Lehmann IA SSM Ch6

8/20/2019 Lehmann IA SSM Ch6

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150

Chapter 6

Polynomial Functions

Homework 6.1

1. Vertex: (0, 0)

3. Vertex: (0, 0)

5. Vertex: (0, -1)

7. Vertex: (0, 5)

9. Vertex: (-5, 0)

11. Vertex: (1, 0)

13. Vertex: (4, -6)

15. Vertex: (6, 0)


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SSM: Intermediate Algebra Homework 6.1

151

17. Vertex: (-5, -7)

19. Vertex: (-7, -3)

21. Vertex: (-4, -1)

23. Vertex: (-6, 3)

25.

The Domain is the set of all real numbers. Since (0,

-4) is the minimum point, the range is the set of

numbers where 4 y ≥ − .

27.



numbers where 4 y ≥ .

29.



numbers where 4 y ≥ .

31. a. Since the function has a minimum point in

quadrant III, then 0a > , 0h < , and 0k <

b. Since the function has a maximum point in

quadrant II, then 0a < , 0h < , and 0k >


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Homework 6.1 SSM : Intermediate Algebra

152

c. Since the function has a minimum point that

lies on the x-axis when x > 0, then 0a > ,

0h > , and 0k =

d. Since the function has a minimum point that

lies on the y-axis when y > 0, then 0a < ,0h = , and 0k <

33. Answers may vary: Example:

2( 5) 3 y a x= + +

Where a = -3, -2, -1, -½, ½, 1, 2, 3

35. From the vertex (5, -6), we know2( ) ( 5) 6 f x a x= − − . Since (1, 4) lies on the

parabola, substitute 1 for x and 4 for f ( x) to solve

for a:

2

2

4 (1 5) 6

10 ( 4)

10 (16)

5

8

0.625

a

a

a

a

a

= − −

= −

=

=

=

The equation of the function is:

2( ) 0.625( 5) 6 f x x= − −

37. The value of a for the function f is the opposite of

the value of a for the function g since g has a

maximum point and f has a minimum point and we

can assume that the graphs of f and g have the

same “shape”. Since the vertex ( ),h k of g is

( )7,3.71− and a = -2.1, an equation for g is:

2( ) 2.1( 7) 3.71g x x= − + +

39. It is possible. Example: 2 2 y x= +

41. It is possible. Example:2

2 y x= −

43. a.

Quadratic Function because the points bend in

a parabola shape.

b.

Yes, the parabola comes close to the points in

the scattergram.

c. ( )8.86,47.49 The percentage of Americans

who were pro-choice was at a minimum of

48% in 1999, according to the model.

d. 1995, 2003

e. 94%

45. Both equations have the same vertex (2, 5). From

the graph notice this is the only point that lies on

both graphs.


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153

47.

a. Answers may vary. Example: ( )1, 2− , ( )0,1 ,

and ( )1, 2 .

(-1, 2) is a solution:

2

?2

?

( ) 1

2 ( 1) 1

2 2 TRUE

f x x= +

= − +

=

(0, 1) is a solution:

2

?2

?

( ) 1

1 (0) 1

1 1 TRUE

f x x= +

= +

=

(1, 2) is a solution:

2

?2

?

( ) 1

2 (1) 1

2 2 TRUE

f x x= +

= +

=

b. Answers may vary. Example: ( )0,0 , ( )1,0 ,

and ( )1,0− .

(0, 0) is NOT a solution:

2

? 2

?

( ) 1

0 (0) 1

0 1 FALSE

f x x= +

= +

=

(1, 0) is NOT a solution:

2

?2

?

( ) 1

0 (1) 1

0 2 FALSE

f x x= +

= +

=

(-1, 0) is NOT a solution:

2

?2

?

( ) 1

0 ( 1) 1

0 2 FALSE

f x x= +

= − +

=

49. a. Answers may vary. Example:

2 2( 1) and ( 1) y x y x= − = − +

b. Answers may vary. Example:

2 2 and y x y x= = −

c. Answers may vary. Example:

2 21 and 1 y x y x= − = − +

d. No, it is not possible.

51. a.

b.


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154

c.

d. Explanations may vary. In part b, the y-axis is

stretched vertically so that the parabola is

stretched vertically compared to a. Also, in

part c, the x-axis is stretched horizontally so

that the parabola is stretched horizontally

compared to part a.

53. Adjust the WINDOW settings. Make your x-minand x-max much larger.

55. a.

b.

c.

d.

57. Step 1: Sketch a graph of 2 y ax= .

Step 2: Translate the graph of 2 y ax= right h

units if h > 0 and left |h| units if h < 0.

Step 3: Lastly, translate the graph from step 2 up k

units if k > 0 and down |k | units if k < 0.

Homework 6.2

1. 2( 1) ( ) (1) x x x x x x x− = − = −

3. 7 ( 2) 7 ( ) ( 7 )(2) x x x x x− − = − − −

27 14 x x= − +

5. 3 (8 5) 3 (8 ) ( 3 )(5) x x x x x− + = − + −

224 15 x x= − −

7.2 4 ( ) (4) ( 4) x x x x x x x+ = + = +

9. 227 36 9 (3 ) ( 9 )(4) x x x x x− + = − − −

9 (3 4) x x= − −

11. 225 35 5 (5 ) ( 5 )(7) x x x x x− + = − − −


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155

5 (5 7) x x= − −

13. 24 8 4 ( ) ( 4 )(2) x x x x x− − = − + −

4 ( 2) x x= − +

15. 23.7 4.7 (3.8 ) (4.7) x x x x x+ = +

(3.8 4.7) x x= +

17. ( ) ( 7) f x x x= −

19. ( ) 6 (4 5)h x x x= −

21. ( ) 2 ( 4) f x x x= − +

23. ( ) (2.5 6.2)h x x x= −

25. Yes, expanding 2 (6 9) x x + , find that

22 (6 9) 2 (6 ) 2 (9) 12 18 x x x x x x x+ = + = +

27. All three students did the problem correctly,

although we usually write the result as Student 2

and Student 3 did.

