(112).

( ) ( )112 0.2008 112 53.8983.

22.4896 53.8983

76.3879 76.4

M = += += ≈

This estimate is 76.4 – 76.45 = 0.05,− or 0.05 of a year lower, than the previous estimate.



69. a) The problem says to let t = the number of years since 2000, and N = the number of registered nurses, in millions, employed in the year t. Enter the data with the number of years since 2000 in 1L and the

number of registered nurses in 2L .


The second screen indicates that the equation is 0.0433 2.1678.N t= +

b) In 2012, 2012 2000 12.t = − = To estimate the number of registered nurses in 2012, find N(12).

( ) ( )12 0.0433 12 2.1678

0.5196 2.1678

2.6874 2.69

N = += += ≈

The model predicts there will be approximately 2.69 million registered nurses in the U.S. in 2112.


of years since 2005, and A = the amount spent, in millions of dollars, on Cyber Monday in the year x. Enter the data with the number of years since 2005 in 1L and the amount spent in 2L .


The second screen indicates that the equation is 102.6 504.2A x= +

b) In 2011, 2011 2005 6.x = − = To estimate the amount spent on Cyber Monday in 2011, find A(6). ( ) ( )6 102.6 6 504.2

615.6 504.2

1119.8

A = += +=

The model predicts that approximately $1119.8 million, or $1.1198 billion, was spent on Cyber Monday 2111.

71. Thinking and Writing Exercise. If the

y-intercept is given, then it is easy to use the slope-intercept form of a line. Whether the point-slope form of a line is used, or the slope-intercept form is used, the two given points must be used to calculate the slope of the line. But once that is done, one can immediately use the slope and the y-intercept to determine the equation of the line.

72. Thinking and Writing Exercise. The equation

for the life expectancy for males has a greater positive slope than the equation for females. Therefore, the lines are not parallel and will intersect at some point. Beyond the time corresponding to that intersection point, the life expectancy for males will be higher than for females.

73. ( ) ( )( )

2 2

2

2

2 3 5 2 3 5

2 1 3 5

2 2 5

x x x x x x

x x

x x

− + − = − + −

= + − + −= + −

74. ( ) ( )

( ) ( )4 3 6 7 4 3 6 7

4 6 3 7

2 4

t t t t

t

t

+ − + = + − −= − + −= − −

75. ( ) ( )

( ) ( )2 1 3 2 1 3

2 1 1 3

2

t t t t

t

t

− − − = − − += − + − += +

76. ( ) ( )( )

2 2 2 2

2

2

5 4 9 7 5 4 9 7

5 9 7 4

4 7 4

x x x x x x

x x

x x

− − − = − − +

= − + −= − + −



77. ( )3

xf x

x=

−

3

x

x − is undefined when its denominator is 0.

Find the value(s) of x that make 3 0.x − = 3 0

3

x

x

− ==

3 is not in the domain of f(x). The domain of

f(x) is { } is a real number and 3x x x ≠ . 78. ( ) 2 1g x x= −

2 1x − is well-defined for any real number x. The domain is the set of all real numbers.

79. ( ) 6 11g x x= +

6 11x + is well-defined for any real number x. The domain is the set of all real numbers.

80. ( ) 7

2

xf x

x

−=

7

2

x

x

− is undefined when its denominator is 0.

Therefore, find the value(s) of x that make 2 0.x = 2 0

00

2

x

x

=

= =

0 is not in the domain of f(x). The domain of

f(x) is { } is a real number and 0 .x x x ≠ 81. Thinking and Writing Exercise. The slope-

intercept form is more useful when using a graphing calculator because to graph using the calculator requires the equation be solved for y. However, the point-slope form of a line is easily solved for y by adding 1y to both

sides of the equation:

( )( )

1 1

1 1.

y y m x x

y m x x y

− = −

= − +

82. Thinking and Writing Exercise. Any

nonvertical line contains an infinite number of points, any one of which can be used in the point-slope form of the line. But any nonvertical line only passes through the y-axis in one place. Therefore, there is only one value of b that can be used in the y mx b= + (slope-intercept) form of the line.

83. First simplify the equation. ( )3 0 52

3 0

3

y x

y

y

− = −− =

=

y c= is the form of a horizontal line that

passes through the y-axis at ( )0, .c In this

case, 3,c = so the graph is:

84. First simplify the equation.

( )4 0 93

4 0

4

y x

y

y

+ = ++ =

= −

y c= is the form of a horizontal line that

passes through the y-axis at ( )0, .c In this

case, 4,c = − so the graph is:

85. Two points with coordinates ( )4, 1− and

( )3, 3− are shown on the graph. First

determine the slope of the line. ( )1 3 1 3 2

24 3 4 3 1

m− − − − +

= = = =− −


( ) ( )1 1, 4, 1x y = − (or ( ) ( )1 1, 3, 3x y = − ) to

determine the required equation. ( ) ( )1 2 4

1 2 8

2 9

y x

y x

y x

− − = −+ = −

= −



86. Two points with coordinates ( )2,5 and ( )5,1

are shown on the graph. First determine the slope of the line.

1 5 4 4

5 2 3 3m

− −= = = −

−

Use the point-slope equation with4

3m = −

and ( ) ( )1 1, 2,5x y = (or ( ) ( )1 1, 5,1x y = ) to

determine the required equation. ( ) ( )4

3

843 3

843 3

8 1543 3 3

2343 3

5 2

5

5

y x

y x

y x

y x

y x

− = − −

− = − += − + += − + += − +

87. First solve the equation for y and determine

the slope of the given line. 2 6 Given line

2 6

1 13

2 2

x y

y x

y x m

+ == − +

= − + = −


2− . Every line

parallel to the given line must also have a

slope of 1

2− . Find the equation of the line

with a slope of 1

,2

− containing the point

( )3,7 .

( )

( )1 1 Point-slope equation

17 3

21 3

72 21 17

2 2

y y m x x

y x

y x

y x

− = −

− = − −

− = − +

= − +



3 7 3

x y

y x m

− == − =

The slope of the given line is 3. Every line parallel to the given line must also have a slope of 3. Find the equation of the line with a slope of 3, containing the point ( )1,4 .−

( )( )( )

( )

1 1 Point-slope equation

4 3 1

4 3 1

4 3 3

3 7

y y m x x

y x

y x

y x

y x

− = −

− = − −

− = +− = +

= +



2 3 2

x y

y x m

+ = −= − − = −

The slope of the given line is 2.− Every line perpendicular to the given line must have a

slope of 1 1 .2 2

− =−

Find the equation of the

line with a slope of 1 ,2

containing the point

( )2,5 .

( )15 2 Substituting

21

5 121

42

y x

y x

y x

− = −

− = −

= +



3

1 1

3 3

x y

x y

x y m

− ==

= =


3. Every line

perpendicular to the given line must have a

slope of 3 3.1

− = − Find the equation of the

line with a slope of 3,− containing the

point ( )4,0 .