29.22 (7 4) 14 8 x x x x− = −

31.2 2( 2)( 4) 4 2 8 6 8 x x x x x x x+ + = + + + = + +

33.2 2( 3)( 6) 6 3 18 3 18 x x x x x x x− + = + − − = + −

35. 2 2( 8)( 8) 8 8 64 16 64 x x x x x x x+ + = + + + = + +

37. 2( 7) ( 7)( 7) x x x− = − −

2

2

7 7 49

14 49

x x x

x x

= − − +

= − +

39. 2( 5)( 5) 5 5 25 x x x x x+ − = + − −

2 25 x= −

41.2(3 5)(4 1) 12 3 20 5 x x x x x+ + = + + +

212 23 5 x x= + +

43. 2(4 1)(6 5) 24 20 6 5 x x x x x− − = − − +

224 26 5 x x= − +

45. 2 3 2( 4)( 20) 2 4 8 x x x x x− + = + − −

47.3 2 5 3 2

( 6)( 3) 3 6 18 x x x x x+ + = + + +

49. 2(2 5) (2 5)(2 5) x x x+ = + +

2

2

4 10 10 25

4 20 25

x x x

x x

= + + +

= + +

51. 2(3 2)(3 2) (3 )(3 ) 6 6 4 9 4 x x x x x x x+ − = − + − = −

53.2 2

(1 ) x x x x x x− − = − + = −

55.2( 1) [( 1)( 1)] x x x− − = − − −

2

2

2

( 1)

( 2 1)

2 1

x x x

x x

x x

= − − − +

= − − +

= − + −

57.23 (5 1) 3 (5 1)(5 1) x x x x x− = − −

2

2

3 2

3 (25 5 5 1)

3 (25 10 1)

63 30 3

x x x x

x x x

x x x

= − − +

= − +

= − +

59. 22 (4 5) 2 (4 5)(4 5) x x x x x− + = − + +

2

2

3 2

2 (16 20 20 25)

2 (16 40 25)

32 80 50

x x x x

x x x

x x x

= − + + +

= − + +

= − − −

61. (2.1 3.8)(1.8 5.6) x x− +

2

2

3.78 11.76 6.84 21.28

3.78 4.92 21.28

x x x

x x

= + − −

= + −

63.

2

( 5)( 4 1) x x x+ + +

3 2 2

3 2

4 5 20 5

9 21 5

x x x x x

x x x

= + + + + +

= + + +

65. 2(2 3)(3 4) x x x− + −


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156

3 2 2

3 2

6 2 8 9 3 12

6 7 11 12

x x x x x

x x x

= + − − − +

= − − +

67.2( 4)( 4 16) x x x+ − +

3 2 2

3

4 16 4 16 64

64

x x x x x

x

= − + + − +

= +

69. 2 2( 2 3)( 2) x x x x+ + + +

4 3 2 3 2 2

4 3 2

2 2 2 4 3 3 6

3 7 7 6

x x x x x x x x

x x x x

= + + + + + + + +

= + + + +

71. 2 2(2 3)( 2 1) x x x x+ − − +

4 3 2 3 2 2

4 3 2

2 4 2 2 3 6 3

2 3 3 7 3

x x x x x x x x

x x x x= − + + − + − + −= − − + −

73. ( 1)( 2)( 3) x x x+ + +

2

2

3 2 2

3 2

( 1)( 3 2 6)

( 1)( 5 6)

5 6 5 6

6 11 6

x x x x

x x x

x x x x x

x x x

= + + + +

= + + +

= + + + + +

= + + +

75. 1 1

2 25 5 x x

− +

2

2

1 2 24

5 5 5

14

25

x x x

x

= + − −

= −

77.2 2( 5) ( 5) x x− − +

2 2

2 2

( 10 25) ( 10 25)

10 25 10 25

20

x x x x

x x x x

x

= − + − + +

= − + − − −= −

79.2 2( 4) ( 4) x x− + −

2 2

2 2

2

( 8 16) ( 8 16)

8 16 8 16

2 16 32

x x x x

x x x x

x x

= − + + − +

= − + + − +

= − +

81.2( ) ( 6)( 6) 6 6 36 f x x x x x x= + + = + + +

2 12 36 x x= + +

83.2( ) 2( 3) 1 2( 3)( 3) 1h x x x x= + + = + + +

2

2

2

2

2( 3 3 9) 12( 6 9) 1

2 12 18 1

2 12 19

x x x x x

x x

x x

= + + + += + + +

= + + +

= + +

85.2( ) 4( 5) 2 4( 5)( 5) 2 p x x x x= − + = − − +

2

2

2

2

4( 5 5 25) 2

4( 10 25) 2

4 40 100 2

4 40 102

x x x

x x

x x

x x

= − − + +

= − + +

= − + +

= − +

87.2( ) 4( 1) 1 4( 1)( 1) 1g x x x x= − − − = − − − −

2

2

2

2

4( 1) 1

4( 2 1) 1

4 8 4 1

4 8 5

x x x

x x

x x

x x

= − − − + −

= − − + −

= − + − −

= − + −

89.2( ) 1.5( 2.8) 3.7k x x= + −

2

2

2

2

1.5( 2.8)( 2.8) 3.7

1.5( 2.8 2.8 7.84) 3.7

1.5( 5.6 7.84) 3.7

1.5 8.4 11.76 3.7

1.5 8.4 8.06

x x

x x x

x x

x x

x x

= + + −

= + + + −

= + + −

= + + −

= + +

91.2( ) 5 ( 2) 5 10 f x x x x x= − = − This function is

quadratic.

93.2 2( ) ( 6)( 6) 6 6 36 36h x x x x x x x= + + = + + − = −

This function is quadratic.

95.2 2( ) ( 6) ( 6)h x x x= + − −

2 2

2 2

( 12 36) ( 12 36)

12 36 12 36

24

x x x x

x x x x

x

= + + − − +

= + + − + −=

This function is linear.


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157

97.2(4) 4 3(4) 16 12 4 f = − = − =

99.2(2.8) (2.8) 3(2.8) 7.84 8.4 0.56 f = − = − = −

101.

2 2

( ) 3( ) 3 f a a a a a= − = −

103.2( 1) ( 1) 3( 1) f a a a+ = + − +

2

2

( 1)( 1) 3( 1)

1 3 3

2

a a a

a a a a

a a

= + + − +

= + + + − −

= − −

105.2(0) 2(0) 5(0) 1 1 f = − + − = −

107.

21 1 1

2 5 12 2 2

f = − + −

1 52 1

4 2

1 51

2 2

41

2

2 1 1

= − + −

= − + −

= −

= − =

109.2 2( ) 2( ) 5( ) 1 2 5 1 f a a a a a= − + − = − + −

111.

2

( 3) 2( 3) 5( 3) 1 f a a a+ = − + + + −

2

2

2

2( 6 9) 5 15 1

2 12 18 5 15 1

2 7 4

a a a

a a a

a a

= − + + + + −

= − − − + + −

= − − −

113. Answers may vary. Your discussion should include

square of a sum, square of a difference, and

difference of two squares

Homework 6.3

1. 2 7 12 ( 3)( 4) x x x x

− + = + +

3. 2 21 20 ( 20)( 1) x x x x− + = − −

5. This expression is prime since there are not two

integers with product -10 and sum 2.

7.2 23 3 18 3( 6) x x x x− − = − −

3( 3)( 2) x x= − +

9.

2 2

5 20 60 5( 4 12) x x x x− + + = − − −

5( 6)( 2) x x= − − +

11. 2 26 36 54 6( 6 9) x x x x+ + = + +

2

6( 3)( 3)

6( 3)

x x

x

= + +

= +

13. 2 2 216 4 ( 4)( 4) x x x x− = − = − +

15. 2 2 21 1 ( 1)( 1) x x x x− = − = − +

17.2 2 225 1( 5 ) ( 5)( 5) x x x x− = − − = − − +

19. 2 2 2100 1 (10 ) 1 (10 1)(10 1) x x x x− = − = − +

21. 2 249 1( 49) ( 7)( 7) x x x x− + = − − = − − +

23. This expression is prime since an expression of the

form 2 2 , 0 x k k + ≠ , is always prime and notfactorable.

25.2 2 264 49 (8 ) 7 (8 7)(8 7) x x x x− = + = − +

27.

2

2 21 1 1 1

9 3 3 3 x x x x

− = − = − +

29.23 10 8 (3 4)( 2) x x x x+ + = + +

31.22 13 15 ( 5)(2 3) x x x x− + = − −

33.29 9 2 (3 4)(3 1) x x x x+ + = − −

35. This expression is prime.

37. 29 15 4 (3 4)(3 1) x x x x− + = − −

39. 2 220 30 140 10(2 3 14) x x x x+ − = + −

10( 2)(2 7) x x= − +

41.2 11 30 ( 5)( 6) x x x x+ + = + +


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158

43. 2 3 2 ( 2)( 1) x x x x− + = − −

45. 215 27 3 (5 9) x x x x− = −

47.

2

10 23 12 (2 3)(5 4) x x x x+ + = + +


51. 2 15 100 ( 20)( 5) x x x x− − = − +

53.2 23 75 3( 25) x x− − = − +

55.2

2 48 ( 6)( 8) x x x x+ − = − +

57.212 7 5 ( 1)(12 5) x x x x− − = − +

59.

2 2

7 14 21 7( 2 3) x x x x+ + = + +

61. 281 16 (9 4)(9 4) x x x− = − +


65. 28 2 15 (2 3)(4 5) x x x x− − = − +

68.2 2 2100 10 ( 10)( 10) x x x x− = − = − +

69. 2 214 49 ( 7)( 7) ( 7) x x x x x− + = − − = −

71. 29 18 8 (3 4)(3 2) x x x x− + = − −

73.24 79 20 ( 20)(4 1) x x x x+ − = + −

75. 2 22 22 48 2( 11 24) 2( 8)( 3) x x x x x x− + = − + = − −

77.210 24 (2 3)(5 8) x x x x+ − = − +

79.2 236 48 16 4(9 12 4) x x x x+ + = + +

2

4(3 2)(3 2)

4(3 2)

x x

x

= + +

= +

81. The student did not factor the expression correctly.

It should be:

24 8 3 (2 3)(2 1) x x x x+ + = + +

83. The student did not factor the expression correctly.

The expression 2 25 x + is prime so2 24 100 4( 25) x x+ = + is the final step.

85. Answers may vary. Your discussion should be

include factoring techniques for factoring2

ax bx c+ + when a = 1 and when a • 1.