( )0 3 4 Substituting

3 12

y x

y x

− = − −= − +



91. a) Use the two points ( )3, 5− and ( )7, 1− to

determine the slope of the linear function. Then use the point-slope form of a line to find ( ).g x :

( )1 5 1 5 4

17 3 4 4

m− − − − += = = =

−

( ) ( )

( )

5 1 3

5 3

88 Function notation

y x

y x

y xg x x

− − = −+ = −

= −= −

b) ( )2 2 8 10g − = − − = −

c) ( ) 8g a a= −

If ( ) 75g a = , we have

8 75

83.

a

a

− ==

Mid-Chapter Review Guided Solutions 1. To find the y-intercept of the line 3 6,y x− =

set x to 0 in the equation and solve for y. : 3 0 6

6

y - intercept y

y

− ⋅ ==

The y-intercept ( )0,6 .

To find the x-intercept of the line 3 6,y x− =

set y to 0 in the equation and solve for x. : 0 3 6

3 6

2

x - intercept x

x

x

− =− =

= −

The x-intercept ( )2,0 .−

2. For the line containing ( )1,5 and ( )3, 1− , the

slope is:

2 1

2 1

1 5

3 1

6

23

y ym

x x

− − −= =− −

−=

= −

Mixed Review 1. 2 5 8x y+ = is in standard form.

2. 2 11

3 3y x= − is in slope-intercept form.

3. 13 5x y− = is in none of these forms.

4. ( )12 6

3y x− = − is in point-slope form.

5. 1x y− = is in standard form.

6. 18 3.6y x= − + is in slope-intercept form.

7. ( )5, 2− − and ( )1,8

( )( )

8 2 8 2 10 5

1 5 1 5 6 3m

− − += = = =

− − +

8. ( )0,0 and ( )0, 2−

( )0 2 2

0 0 0m

− −= =

−

Since division by zero is undefined, the slope of this line is undefined. 9. The line 4y = is a horizontal line. The slope

is 0. 10. The line 7x = − is a vertical line. The slope is undefined. 11. 3 1

3 1

1 1

3 3

x y

y x

y x

− =− = − +

= −

Slope is 1

3; y-intercept is

10,

3⎛ ⎞−⎜ ⎟⎝ ⎠

12. Substitute the values of the given slope

m 3= − and the given y-intercept b 7= into the slope-intercept form of a linear function ( ) .f x mx b= +

( ) 3 7.f x x= − +

13. Substitute the values of the given slope m 5=

and the given point ( ) ( )1 1, 3,7x y = − into the

point-slope form of a line ( )1 1 .y y m x x− = −

( )( ) ( )7 5 3 , or 7 5 3 .y x y x− = − − − = +



14. First use the given points to determine the slope of the line:

( )5 1 5 1 4 2

2 4 6 6 3m

− − − − + −= = = =− − − −

Then substitute the values of the slope and the given point ( ) ( )1 1, 4, 1x y = − into the point-

slope form of a line ( )1 1 .y y m x x− = −

( ) ( )21 4

32 8

13 32 11

3 3

y x

y x

y x

− − = −

+ = −

= −

15. Graph y = 2x – 1.

16. Graph 3x + y = 6.

17. Graph ( )12 1 .

2y x− = −

18. Graph ( ) 4.f x =

19. Graph ( ) 35.

4f x x= − +

20. Graph 3 12.x = First simplify the equation.

3 12

1 13 12

3 34

x

x

x

=

=

=

⋅ ⋅

Exercise Set 2.5 1. difference 2. subtract 3. evaluate 4. common to 5. excluding 6. sum 7.

2

(2) 3 2 1 5

(2) 2 2 6

(2) (2) 5 6 1

f

g

f g

= − ⋅ + = −

= + =+ = − + =

8.

2

( 1) 3( 1) 1 3 1 4

( 1) ( 1) 2 1 2 3

( 1) ( 1) 4 3 7

f

g

f g

− = − − + = + =

− = − + = + =− + − = + =

9.

2

(5) 3 5 1 15 1 14

(5) 5 2 25 2 27

(5) (5) 14 27 41

f

g

f g

= − ⋅ + = − + = −

= + = + =− = − − = −



10. 2

(4) 3 4 1 12 1 11

(4) 4 2 16 2 18

(4) (4) 11 18 29

f

g

f g

= − ⋅ + = − + = −

= + = + =− = − − = −

11.

2

( 1) 3( 1) 1 3 1 4

( 1) ( 1) 2 1 2 3

( )( 1) 4 3 12

f

g

f g

− = − − + = + =

− = − + = + =⋅ − = ⋅ =

12.

2

( 2) 3( 2) 1 6 1 7

( 2) ( 2) 2 4 2 6

( 2) ( 2) 7 6 42

f

g

f g

− = − − + = + =

− = − + = + =− ⋅ − = ⋅ =

13.

2

( 4) 3( 4) 1 12 1 13

( 4) ( 4) 2 16 2 18

13( 4) / ( 4)

18

f

g

f g

− = − − + = + =

− = − + = + =

− − =

14.

2

(3) 3(3) 1 9 1 8

(3) 3 2 9 2 11

8(3) / (3)

11

f

g

f g

= − + = − + = −

= + = + =

= −

15. 2(1) 1 2 1 2 3

(1) 3 1 1 3 1 2

(1) (1) 3 ( 2) 5

g

f

g f

= + = + == − ⋅ + = − + = −

− = − − =

16. (2) / (2)g f

2(2) 2 2 4 2 6

(2) 3 2 1 6 1 5

6(2) / (2)

5

g

f

g f

= + = + == − ⋅ + = − + = −

= −

17.

( )2

2

( )( ) ( ) ( )

( 3 1) 2

3 3

f g x f x g x

x x

x x

+ = +

= − + + +

= − +

18.

( )2

2

2

( )( ) ( ) ( )

2 ( 3 1)

2 3 1

3 1

g f x g x f x

x x

x x

x x

− = −

= + − − +

= + + −

= + +

19. ( )( ) ( ) ( )( ) ( )2

2

2

3 1 2

3 1 2

3 1

f g x f x g x

x x

x x

x x

− = −

= − + − +

= − + − −

= − − −

20. ( )2 2

( )( / )

( )

2 2, or

3 1 3 1

g xg f x

f x

x x

x x

=

+ += −− + −

21.

( )2

2

( )( ) ( ) ( )

2 (5 )

3

F G x F x G x

x x

x x

+ = +

= − + −

= − +

22.

( )2

2

( )( ) ( ) ( )

2 (5 )

3

F G a F a G a

a a

a a

+ = +

= − + −

= − +

23. ( )( ) ( )

( )( ) ( ) ( )

2( 4) 4 2 16 2 14

4 5 4 5 4 9

4 4 4 14 9 23

F

G

F G F G

− = − − = − =

− = − − = + =

+ − = − + − = + =

24. ( )( ) ( )

( )( ) ( ) ( )

2( 5) 5 2 25 2 23

5 5 5 5 5 10

5 5 5 23 10 33

F

G

F G F G

− = − − = − =

− = − − = + =

+ − = − + − = + =

25. 2(3) 3 2 9 2 7

(3) 5 3 2

( )(3) (3) (3) 7 2 5

F

G

F G F G

= − = − == − =

− = − = − =

26.