Homework 6.4

1. 3 23 75 3 ( 25) x x x x− = −

3 ( 5)( 5) x x x= − +

3. 3 263 7 7 (9 1) x x x x− = −

7 (3 1)(3 1) x x x= − +

5.5 3 3 28 72 8 ( 9) x x x x− = −

38 ( 3)( 3) x x x= − +

7.4 2 2 275 27 3 (25 9) x x x x− + = − −

23 (5 3)(5 3) x x x= − − +

9. 3 2 22 8 6 2 ( 4 3) x x x x x x+ + = + +

2 ( 1)( 3) x x x= + +

11. 4 3 2 2 24 12 40 4 ( 3 10) x x x x x x+ − = − −

24 ( 5)( 2) x x x= − +

13. 3 2 22 20 50 2 ( 10 25) x x x x x x− + − = − − +

2

2 ( 5)( 5)

2 ( 5)

x x x

x x

= − − −

= − −

15. 5 4 3 3 26 33 45 3 (2 11 15) x x x x x x+ + = + +

33 (2 5)( 3) x x x= + +

17.3 2 280 140 150 10 (8 14 15) x x x x x x+ − = + −

10 (2 5)(4 3) x x x= + −

19. 6 5 4 4 230 5 5 5 (6 1) x x x x x x− − + = − + −

4 2

4

5 (6 1)

5 (3 1)(2 1)

x x x

x x x

= − + −

= − − +


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159

21. 3 2 23 4 12 ( 3) 4( 3) x x x x x x + + + = + + +

2( 3)( 4) x x = + +

23. 3 2 2

4 3 12 ( 4) 3( 4) x x x x x x − + − = − + −

2( 4)( 3) x x = − +

25.3 2 23 15 4 20 3 ( 5) 4( 5) x x x x x x − − + = − − −

2( 5)(3 4) x x = − −

27. 3 2 22 32 16 (2 1) 8(2 1) x x x x x x + − − = + − +

2(2 1)( 8) x x = + −

29. 3 2 25 4 20 ( 5) 4( 5) x x x x x x − − + = − − −

2( 5)( 4)

( 5)( 2)( 2)

x x

x x x

= − −

= − − +

31.3 2 24 12 9 27 4 ( 3) 9( 3) x x x x x x − − + = − − −

2( 3)(4 9)

( 3)(2 3)(2 3)

x x

x x x

= − −

= − − +

33.3 2 212 4 3 1 4 (3 1) 1(3 1) x x x x x x + − − = + − +

2(3 1)(4 1)

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

x x

x x x

= + −

= + − +

35. 3 22 50 2 ( 25) x x x x − = −

2 ( 5)( 5) x x x = − +

37. 3 2 24 7 28 ( 4) 7( 4) x x x x x x + − − = + − +

2( 4)( 7) x x = + −

39. 2

81 ( 9)( 9) x x x − = + −

41.28 10 3 (4 3)(2 1) x x x x − + = − −

43. 29 15 3 (3 5) x x x x − = −

45. 4 3 2 2 23 18 24 3 ( 6 8) x x x x x x + + = + +

23 ( 4)( 2) x x x = + +

47.2 214 49 14 49 x x x x + + = + +

2

( 7)( 7)

( 7)

x x

x

= + +

= +

51. 2 ( 1) x x x x + = +

53. 3 216 ( 16) ( 4)( 4) x x x x x x x − = − − = − − +

55.3 2 24 20 5 4 ( 5) 1( 5) x x x x x x + − − = + − +

2( 5)(4 1)

( 5)(2 1)(2 1)

x x

x x x

= + −

= + − +

57. 3 2100 ( 100) ( 10)( 10) x x x x x x x − = − = + −

59.2 3 3 215 6 5 6 5 15 x x x x x x − + − = − − + This

expression is prime.

61. 4 3 2 2 25 35 60 5 ( 7 12) x x x x x x − + = − +

25 ( 4)( 3) x x x = − −

63. 2 3 3 226 12 12 12 26 12 x x x x x x + + = + +

22 (6 13 6)

2 (2 3)(3 2)

x x x

x x x

= + +

= + +

65. 2 24 16 16 4( 4 4) x x x x − + = − +

2

4( 2)( 2)

4( 2)

x x

x

= − −

= −

67. 2 22 1 1( 2 1) x x x x − − − = − + +

2

( 1)( 1)

( 1)

x x

x

= − + +

= − +

69. 4 5 48 6 2 (3 4) x x x x − = − −


73. 5 4 3 3 220 20 15 5 (4 4 3) x x x x x x + − = + −

35 (2 1)(2 3) x x x = − +


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75. The student hasn’t finished the problem, only done

the first step. To finish the problem factor ( 5) x +

out of both terms to give a final answer of:2( 5)( 3) x x + −

77. The student’s answer of (4 6)(4 6) x x + − is not

correct because it is not factored completely (2 can

be factored out). The student should have factored

out the GCF first. The answer would be:

2 216 36 4(4 9) 4(2 3)(2 3) x x x x − = − = − +

79. Answers may vary. See box titled “Five-Step

Factoring Strategy” on page 281 of the text or look

under the “Key Points of This Section” which

precedes this homework section.

Homework 6.5

1. 2 3 0 x x − =

( 3) 0

0 or 3 0

0 or 3

x x

x x

x x

− =

= − =

= =

3. 24 6 0 x x − =

2 (2 3) 0

2 0 or 2 3 0

0 or 2 33

0 or2

x x

x x

x x

x x

− =

= − =

= =

= =

5.2 8 15 0 x x + + =

( 3)( 5) 0

3 0 or 5 0

3 or 5

x x

x x

x x

+ + =

+ = + =

= − = −

7.2 4 0 x − =

( )( )

2 2

2 02 2 0

2 0 or 2 0

2 or 2

x

x x

x x

x x

− =

+ − =

+ = − =

= − =

9. 27 6 1 1 x x + + =

( )

( )

27 6 0

7 6 0

7 6 0

7 0 or 6 0

0 or 6

x x

x x

x x

x x

x x

+ =

+ =

+ =

= + =

= = −

11. 23 6 x x =

( )

( )

23 6 0

3 2 0

3 2 0

3 0 or 2 0

0 or 2

x x

x x

x x

x x

x x

− =

− =

− =

= − =

= =

13. 29 2 5 x x = − +

( )( )

22 9 5 0

2 1 5 0

2 1 0 or 5 0

2 1 or 5

1 or 5

2

x x

x x

x x

x x

x x

+ − =

− + =

− = + =

= = −

= = −

15. 29 12 4 0 x x − + =

( )( )3 2 3 2 0

3 2 0

3 2

2

3

x x

x

x

x

− − =

− =

=

=

17.23 12 x =

( )

( )( )

( )( )

2

2

3 12 0

3 4 0

3 2 2 0

2 2 0

2 0 or 2 0

2 or 2

x

x

x x

x x

x x

x x

− =

− =

− + =

− + =

− = + =

= = −

19. 216 25 x =


12/34


161

( )( )

2

2 2

16 25 0

(4 ) 5 0

4 5 4 5 0

4 5 0 or 4 5 0

4 5 or 4 5

5 5 or

4 4

x

x

x x

x x

x x

x x

− =

− =

− + =

− = + == = −

= = −

21. (5 3) 4 4 x x + − = −

( )

2

2

5 3 4 4

5 3 0

5 3 0

0 or 5 3 0

0 or 5 33

0 or5

x x

x x

x x

x x

x x

x x

+ − = −

+ =

+ =

= + =

= = −= = −

23. 3 ( 2) 4 x x x − =

( )

2

2

3 6 4

3 10 0

3 10 0

0 or 3 10 0

0 or 3 10

100 or

3

x x x

x x

x x

x x

x x

x x

− =

− =

− =

= − == =

= =

25.2

9 49 0 x − =

( )( )

2 2(3 ) 7 0

3 7 3 7 0

3 7 0 or 3 7 0

3 7 or 3 7

7 7 or

3 3

x

x x

x x

x x

x x

− =

− + =

− = + == = −

= = −

27. 2

1 x =

( )( )

2

2 2

1 0

1 0

1 1 0

1 0 or 1 0

1 or 1

x

x

x x

x x

x x

− =

− =

− + =

− = + == = −

29.2

64 28 9 x x = −

( )( )

29 28 64 0

3 8 3 8 0

3 8 0

3 8

8

3

x x

x x

x

x

x

− + =

− − =

− ==

=

31.2

4 8 32 x x − =

( )

( )( )

2

2

2

4 8 32 0

4 2 8 0

2 8 0

4 2 0

4 0 or 2 0

4 or 2

x x

x x

x x

x x

x x

x x

− − =

− − =

− − =

− + =

− = + == = −

33.2

12 36 0 x x − + =

( )( )6 6 0

6 0

6

x x

x

x

− − =

− ==

35. 2 1

025

x − =

2

2 10

5

1 10

5 5

1 10 or 0

5 5

1 1 or

5 5

x

x x

x x

x x

− =

− + =

− = + =

= = −

37. 2

12 2 2 0 x x − − =

( )

( )( )

2

2

2 6 1 0

6 1 0

2 1 3 1 0

2 1 0 or 3 1 0

2 1 or 3 1

1 1 or

2 3

x x

x x

x x

x x

x x

x x

− − =

− − =

− + =

− = + == = −

= = −


13/34


162

39. 21 1

64 2

x x − =

( )

( )( )

2

2

2

1 14 4 6

4 22 24

2 24 0

6 4 0

6 0 or 4 0

6 or 4

x x

x x

x x

x x

x x

x x

− = − =

− − =

− + =

− = + == = −

41.3

6 24 0 x x − =

( )

( )

( )( )

2

2 2

6 4 0

6 2 0

6 2 2 0

6 0 or 2 0 or 2 0

0 or 2 or 2

x x

x x

x x x

x x x

x x x

− =

− =

− + =

= − = + == = = −

43. 3 2

4 2 36 18 0 x x x − − + =

( ) ( )

( )( )

( )( )

( )( )

( )( )( )

( )( )( )

2

2

2

2 2

2 2 1 18 2 1 0

2 1 2 18 0

2 2 1 9 0

2 2 1 3 0

2 2 1 3 3 0

2 1 3 3 0

2 1 0 or 3 0 or 3 0

2 1 or 3 or 3

1 or 3 or 3

2

x x x

x x

x x

x x

x x x

x x x

x x x

x x x

x x x

− − − =

− − =

− − =

− − =

− − + =

− − + =

− = − = + == = = −

= = = −

45. 3 2

4 7 x x =

( )

3

2

2

4 7 0

4 7 0

0 or 4 7 0

0 or 4 7

70 or

4

x

x x

x x

x x

x x

− =

− == − =

= =

= =

47.3 2

4 9 20 45 x x x − = −

( ) ( )

( )( )( )( )

( )( )( )

3 2

2

2

2 2

4 20 9 45 0

4 5 9 5 0

5 4 9 0

5 (2 ) 3 0

5 2 3 2 3 0

5 0 or 2 3 0 or 2 3 0

5 or 2 3 or 2 3

3 35 or or

2 2

x x x

x x x

x x

x x

x x x

x x x

x x x

x x x

− − + =

− − − =

− − =

− − =

− − + =

− = − = + == = = −

= = = −

49. 3 2

9 12 4 27 x x x − = −

( ) ( )

( )( )

( )( )

( )( )( )