( )

2(2) 2 2 4 2 2

(2) 5 2 3

( )(2) 2 (2) 2 3 1

F

G

F G F G

= − = − == − =

− = − = − = −

27. 2( 3) ( 3) 2 9 2 7

( 3) 5 ( 3) 5 3 8

( )( 3) ( 3) ( 3) 7 8 56

F

G

F G F G

− = − − = − =− = − − = + =

⋅ − = − ⋅ − = ⋅ =

28. 2( 4) ( 4) 2 16 2 14

( 4) 5 ( 4) 5 4 9

( )( 4) ( 4) ( 4) 14 9 126

F

G

F G F G

− = − − = − =− = − − = + =

⋅ − = − ⋅ − = =⋅



29. ( )( ) ( ) ( )( ) ( )2

2

2

2 5

2 5

7

F G a F a G a

a a

a a

a a

− = −

= − − −

= − − +

= + −

30. ( )( )

2

( / )( )

2

5

F xF G x

G x

x

x

=

−=−

31. ( )( ) ( ) ( )

( ) ( )2

2

2

2 5

2 5

7

F G x F x G x

x x

x x

x x

− = −

= − − −

= − − +

= + −

32.

( )2

2

2

( )( ) ( ) ( )

(5 ) 2

5 2

7

G F x G x F x

x x

x x

x x

− = −

= − − −

= − − +

= − − +

33.

( )( )

2( 2) ( 2) 2 4 2 2

( 2) 5 ( 2) 5 2 7

2 2( / )( 2)

2 7

F

G

FF G

G

− = − − = − =− = − − = + =

−− = =

−

34.

( )( )

2( 1) ( 1) 2 1 2 1

( 1) 5 ( 1) 5 1 6

1 1 1( / )( 1)

1 6 6

F

G

FF G

G

− = − − = − = −− = − − = + =

− −− = = = −−

35. Answers may vary slightly. Locate 2 on the

horizontal axis then move vertically to the graph of P. (2) 26.5.P ≈ Locate 2 on the

horizontal axis then move vertically to the graph of L. (2) 22.5.L ≈

( )(2) (2) (2) 26.5 22.5 4%.P L P L− = − ≈ − =

36. Answers may vary slightly. Locate 1 on the

horizontal axis then move vertically to the graph of P. (1) 20.P ≈ Locate 1 on the

horizontal axis then move vertically to the graph of L. (1) 18.L ≈

( )(1) (1) (1) 20 18 2%.P L P L− = − ≈ − =

37. Answers may vary slightly. Using the graph,

(2004) 1.2 and (2004) 2.9.C B≈ ≈

(2004) (2004) (2004)

1.2 2.9

4.1 million

N C B= +≈ +=

We estimate that 4.1 million U.S. women had children in 2004. 38. Answers may vary slightly. Using the graph,

(1985) 0.8 and (1985) 2.9.C B≈ ≈

(1985) (1985) (1985)

0.8 2.9

3.7 million

N C B= +≈ +=

We estimate that 3.7 million U.S. women had children in 1985. 39. Answers may vary slightly. Since the p and r bands are stacked on top of

one another, the value of ( )('05)p r+ can be

found by subtracting the value at the bottom of the r band from the value at the top of the p band for ’05. Thus:

( )('05) 256 162 94p r+ ≈ − = million tons.

This value represents the amount of trash that was either composted or recycled in the U.S. in 2005.

40. Answers may vary slightly. Since the p, r, and b bands are stacked on top

of one another, the value of ( )('05)p r b+ +

can be found by subtracting the value at the bottom of the b band from the value at the top of the p band for ’05. Thus:

( )('05) 256 135 121p r b+ + ≈ − = million

tons. This value represents the amount of trash that was either composted, recycled or burned for energy in the U.S. in 2005.

41. Answers may vary slightly. ('96) 215F ≈ million tons. This value

represents the total amount of trash generated in the U.S. in 1996.

42. Answers may vary slightly. ('06) 260F ≈ million tons. This value

represents the total amount of trash generated in the U.S. in 2006.



43. Answers may vary slightly. ( )('04)F p− is found graphically by

determining the value at the bottom of the p band for 2004. Thus, ( )('04) 231F p− ≈

million tons. This value represents the amount of trash generated in the U.S. in 2004 that was not composted.

44. Answers may vary slightly. ( )('03)F l− is found graphically by

subtracting the value the top of the l band, from F, for 2003. Thus, ( )('03) 252 131 121F l− ≈ − = million tons.

This value represents the amount of trash generated in the U.S. in 2003 that was not put in a landfill.

45. 2( ) ; ( ) 7 4f x x g x x= = −

The domain of f is R. The domain of g is R.

The domains of , , and f g f g f g+ − ⋅ are all

equal to the set of elements common to the domains of f and g. The domain of

,f g+ ,f g− and f g⋅ is the set of

elements common to the domains of f and g:

R { | is a real number}.x x=

46. 2( ) 5 1; ( ) 2f x x g x x= − =


The domains of , , and f g f g f g+ − ⋅ are

equal to the set of elements common to the domains of f and g. The domain of

,f g+ ,f g− and f g⋅ is the set of


R { | is a real number}.x x=

47. 31( ) ; ( ) 4

3f x g x x

x= =

−

The domain of ( ) is { |f x x x is a real number,

and 3}.x ≠ The domain ( )g x is R. The

domain of , ,f g f g+ − and f g⋅ is the set of

elements common to the domains of f and g: { |x x is a real number, and 3}.x ≠

48. 2 1( ) 3 ; ( )

9f x x g x

x= =

−

The domain of ( ) isf x R. The domain of

( ) is { |g x x x is a real number, and 9}.x ≠

The domain of , ,f g f g+ − and f g⋅ is the

set of elements common to the domains of f and g: { |x x is a real number, and 9}.x ≠

49. 22( ) ; ( ) 4f x g x x

x= = −

The domain of is { |f x x is a real number,

and 0}.x ≠ The domain of g is R. The



50. 3 5( ) 1; ( )f x x g x

x= + =





51. 32( ) ; ( ) 3

1f x x g x x

x= + =

−

The domain of ( ) is { |f x x x is a real number,

and 1}.x ≠ The domain ( )g x is R. The



52. 2 3( ) 9 ; ( ) 2

6f x x g x x

x= − = +

−







53. 5

( ) ; ( )2 9 1

xf x g x

x x= =

− −

The domain of is { |f x x= in R, 9

}.2

x ≠ The

domain of is { |g x x= in R, 1}.x ≠ The



{ |x x in R, 9

, or 1}.2

x ≠

54. 5

( ) ; ( )3 4 1

xf x g x

x x= =

− −

The domain of is { |f x x= in R, 3}x ≠

The domain of is { |g x x= R, 1

}.4

x ≠


set of elements common to both the domains

of f and g:{ |x x in R, 1

or 3}.4

x ≠

55. 4( ) ; ( ) 3f x x g x x= = −


( ) 0g x = has only one solution, 3.x =

The domain of4

/3

xf g

x=

−is:

{ |x x is a real number, 3}.x ≠

56. 3( ) 2 ; ( ) 5f x x g x x= = −



The domain of4

/3

xf g

x=

−is:

{ |x x is a real number, 5}x ≠

57. ( ) 3 2; ( ) 2 8f x x g x x= − = −


2 8 0

2 8

4 So, 4.

x

x

x x

− === ≠


The domain of3 2

/2 8

xf g

x

−=−

is:


58. ( ) 5 ; ( ) 6 2f x x g x x= + = −


6 2 0

6 2

3 So, 3

x

x

x x

− === ≠


The domain of5

/6 2

xf g

x

+=−

is:


59. 3

( ) ; ( ) 54

f x g x xx

= = −−

The domain of is { |f x x= is a real number

and 4}.x ≠ The domain of g is R.