3 2

2

2

2 2

9 27 4 12 09 3 4 3 0

3 9 4 0

3 (3 ) 2 0

3 3 2 3 2 0

3 0 or 3 2 0 or 3 2 0

3 or 3 2 or 3 2

2 25 or or

3 3

x x x

x x x

x x

x x

x x x

x x x

x x x

x x x

+ − − =+ − + =

+ − =

+ − =

+ − + =

+ = − = + == = = −

= = = −

51.3 2

18 3 6 x x x + =

( )

( )( )

3 2

2

18 3 6 0

3 6 2 0

3 3 2 1 0

3 0 or 3 2 0 or 1 0

0 or 3 2 or 1

20 or or 1

3

x x x

x x x

x x x

x x x

x x x

x x x

+ − =

− − =

+ − =

= + = − == = − =

= = − =

53. 2

17 28 3 x x − = −

( )( )

23 17 28 0

3 4 7 0

3 4 0 or 7 0

3 4 or 7

4 or 7

3

x x

x x

x x

x x

x x

− − =

+ − =

+ = − == − =

= − =


14/34


163

55. ( )( )2 5 40 x x+ + =

( ) ( )

2

2

7 10 40

7 30 0

3 10 0

3 0 or 10 0

3 or 10

x x

x x

x x

x x

x x

+ + =

+ − =

− + =− = + == = −

57. ( ) ( )4 1 24 3 2 x x x x− − = −

( )( )

2 2

2

4 4 24 3 6

2 24 0

4 6 0

4 0 or 6 0

4 or 6

x x x x

x x

x x

x x

x x

− − = −

+ − =

− + =

− = + =

= = −

59. ( )( )2 5 6 2 3 x x x x+ + = + +

61. 2 5 6 0 x x+ + =

( )( )2 3 0

2 0 or 3 0

2 or 3

x x

x x

x x

+ + =

+ = + == − = −

63. 2 7 10 0 x x− + =

( )( )2 5 02 0 or 5 0

2 or 5

x x x x

x x

− − =− = − == =

65. ( )( )2 7 10 2 5 x x x x− + = − −

67. False, a and b can have values other than a = 2 and

b = 10. (For example a = 4 and b = 5)

69. The student did not do factor by grouping correctly

(step 3 has the error). The correct solution is:

( ) ( )

( )( )

( )( )

( )( )( )

3 2

2

2

2 2

4 9 36 0

4 9 4 0

4 9 0

4 ( ) 3 0

4 3 3 0

4 0 or 3 0 or 3 0

4 or 3 or 3

x x x

x x x

x x

x x

x x x

x x x

x x x

+ − − =

+ − + =

+ − =

+ − =

+ − + =

+ = − = + == − = = −

71. Solve for x when ( ) 0 f x =

( )( )

20 9 20

0 5 4

5 0 or 4 0

5 or 4

x x

x x

x x

x x

= − +

= − −

− = − == =

The x-intercepts are (5, 0) and (4, 0).


( )( )

2

2 2

0 16

0 4

0 2 2

2 0 or 2 0

2 or 2

x

x

x x

x x

x x

= −

= −

= + −

+ = − =

= − =

The x-intercepts are (-2, 0) and (2, 0).


( )

( )( )

2

2

2

0 2 20 32

0 2 10 16

0 10 16

0 8 2

8 0 or 2 0

8 or 2

x x

x x

x x

x x

x x

x x

= − +

= − +

= − +

= − −

− = − =

= =



( )( )

20 36 25

0 6 5 6 5

6 5 0 or 6 5 0

6 5 or 6 5

5 5 or

6 6

x

x x

x x

x x

x x

= −

= − +

− = + == = −

= = −

The x-intercepts are5

,06

and5

,06

−

.



15/34


164

( )( )

20 12 7 10

0 3 2 4 5

3 2 0 or 4 5 0

3 2 or 4 5

2 5 or

3 4

x x

x x

x x

x x

x x

= − −

= + −

+ = − == − =

= − =

The x-intercepts are2

,03

−

and5

,04

.


( )

20 3 30

0 3 10

3 0 or 10 0

0 or 10

x x

x x

x x

x x

= −

= −

= − =

= =



( )

( )( )

3 2

2

0 2 2 84

0 2 42

0 2 7 6

2 0 or 7 0 or 6 0

0 or 7 or 6

x x x

x x x

x x x

x x x

x x x

= − −

= − −

= − +

= − = + == = = −

The x-intercepts are (0, 0), (7, 0) and (-6, 0).


( ) ( )

( )( )

( )( )

( )( )( )

3 2

2

2

2 2

0 2 2

0 2 1 2

0 2 1

0 2 1

0 2 1 1

2 0 or 1 0 or 1 0

2 or 1 or 1

x x x

x x x

x x

x x

x x x

x x x

x x x

= + − −

= + − +

= + −

= + −

= + − +

+ = − = + == − = = −

The x-intercepts are (-2, 0), (1, 0) and (-1, 0).

87. a.

The scattergram suggests that the data can be

best modeled by a quadratic function since the

points are making a parabola shape.

b.

It appears that f models the data quite well.

c. Solve for t when f (t ) = 109

( )( )

2

2

109 22 130

0 22 21

0 21 1

21 0 or 1 0

21 or 1

t t

t t

t t

t t

t t

= − +

= − += − −

− = − == =

Sales will be 109 million cases in the years

1981 or 2001.

d. Solve for f (t ) when t = 23 (year 2003)

2(23) (23) 22(23) 130 153 f = − + =

In 2003, the sales will be 153 million cases.


16/34


165

89. a.