The domain of

( ) ( )( )3 3

/ / 5 =4 4 5

f g xx x x

⎛ ⎞= −⎜ ⎟⎝ ⎠− − − is:

{ |x x is a real number, 4 and 5}.x x≠ ≠

60. 1

( ) ; ( ) 72

f x g x xx

= = −−




The domain of

( ) ( )( )1 1

/ / 7 =2 2 7

f g xx x x

⎛ ⎞= −⎜ ⎟⎝ ⎠− − − is:

{ |x x is a real number, 2x ≠ and 7}.x ≠

61. 2

( ) ; ( ) 2 51

xf x g x x

x= = +

+


and 1}.x ≠ − The domain of g is R.

2 5 0

2 5

52

x

x

x

+ == −

= −

( ) 0g x = has only one solution, 5 .2

x = −



The domain of ( )2/ / 2 5

1

xf g x

x⎛ ⎞= +⎜ ⎟⎝ ⎠+

( )( )2

=1 2 5

x

x x+ + is:

{ |x x is a real number, 1x ≠ − and5

}.2

x ≠ −

62. 7

( )2

( ) 3 7

xf x

xg x x

=−

= +



3 7 0

3 7

7

3

x

x

x

+ == −

= −

( ) 0g x = has only one solution, 7 .3

x = −

The domain of ( )7/ / 3 7

2

xf g x

x⎛ ⎞= +⎜ ⎟⎝ ⎠−

( )( )7

=2 3 7

x

x x− +is:

{ |x x is a real number,7

2 and }.3

x x≠ ≠ −

63. From the graph:

( )( )

( )( )

(5) 1 and (5) 3.

5 (5) (5) 1 3 4

(7) 1 and (7) 4.

7 (7) (7) 1 4 3

F G

F G F G

F G

F G F G

= =+ = + = + == − =

+ = + = − + =

64. From the graph:

( )

( )

1(6) 0 and (6) 3

21

(6) (6) (6) 0 3 02

(9) 1 and (9) 2

(9) (9) (9) 1 2 2.

F G

F G F G

F G

F G F G

= =

⋅ = ⋅ = ⋅ =

= =⋅ = ⋅ = ⋅ =

65. From the graph:

( )

( )

(7) 4 and (7) 1

(7) (7) (7)

4 ( 1)

4 1 5

(3) 1 and (3) 2

(3) (3) (3)

1 2 1

G F

G F G F

G F

G F G F

= = −− = −

= − −= + =

= =− = −

= − = −

66. From the graph:

( )

( )

(3) 2 and (3) 1

2/ (3) (3) / (3) 2

1(7) 1 and (7) 4

1/ (7) (7) / (7)

4

F G

F G F G

F G

F G F G

= =

= = =

= − =

= = −

67. The x-values of f are from 0 to 9, so the

domain of { | 0 9}.F x x= ≤ ≤

The x-values of g are from 3 to 10, so the domain of { | 3 10}.G x x= ≤ ≤

The domain of F G+ is the set of numbers common to the domains of F and G, so the domain of { | 3 9}.F G x x+ = ≤ ≤

The domain of /F G is the set of numbers common to the domains of F and G, excluding any number(s) where G = 0. There are no exclusions. The domain of

/ { | 3 9}.F G x x= ≤ ≤

68. The x-values of f are from 0 to 9, so the

domain of { | 0 9}.F x x= ≤ ≤

The x-values of g are from 3 to 10, so the domain of { | 3 10}.G x x= ≤ ≤ The domain

of F G− and F G⋅ is the set of numbers common to the domains of F and G. The domain of F G− and the domain of

{ | 3 9}.F G x x⋅ = ≤ ≤ The domain of /G F

is the set of numbers common to the domains of F and G, excluding any number(s) where F = 0. From the graph 0F = at 6x = and

8.x = The domain of /G F is: { | 3 9; 6 and 8}.x x x x≤ ≤ ≠ ≠



69. Determine the values of F G+ at some values in its domain and graph these points. Then join these points with a smooth curve.

F G+ 3 2 1 3+ = 5 1 3 4+ = 7 1 4 3− + = 9 1 2 3+ =

70. Determine the values of G F− at some values

in its domain and graph these points. Then join these points with a smooth curve.

x G F− 3 1 2 1− = − 5 3 1 2− = 7 4 1 5+ = 9 2 1 1− =

71. Thinking and Writing Exercise. To determine the values where S M W J> + + one must estimate where the sum on the right is equal to the value of S. This appears to be the case around 1985.

Check:

(1985) 41,

(1985) (1985) (1985)

27 8 5 40

S

M W J

≈+ +≈ + + =

The values are close. Similarly, it appears to be the case around 2003.

Check: (2003) 52

(2003) (2003) (2003)

21 21 9 51

S

M W J

≈+ +≈ + + =

Again, the values are close. Since it is difficult to do better than a rough estimate

with a graph of this type. Any answer near (1985, 2003) would be acceptable.

72. Thinking and Writing Exercise. The total

number of births, found by adding the values of C and B, appear to have increased from about 3.7 million to 4.1 million from 1970 to 2004. However, there were some time periods where the number of births remained approximately constant, and some periods where the number of births even decreased (such as from 1970 to 1975 or 1990 to 1995). The percent of Caesarean section births is determined by dividing the number of Caesarean section births by the total number of births. This value definitely increased from 1970 to 2004 as a result of the overall increase in C and decrease in B.

73. 6 3

6 3

1 1

6 2

x y

y x

y x

− =− = − +

= −

74. 3 8 5

8 3 5

3 5

8 8

x y

y x

y x

− =− = − +

= −

75. 5 2 3

2 5 3

5 3

2 2

x y

y x

y x

+ = −= − −

= − −

76. 8 4

8 4

1 4

8 81 1

8 2

x y

y x

y x

y x

+ == − +

= − +

= − +

77. Let n represent the number. 2 5 49.n + =

78. Let n represent the number. 1

3 57.2

n − =

79. Let n represent the smaller integer. ( )1 145.n n+ + =



80. Let n represent the number. ( ) 20n n− − =

81. Thinking and Writing Exercise. We would

change the vertical axis to correspond to absorption of Advil and graph the points accordingly. If taken 4 times a day, we would expect to see 4 peaks over the course of 1 day.

82. Thinking and Writing Exercise. The graph of

( )( )f g x+ will have the same shape as the

graph of g, but will be shifted upward (because c is positive) by c units.

83. 43 1

( ) ; ( )2 5 3 9

x xf x g x

x x

−= =+ +

: 2 5 0

2 5

5 5, So

2 2

f x

x

x x

+ == −− −= ≠

Domain of { |f x x= is a real number, }52x −≠

: 3 9 0g x + =

3 9

3; So 3

x

x x

= −= − ≠ −

Domain of { |g x x= is a real number,

3}.x ≠ −

( )

( )( )( )

4

2

0

1 0

1 1 1 0

1

g x

x

x x x

x

=

− =

+ + − =

= ±

( )

( )( )4

4

3 3 93 1/

2 5 3 9 2 5 1

x xx xf g

x x x x

+−= =+ + + −

Therefore, the domain of / { |f g x x= is a

real number, 52 , 3, and 1}.x x x−≠ ≠ − ≠ ±

84. 21 4

( ) and G( )4 3

xF x x

x x

−= =− −

Domain of { |F x x= is a real number and

4}x ≠

Domain of { |G x x= is a real number and

3}x ≠

( )

( )( )2

0

4 0

2 2 0

2

G x

x

x x

x

=

− =+ − =

= ±

( )( )

2

2

31 4/

4 3 4 4

xxF G

x x x x

−−= =− − − −

Therefore, the domain of / { |F G x x= is a

real number, 4, 3, and 2}.x x x≠ ≠ ≠ ±

85. Answers may vary. The two functions must

be defined over the intervals [ ]2,3− , but must

not both be defined anywhere else. To remove the value x = 1 from the domain of f / g, make ( )1 0.g =

86. The domain of , , and f g f g f g+ − ⋅ is the

set of numbers common to the domains of f and g:{ 2, 1, 0,1}.− − The domain of /f g is

the same set, excluding any x values ( ) 0.g x = Since ( )1 0,g − = it is excluded

from the set. The domain of /f g is:

{ 2, 0,1}.−

87. The domain of m is { | 1 5}.x x− < <

( ) 0 when 2 3 0

2 3

3

2

n x x

x

x

= − ==

=

We exclude this value from the domain of /m n . The domain of /m n is:

3| 1 5 and .

2x x x

⎧ ⎫− < < ≠⎨ ⎬⎩ ⎭


Chapter 2 Study Summary 141

88. ( )( 2); ( )(0); ( / )(1)f g f g f g+ − ⋅

( )

( 2) 1 and ( 2) 4

( )( 2) 1 4 5

(0) 3 and (0) 5

( ) 0 3 5 15

(1) 4 and (1) 6

4 2( / ) 1

6 3

f g

f g

f g

f g

f g

f g

− = − =+ − = + =

= =⋅ = ⋅ =

= =

= = =

89. Answers may vary. Since the domain of f + g

is the intersection of the individual domains of f and g, at least one of the functions needs to be undefined at 2,x = − and at least one needs to be undefined at 5.x = Both functions should be well-defined everywhere else.

1 1

( ) ; ( )2 5

f x g xx x

= =+ −

90. For 1 5y = , 2 2y x= + , and 3 ,y x= the

graph of 1 2y y+ will be the same as the

graph of 2y , but shifted up 5 units.

The graph of 1 3y y+ will be the same as the

graph of 3y but shifted up 5 units. The graph

of 2 3y y+ is similar to the graph of 3y , but it

is shifted up 2 units and rises more steeply as x increases.

91. 1 2.5 1.5y x= + and 2 3y x= −

1

2

2.5 1.5

3

y x

y x

+=−

Domain: { }| is any real number 3x and x ≠

The CONNECTED mode graph contains values that cross the line 3x = , whereas, the DOT mode graph contains no points having 3 as the first coordinate. Thus, the DOT mode graph represents 3y more accurately.

92. Think of adding, subtracting, multiplying, or dividing the y-values for various x-values. a) IV b) I c) II d) III Chapter 2 Study Summary 1. ( ) 2 3

( 1) 2 3( 1) 2 3 5

f x x

f

= −− = − − = + =

2. Use the vertical line test by visualizing a

vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function.

3. The domain is the set of all real numbers, or

.The range is the set of all y-values in the graph. It is { }| 2y y ≥ − or [ )2,− ∞

4. ( ) 15

4f x x= − is well-defined for all real

numbers. The domain is, therefore, the set of all real numbers.

5. ( )1,4 and ( )9,3−

3 4 1 1

9 1 10 10m

− −= = =− − −

6. 2

54y x

y mx b

= − += +

The slope m is –4, and the y-intercept (0, b) is

( )250, .

7. 1

2 2y x= +

The slope m is 1

2 and the y-intercept (0, b) is

( )0, 2 . First plot the y-intercept ( )0, 2 .

Then, starting at the point (0, 2), use the

slope, 1

2, to find another point by moving up

1 unit and right 2 units to ( )2,3 . Then draw

the line through the two points.



8. The line 2y = − is a horizontal line with y-

intercept ( )0, 2− .

9. The line 3x = is a vertical line with an x-

intercept of ( )3,0 .

10. 10 10x y− =

Find the x-intercept, by letting 0y = and

solving for x. 10 0 10

10 10

1

x

x

x

− ===


Find the y-intercept by letting 0x = and solving for y.

10 0 10

10

10

y

y

y

⋅ − =− =

= −


11. Two lines are parallel if their slopes are equal.

The equation 4 12y x= − is in slope-intercept

form, and its slope is 4. Rewrite the second equation in slope-intercept form:

4 9

1 9

4 4

y x

y x

= −

= −

Its slope is 1

4. Since the lines have different

slopes, their graphs are not parallel. 12. Two lines are perpendicular if the product of

their slopes is 1.− The equation 7y x= − is

in slope-intercept form. It has a slope of 1. Rewrite the second equation in slope-intercept form.

3

3

x y

y x

+ == − +

Its slope is –1. Since ( )1 1 1− = − , their graphs

are perpendicular lines. 13. ( )1

4 ; 1,6m = −

The point-slope equation is ( )1 1y y m x x− = − .

Substitute 14 for m, –1 for 1x , and 6 for 1y :

( )( )146 1y x− = − − .

14. ( )( ) ( ) ( )

( 2) ( 7)

2 7

2 9

f g x f x g x

x x

x x

x

+ = += − + −= − + −= −

15. ( )( ) ( ) ( )

( 2) ( 7)

2 7

5

f g x f x g x

x x

x x

− = −= − − −= − − +=

16.

( )

( )(5) (5) (5)

(5 2)(5 7)

3 2 6

f g f g⋅ = ⋅= − −= − = −

17. ( ) 2/ ( ) ( ) / ( ) .7

xf g x f x g xx−= =−

Chapter 2 Review Exercises 1. True 2. False 3. False


Chapter 2 Review Exercises 143

4. False 5. False 6. True 7. True 8. True 9. False 10. True 11. a) To use the vertical line test visualize a

vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus, the graph is the graph of a function.

b) The domain is the set of all x values: { }| 4 5 .x x− ≤ ≤

The range is the set of all y-values:

{ }| 2 4y y− ≤ ≤ .

12. a) To use the vertical line test visualize a

vertical line moving across the graph. No vertical line will intersect the graph more than once. Thus the graph is the graph of a function.

b) The domain is the set of all x values: ll real numbers or

{ }| is a real number .x x

The range is the set of all y values:

{ }| 1y y ≥ .

13. a) To use the vertical line test visualize a

vertical line moving across the graph. It is possible for a vertical line to intersect the graph more than once. Thus this is not the graph of a function.