It appears that f models the data quite well.

b. Solve for t when f (t ) = 63

( )

( )

( )( )

2

2

2

2

2

2

463 10 91

7

4

7 63 10 91 77

441 4 70 637

0 4 70 196

0 2 2 35 98

0 2 35 98

0 2 7 14

2 7 0 or 14 0

2 7 or 14

7 or 14

23.5 or 14

t t

t t

t t

t t

t t

t t

t t

t t

t t

t t

t t

= − +

= − +

= − +

= − +

= − +

= − +

= − −

− = − == =

= =

= =

The model predicts that 63% of Americans

will be pro-choice in 2004 and in 1994.

c. Solve for t when f (t ) = 91

( )

( )

2

2

2

2

491 10 91

7

47 91 10 91 7

7

637 4 70 637

0 4 70

0 2 2 35

2 0 or 2 35 0

0 or 2 35

350 or

2

0 or 17.5

t t

t t

t t

t t

t t

t t

t t

t t

t t

= − +

= − +

= − +

= −

= −

= − == =

= =

= ≈

91% of Americans will be pro-choice in 2008

and in 1990. It is not likely that these

predictions will be correct since public opinion

has been between 48% and 58% for the past

30 years.

91. ( ) ( ) ( )2

9 9 9 6 66 f = − − =

93. ( ) ( ) ( )2

7 7 7 6 50 f − = − − − − =

95. ( ) 2 6 f x x x= − −

( )( )

2

2

14 6

0 20

0 5 4

5 0 or 4 0

5 or 4

x x

x x

x x

x x

x x

= − −

= − −

= − +

− = + == = −

97. ( ) 2 6 f x x x= − −

( )

2

2

6 6

0

0 1

0 or 1 0

0 or 1

x x

x x

x x

x x

x x

− = − −

= −

= −

= − == =

99. Since ( )2,5− lies on the parabola, ( )2 5 f − = .

101. Since ( )1, 1− lies on the parabola, ( )1 1 f = − .

103. Since ( )1.7,4− and ( )3.7,4 lie on the parabola,

( )1.7 and 3.7 when 4a a f a= − = = .

105. Since ( )1, 1− lies on the parabola,

( )1 when 1a f a= = − .

107. Since ( )0,19 is in the table, ( )0 19 f = .

109. Since ( )4,3 is in the table, ( )4 3 f = .

111. Since ( )0,19 and ( )6,19 are in the table,

( )0 and 6 when 19 x x f x= = = .

113. Since ( )3,1 is in the table, ( )3 when 1 x f x= = .


17/34


166

115. a. Since ( )1,9 and ( )5,9 are in the table,

( )1 and 5 when 9 x x f x= = = .

b. The function f does not have an inverse

function since the output 9 corresponds to notone input, but two (1 and 5).

117. Answers may vary. Example:

2( ) ( 5)( 1) 4 5 f x x x x x= + − = + −


2( 3)( 1) 2 3 y x x x x= + − = + −

121. Answers may vary, but must be of the form:

( ) ( 6)( 3)h x a x x= + +

, where a > 0

because the graph has a minimum point and x-

intercepts ( )6,0− and ( )3,0− .

Example: 2( ) ( 6)( 3) 9 18h x x x x x= + + = + +

123. Student 1’s work is correct. Student 2’s work is

incorrect because we cannot divide by x since x is

possibly 0.

125. Answers may vary. See box titled “Solving

Quadratic Equations” on page 28 8 of the text or

look under the “Key Points of This Section” whichprecedes this homework section.

Homework 6.6

1. Since0 10

52

+= the x-coordinate of the vertex

must be 5.

3. Since0 6

32

+= the x-coordinate of the vertex must

be 3.

5. Since 0 ( 4) 22

+ −= − the x-coordinate of the vertex

must be -2.

7. Since0 7

3.52

+= the x-coordinate of the vertex

must be 3.5.

9. Since2 8

52

+= the x-coordinate of the vertex must

be 5.

11. Since4 6

12

− +

= the x-coordinate of the vertex

must be 1.

13. A symmetric point to the y-intercept has a value of

x that is 2 units to the right of 2 x = (value of x at

the vertex). The value of y is the same as that of the

y-intercept, so another point on the parabola is

( )4,9 .

15. A symmetric point to the given point ( )1,1 has a

value of x that is 2 units to the right of 3 x = (value

of x at the vertex). The value of y is the same as

that of the given point, so symmetric point to ( )1,1

on the parabola is ( )5,1 .

17. First, find the y-intercept by substituting 0 for x in

the function:

( )20 6 0 7 7 y = − + =

The y-intercept is ( )0,7 . Next find the symmetric

point to ( )0,7 . Substitute 7 for y in the function

and solve for x:

( )

2

2

7 6 7

0 6

0 6

0 or 6 0

0 or 6

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =

Therefore the symmetric points are ( )0,7 and

( )6,7 . Since0 6

32

+= , the x-coordinate of the

vertex is 3. To find the y-coordinate of the vertex,

substitute 3 for x and solve for y:( )23 6 3 7 2 y = − + = − . So the vertex is ( )3, 2− .


18/34


167


the function:

( )20 8 0 9 9 y = + + =



and solve for x:

( )

2

2

9 8 9

0 8

0 8

0 or 8 0

0 or 8

x x

x x

x x

x x

x x

= + +

= +

= +

= + =

= = −


( )8,9− . Since0 ( 8)

42

+ −= − , the x-coordinate of

the vertex is -4. To find the y-coordinate of thevertex, substitute -4 for x and solve for y:

( ) ( )2

4 8 4 9 7 y = − + − + = − . So the vertex is

( )4, 7− − .


the function:

( ) ( )2

0 8 0 10 10 y = − + − = −

The y-intercept is ( )0, 10− . Next find the

symmetric point to ( )0, 10− . Substitute -10 for y in

the function and solve for x:

( )