14. a) On the graph provided, find the point

with an x value of 3. The point is (3,0) .

Therefore ( )3 0f =

b) Find the point or points with a y value of 2− . The y value is 2− at the point

( 4, 2)− − : ( 4) 2.f − = −

15. a) 4 4

(1)2(1) 1 3

g = =+

b) Since 4

2 1x + is undefined when the

denominator is 0, find the x-value(s) where 2 1 0.x + =

2 1 0

2 1 Adding 1 to both sides.

1 Dividing both sides by 2

2

x

x

x

+ == − −

= −

Thus, 1

2− is not in the domain of g. The

domain of g is: 1


x x and x⎧ ⎫≠ −⎨ ⎬⎩ ⎭

.

16. ( )2 , if 0,

3 5, if 0 2

7, if 2

x x

f x x x

x x

⎧ <⎪= − ≤ ≤⎨⎪ + >⎩

a) Since 0 0 2≤ ≤ , ( )0 3 0 5 5f = − = −⋅ .

b) Since 3 2> , ( )3 3 7 10.f = + =

17. The equation ( ) 4 9g x x= − − is in slope-

intercept form. Slope is –4; y-intercept is (0, –9)

18. Convert the equation to slope-intercept form.

6 2 14

6 2 14

2 14 1 7

6 6 3 3

y x

y x

y x x

− + =− = − +

= − = −

Slope is 1

3; y-intercept is

70,

3⎛ ⎞−⎜ ⎟⎝ ⎠ .



19. To find the rate of change, or slope, of the graph, select any two points, say ( )2,75 and

( )8,120 and use the slope formula.

2 1

2 1

120 75 457.5

8 2 6

y ym

x x

− −= = = =− −

The rate of change in the apartment's value is $7500 per year.

20. Let ( ) ( )1 1, 4,5x y = and ( ) ( )2 2, 3,1x y = − .

2 1

2 1

1 5 4 4

3 4 7 7

y ym

x x

− − −= = = =− − − −

21. Let ( ) ( )1 1, 16.4,2.8x y = − and

( ) ( )2 2, 16.4,3.5x y = − .

( )2 1

2 1

3.5 2.8 0.7

16.4 16.4 0

y ym

x x

− −= = =− − − −

The slope of the line containing the points is undefined since division by zero is undefined. 22. ( ) 11 1542C t t= +

The slope 11 signifies the rate of change in the average number of calories consumed each day. The average number of calories consumed each day has increased by 11 calories per year since 1971. The intercept 1542 signifies the average number of calories consumed each day in 1971.

23. Simplify the equation.

3 7

4

y

y

+ ==

The graph of y = 4 is a horizontal line. Its slope is 0.


2 9

9

2

x

x

− =

= −

The graph of 9

2x = − is a vertical line. Its

slope is undefined. 25. 3 2 8x y− =

To find the y-intercept, let x = 0 and solve for y.

3 0 2 8

2 8

4

y

y

y

⋅ − =− =

= −

The y-intercept is (0, –4). To find the x-intercept, let y = 0 and solve for x. 3 2 0 8

3 8

8

3

x

x

x

− ⋅ ==

=

The x-intercept is ( )8 , 03

.

26. The equation 3 2y x= − + is in slope-intercept

form. The slope is –3 = 3.1− The y-intercept

is (0, 2). From the y-intercept, go down 3 units and to the right 1 unit to (1, –1). Draw the line through the two points.

27. Graph 2 4 8x y− + = using the intercepts.

To find the y-intercept, let x = 0 and solve for y.

2 0 4 8

2

y

y

− ⋅ + ==

The y-intercept is (0, 2). To find the x-intercept, let y = 0 and solve for x.

2 4 0 8

2 8

4

x

x

x

− + ⋅ =− =

= −

The x-intercept is (–4, 0). Plot these points and draw the line.



28. The graph of y = 6 is a horizontal line that crosses the y-axis at ( )0, 6 .

29. ( )

( ) ( )

31 5

43

1 54

y x

y x

+ = −

− − = −

The equation is in point-slope form. Starting at the point ( )5, 1− , use the slope to determine

a second point. ( ) ( ), (5 4, 1 3) 1, 4x y = − − − = −

Use the two points to draw the line.


8 32 0

4.

x

x

+ == −

This is a vertical line that crosses the x-axis at (–4, 0).

31. ( ) 15 15.g x x x= − = − + .

The slope is 1

11

−− = . The y-intercept is

(0, 15). From the y-intercept, go down 1 unit and right 1 unit to the point (1, 14). Use the

two points to draw the graph.

32. The function ( ) 13

2f x x= − is in slope-

intercept form. The slope is 1

2. The

y-intercept is (0, –3). From the y-intercept, go up 1 unit and right 2 units to the point (2, –2). Use the two points to draw the graph.

33. The graph of ( ) 0f x = is a horizontal line that

crosses the vertical axis at (0, 0).

34. The window should show both intercepts:

(98, 0) and (0, 14). One possibility is

[ ]10,120, 5,15− − , with Xscl = 10.

35. First solve for y and determine the slope of

each line. 5

5

y x

y x

+ = −= − −

The slope of y + 5 = –x is –1. 2

2

x y

y x

− == −

The slope of x – y = 2 is 1. The product of their slopes is (–1)(1) = –1. Therefore, the lines are perpendicular.



36. First solve for y and determine the slope of each line. 3 5 7

3 5

7 7

x y

y x

− =

= −

The slope of 3 5 7x y− = is 3

7.

7 3 7

31

7

y x

y x

− =

= +

The slope of 7 3 7y x− = is 3

7.

The slopes are the same, so the lines are parallel.

37. Use the slope-intercept equation

( ) ,f x mx b= + with 2

9m = and b = –4.

( )

( ) 24

9

f x mx b

f x x

= +

= −

38. ( )

( )1 1

1 1


10 5 1 Substituting 5 for ,

1 for , and 10 for

y y m x x

y x m

x y

− = −

− = − − −

39. First find the slope using the pair of given

points. 5 6 1 1

2 ( 2) 4 4m

− −= = = −− −

Use the point-slope equation.

( )

( )

15 2

41 1

54 21 11

4 21 11

4 2

y x

y x

y x

f x x

− = − −

− = − +

= − +

= − +

40. The slope in the given equation is 1. The

slope of any perpendicular line is, therefore, 1 1.1

− = − Use the slope and the given

y-intercept (0, 3)− to write the equation in

function form. ( ) 3.f x x= − −

41. Plot and connect the points, using the years as the first coordinate and the corresponding cost per gallon as the second coordinate.

To estimate the cost in 2004, locate the point that is directly above 2004. Then move horizontally from the point to the vertical axis and read the approximate function value. The approximate cost per gallon in 2004 was $1.50.

42. To estimate the cost per gallon in 2008,

extend the graph and extrapolate. The approximate cost per gallon in 2008 was $2.40.

43. a) The equation says to let t = the number of

years since 1980, and R(t) = the record, in seconds, for the 200-m run. Use the given data to form the pairs (3, 19.75) and (27, 19.32). Then find the slope of the function that fits the data.

19.32 19.75 0.43 43

27 3 24 2400m

− −= = = −−

Substitute 43

2400m = − , 1 3,t = and

1 19.75R = into the point-slope equation.

( )( )

( )

1 1

4319.75 3

240043 15,843

2400 80043 15,843

2400 800

R R m t t

R t

R t

R t t

− = −

− = − −

= − +

= − +

b) In 2013, t = 2013 – 1980 = 33

( ) ( )43 15,84333 33

2400 80019.21 sec

R = − +

≈

In 2020, t = 2020 – 1980 = 40

( ) ( )43 15,84340 40

2400 80019.09 sec

R = − +

≈



44. 2 7 0

7

2

x

x

− =

=

can be written in standard form for a linear equation, Ax By C+ = . Thus, it is a linear

equation. The graph is a vertical line.

45. 3 78

24 56

yx

x y

− =

− =

This equation can be written in standard form for a linear equation, Ax By C+ = . Thus, it is

a linear equation. 46. 32 7 5x y− =

The equation is not linear, because it contains an 3 term.x

47. 2

2 Multiplying by

yx

xy x

=

=

The equation is not linear, because it has an xy term.

48. Enter the data with the number of years since

2000 as 1L and the revenue, in billions of

dollars, as 2L . The data appear to be linear.

49. Enter the data with the number of years since

2000 as 1L and the revenue, in billions of

dollars, as 2L .


The second screen indicates that the equation is 0.96 12.85F t= + .

50. 2012 is 12 years after year 2000. We

substitute 12 for t and solve for F.

( )0.96 12.85

0.96 12 12.85

11.52 12.85

24.37

F t

F

F

F

= += += +=

The estimated revenue in 2012 will be about $24.4 billion. 51. ( )5 3( 5) 6

3 15 6

3 9

g a a

a

a

+ = + −= + −= +

52. ( )( ) ( ) ( )

( )( )( )( )

2

4 4 4

3 4 6 4 1

12 6 16 1

6 17

102

g h g h⋅ = ⋅

= ⋅ − +

= − += ⋅=

53. ( )( ) ( ) ( )

( )( )2

/ 1 1 / 1

3 1 6

1 1

3 6

1 19 9

2 2

g h g h− = − −

− −=

− +− −=+

−= = −

54. ( )( ) ( ) ( )2

2

2

3 6 1

3 6 1

3 5

g h x x x

x x

x x

+ = − + +

= − + += + −

55. To find the zeros of ( )g x , solve the equation

( ) 0g x = for x.

3 6 0

3 6

2

x

x

x

− ===



56. g(x) is well-defined for all real numbers. Its domain is { }| is any real number .x x

57. The domains of g and h are both

{ }| is any real numberx x = . Thus, the

domain of g h+ = .

58. The domains of g and h are

{ }| is a real numberx x . ( ) 0g x = when

3 6 0x − = , when 2x = . Therefore the domain of /h g =

{ }| is a real number 2x x and x ≠ .

59. Thinking and Writing Exercise. For a

function, every member of the domain corresponds to exactly one member of the range. Thus, for any function, each member of the domain corresponds to at least one member of the range and the function is, therefore, a relation. In a relation, every member of the domain corresponds to at least one, but not necessarily exactly one, member of the range. Therefore, a relation may or may not be a function.

60. Thinking and Writing Exercise The slope of a

line is the change in y between any two points on the line divided by the change in x between those points. For a vertical line, there is no change in x between any two points, or, rather, the change in x is 0. Since division by 0 is undefined, the slope is also undefined. For a horizontal line, the change in y between any two points is 0, so the slope is 0/(change in x) = 0.

61. The y-intercept is found by determining

(0).f

( ) ( )( ) ( )( )( )

( )

02

0

0 3 0.17 0 5 2 0 7

0 3 0 5 7

0 3 0 1 7

0 3 6

0 6 3 9

f

f

f

f

f

+ = + − −

+ = + −

+ = + −

+ = −

= − − = −

⋅ ⋅


62. Parallel lines have the same slope. Begin by writing each equation in slope-intercept form. 3 4 12

3 12 4

33

4

x y

x y

x y

− =− =

− =

6 9

9 6

9

6 6

ax y

ax y

ax y

+ = −+ = −

+ =− −

The two slopes are 34

and .6a− Solve the

equation 3 .4 6

a= −

34 64 18

18 94 2

a

a

a

= −

= −

= − = −

63. Each package costs $7.99 and costs $2.95 to

ship. This makes a total charge of $7.99 + $2.95 = $10.94 per package. There is also a flat fee of $20 for overnight delivery no matter how many packages are purchased. Let x = the number of packages purchased and ( )f x represent the total cost. Then

( ) cost per pkg # of pkgs + flat fee

10.94 20

f x

x

== +

⋅

64. a) The slope of the line represents the

distance covered per unit time, or speed. So, the steeper the line, the greater the speed. Graph III matches this situation since a slower (walking) speed is followed by a faster (train) speed, followed by a slower (walking) speed.

b) Graph IV matches this situation since biking would be faster than running, and running would be faster than walking. c) Graph I matches this situation since sitting in one spot to fish would correspond to a speed of 0. d) Graph II matches this situation since

waiting corresponds to a speed of 0, and riding a train would be faster than running.


Chapter 2 Test 149

Chapter 2 Test 1. a) To determine ( )2f − locate the point

directly above –2 on the x-axis. Then estimate the second coordinate by moving horizontally to the y-axis. From the graph, ( )2 1f − = .

b) The domain is the set of all x-values for which f is defined. From the graph the domain of f appears to be

{ }| 3 4 .x x− ≤ ≤

c) To determine any x-value(s) for which ( ) 1

2f x = , locate the point 12 on the

y-axis and move left or right to locate any points on the graph with a y value of 12 . Then estimate the x-coordinate by

moving down from the point to the x-axis. From the graph ( ) 1

23 .f =

d) The range of f is the set of all outputs or second values of the function. From the graph it appears that the range of f = { }| 1 2y y− ≤ ≤ .

2. Use the indicated two points on the graph to

determine the rate of change in y. It is the same as the slope of the line. For ( )2005,100 and ( )2007,150 :

2 1

2 1

150 100 5025

2007 2005 2

y ym

x x

− −= = = =− −

3. ( )( )

5change in 3 2 3 2Slopechange in 6 26 2 8

yx

− − += = = =+− −

4.

( )

change in Slope

change in

5.2 5.2 0 04.4 3.14.4 3.1

yx

=

−= = =− +− − −

5. The function ( ) 312

5f x x= − + is in slope-

intercept form.

The slope is 3

5− . The y-intercept is (0, 12).

6. Convert the equation to slope-intercept form. 5 2 7

5 2 7

2 7

5 5

y x

y x

y x

− − =− = +

= − −

The slope is 2

5− . The y-intercept is ( )70,

5− .

7. ( ) 3f x = −

This is a horizontal line that crosses the vertical axis at ( )0, 3− . The slope is 0.

8. 5 11

16

x

x

− ==

The graph of x = 16 is a vertical line that crosses the horizontal axis at ( )16,0 . The

slope is undefined. 9. 5 15x y− =

To find the y-intercept, let x = 0 and solve for y. 5 0 15

15

y

y

⋅ − == −

The y-intercept is (0, –15). To find the x-intercept, let y = 0 and solve for x.

5 0 15

3

x

x

− ==

The x-intercept is (3, 0). 10. Graph ( ) 3 4f x x= − + .

Slope is –3 or 31− ; y-intercept is (0, 4).

From the y-intercept, go down 3 units and to the right 1 unit to get to (1, 1). Use the two points to draw the graph.



11. ( )11 4

2y x− = − + is in point-slope form.

The slope is 1

2− . The point (–4, 1) is on the

graph. From the point (–4, 1), go down 1 unit and right 2 units to get to (–2, 0). Use the two points to draw the graph.

12. Find the x-intercept.

2 5 0 20

20 102

x

x

− + =

= = −−

⋅

The x-intercept is ( )10,0 .−

Find the y-intercept. 2 0 5 20

20 45

y

y

− + =

= =

⋅


Use the two points to draw the graph.


3 9

9 3 6

6

x

x

x

− =− = − =

= −

This is a vertical line that crosses the x-axis at (–6, 0).

14 Yes, the standard viewing window of

[ ]10,10, 10,10− − , with Xscl = 1 and Yscl = 1,

works well, because 10 (0) 10f− < < and

9 90 and 10 10 also.2 2

f −⎛ ⎞ = − < − <⎝ ⎠

15. a) 8 7 0

7

8

x

x

− =

=

The equation is linear. (Its graph is a vertical line.) b) 24 9 12x y− =

The equation is not linear, because it has a 2 term.y

c) 2 5 3x y− =

The equation is linear. 16. Write both equations in slope-intercept form.

4 2 3 3 4 12

3 1 33

4 2 43 3

4 4

y x x y

y x y x

m m

+ = − + = −

= − = −

= =

The slopes are the same, so the lines are parallel.

17. Write both equations in slope-intercept form.

2 5 2 6

2 13

21

2

y x y x

my x

m

= − + − == − = +

=

The product of the slopes is ( ) 12 1,

2⎛ ⎞− = −⎜ ⎟⎝ ⎠ so

the lines are perpendicular. 18. Use the slope-intercept equation,

( )f x mx b= + , with m = –5 and b = –1.

( ) 5 1f x x= − −

19. Use the point-slope equation,

( ) ( )( ) ( )4 4 2 , or 4 4 2y x y x− − = − − + = +


Chapter 2 Test 151

20. Determine the slope of the linear function. ( )2 1 1

14 3 1

m− − − −= = = −

−

Use the slope and the given point in the point-slope equation. ( ) ( ) ( )

( )( )

2 1 4

2 4

2

f x x

f x x

f x x

− − = − −+ = − +

= − +


years after 1982, and c = the average number of hours commuters sit in traffic in the year t. Use the given information to generate the two points ( )0,16 and

( )25,41 . Determine the slope of the line

containing these points.

41 16 25

125 0 25

m−

= = =−

Use the point-slope equation with 1m =

and ( ) ( )1 1, 0,16t c = to obtain the

equation. ( )16 1 0

16

16

c t

c t

c t

− = −− =

= +

b) 2000 is 2000 – 1982 = 18 years after 1982. Substitute 18 for t and solve for c.

16

18 16

34

c t

c

c

= += +=

Commuters spent an average of 34 hours sitting in traffic in 2000.

c) 2012 is 2012 – 1982 = 30 years after 1982. Substitute 30 for t and solve for c.

16

30 16

46

c t

c

c

= += +=

By the year 2012, the average commuter will spend 46 hours sitting in traffic.

22. Let t = the number of years since 1980 and

B = the number of twins, in thousands, born in the year t. Enter the data with the number of years since 1980 in 1L , and the number of

births, in thousands, in 2L .


The second screen indicates that the equation is 2.6718 65.0212B t= + .

23. 2012 is 2012 – 1980 = 32 years after 1980.

Substitute 32 for t and solve for B.

( )2.6718 65.0212

2.6718 32 65.0212

85.4976 65.0212

150.5188

B t

B

B

B

= += += +=

The predicted number of twin births in 2012 is about 151,000. 24. ( ) ( )5 2 5 1 9h − = − + = −

25. ( )( ) ( ) ( ) 12 1g h x g x h x x

x+ = + = + +

26. The zeros for h are any values of x such that

h(x) = 0. ( ) 2 1

0 2 1

1 2

1

2

h x x

x

x

x

= += +

− =

− =

The only zero of h is x =1

2− .

27. Domain of g:

{ }is a real number and 0x x x ≠ 28. The domain of g is:

{ }is a real number and 0 .x x x ≠

The domain of h is: { } is a real number .x x The domain of g + h is: { }is a real number and 0 .x x x ≠



29. The domain of g is: { } is a real number and 0 .x x x ≠ The domain of h is: { } is a real number .x x

( ) 12 1 0, if

2h x x x= + = = −

The domain of g /h is:

1 is a real number and 0 and

2x x x x

⎧ ⎫≠ ≠ −⎨ ⎬⎩ ⎭

30. a) 1 h and 40 min is equal to

40 2 51 1 h.

60 3 3= =

Find 5

.3

f⎛ ⎞⎜ ⎟⎝ ⎠

5 55 15 5 25 30

3 3f⎛ ⎞ = + ⋅ = + =⎜ ⎟⎝ ⎠

The cyclist will be 30 mi from the starting point 1 h and 40 min after passing the 5-mile mark.

b) In the equation ( ) 5 15f t t= + , 15 is the

slope. The cyclist is traveling 15 mi every hour = 15 mph.

31. Determine the slope of the given line.

2 5 8

2 8

5 5

x y

y x

− =

= −

The slope is 2

5.

Any parallel line must also have a slope of 2

5. The parallel line containing the point

( )3,2− is:

( )22 3

52 6

25 52 16

5 5

y x

y x

y x

− = +

− = +

= +

32. Determine the slope of the given line. 2 5 8

2 8

5 5

x y

y x

− =

= −

The slope is 2

5.

Any perpendicular line must have a slope

given by the opposite of the reciprocal of 2

5,

5

2− . The perpendicular line containing the

point ( )3,2− is:

( )( )52 3

25 15

22 25 11

2 2

y x

y x

y x

− = − − −

− = − −

= − −

33. Determine the slope of the given line.

3 2 7

2 3 7

3 7 3

2 2 2

x y

y x

y x m

− =− = − +

= − =

Since parallel lines have the same slope, the

slope of the function must also be 3

2. An

expression for the slope of the function can also be generated using the two points (r, 3) and (7, s), and the formula.

2 1

2 1

.y y

mx x

−=

−

Therefore, set the expression for the slope in r

and s equal to 3

2 and solve for s in terms of r.

( ) ( )

3 3

7 22 3 3 7

2 6 21 3

2 21 3 6 3 27

3 27 3 27 or

2 2 2

s

rs r

s r

s r r

rs s r

− =−− = −− = −

= − + = − +− += = − +


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aplustestbank.euaplustestbank.eu/...Intermediate-Algebra-Graphs-Models-4-E-4th-Editi… · 78 Chapter 2: Intermediate Algebra: Graphs and Models b) The range is the set of all y-values