2

2

10 8 10

0 8

0 8

0 or 8 0

0 or 8

x x

x x

x x

x x

x x

− = − + −

= − +

= − −

− = − =

= =

Therefore the symmetric points are ( )0, 10− and

( )8, 10− . Since0 8

42



substitute 4 for x and solve for y:

( ) ( )24 8 4 10 6 y = − + − = . So the vertex is ( )4,6 .


the function:

( ) ( )2

3 0 6 0 4 4 y = + − = −

The y-intercept is ( )0, 4− . Next find the symmetric

point to ( )0, 4− . Substitute -4 for y in the function

and solve for x:

( )

2

2

4 3 6 4

0 3 6

0 3 2

3 0 or 2 0

0 or 2

x x

x x

x x

x x

x x

− = + −

= +

= +

= + =

= = −


( )2, 4− − . Since0 ( 2)

12

+ −= − , the x-coordinate of

the vertex is -1. To find the y-coordinate of the

vertex, substitute -1 for x and solve for y:


19/34


168

( ) ( )2

3 1 6 1 4 7 y = − + − − = − . So the vertex is

( )1, 7− − .


the function:

( ) ( )2

3 0 12 0 5 5 y = − + − = −



and solve for x:

( )

2

2

5 3 12 5

0 3 12

0 3 4

3 0 or 4 0

0 or 4

x x

x x

x x

x x

x x

− = − + −

= − +

= − −

− = − =

= =


( )4, 5− . Since0 4

22




( ) ( )2

3 2 12 2 5 7 y = − + − = . So the vertex is ( )2,7 .


the function:

( ) ( )2

4 0 9 0 5 5 y = − − − = −



and solve for x:

( )

2

2

5 4 9 5

0 4 9

0 4 9

0 or 4 9 0

0 or 4 9

9

0 or 2.254

x x

x x

x x

x x

x x

x x

− = − − −

= − −

= − +

− = + =

= = −

= = − ≈ −


( )2.25, 5− − . Since0 ( 2.25)

1.1252

+ −= − , the x-

coordinate of the vertex is -1.125. To find the y-

coordinate of the vertex, substitute -1.125 for x and

solve for y:

( ) ( )2

3 1.125 12 1.125 5 0.0625 y = − − + − − = . So

the vertex is ( )1.125,0.0625− .


the function:

( ) ( )2

2 0 7 0 7 7 y = − + =



and solve for x:


20/34


169

( )

2

2

7 2 7 7

0 2 7

0 2 7

0 or 2 7 0

0 or 2 7

70 or 3.5

2

x x

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =

= = =


( )3.5,7 . Since0 3.5

1.752

+= , the x-coordinate of

the vertex is 1.75. To find the y-coordinate of the

vertex, substitute 1.75 for x and solve for y:

( ) ( )2

2 1.75 7 1.75 7 0.875 y = − + = . So the vertex is

( )1.75,0.875 .

31. First, change the equation to standard form:

2

2

4 6 8

4 8 6

x y x

y x x

− + =

= − +

Next, find the y-intercept by substituting 0 for x in

the function:

( ) ( )2

4 0 8 0 6 6 y = − + =



and solve for x:

( )

2

2

6 4 8 6

0 4 8

0 4 2

4 0 or 2 0

0 or 2

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =


( )2,6 . Since0 2

12



substitute 1 for x and solve for y:( ) ( )

24 1 8 1 6 2 y = − + = . So the vertex is ( )1,2 .

33. First, change the equation to standard form:

( )

( )

2

2

2

2

2 3 15

2 6 9 15

2 12 18 15

2 12 33

y x

y x x

y x x

y x x

= − +

= − + +

= − + +

= − +


the function:

( ) ( )2

2 0 12 0 33 33 y = − + =



and solve for x:

( )

2

2

33 2 12 33

0 2 12

0 2 6

2 0 or 6 0

0 or 6

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =


( )6,33 . Since0 6

32




( ) ( )2

2 3 12 3 33 15 y = − + = . So the vertex is

( )3,15 .


21/34


170

35. The equation is in vertex form. To sketch the graph

of 2 6 y x= − , which has a vertex of ( )0, 6− ,

translate the graph of 2 y x= down 6 units.


the function:

( ) ( )2

2.8 0 8.7 0 4 4 y = − + =

The y-intercept is ( )0,4 . Next find the symmetricpoint to ( )0,4 . Substitute the 4 for y in the

function and solve for x:

( )

2

2

4 2.8 8.7 4

0 2.8 8.7

0 2.8 8.7

0 or 2.8 8.7 0

0 or 2.8 8.7

8.70 or 3.11

2.8

x x

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =

= = ≈


( )3.11,4 . Since0 3.11

1.562

+≈ , the x-coordinate

of the vertex is 1.56. To find the y-coordinate of

the vertex, substitute 1.56 for x and solve for y:

( ) ( )2

2.8 1.56 8.7 1.56 4 2.76 y = − + ≈ − . So the

vertex is ( )1.56, 2.76− .


the function:

( ) ( )2

3.9 0 6.9 0 3.4 3.4 y = + − = −

The y-intercept is ( )0, 3.4− . Next find the

symmetric point to ( )0, 3.4− . Substitute -3.4 for y

in the function and solve for x:

( )

2

2

3.4 3.9 6.9 3.4

0 3.9 6.9

0 3.9 6.9

0 or 3.9 6.9 0

0 or 3.9 6.9

6.90 or 1.773.9

x x

x x

x x

x x

x x

x x

− = + −

= +

= +

= + =

= = −

= = − ≈ −

Therefore the symmetric points are ( )0, 3.4− and

( )1.77, 3.4− − . Since0 ( 1.77)

0.882

+ −≈ − , the x-

coordinate of the vertex is -0.88. To find the y-

coordinate of the vertex, substitute -0.88 for x and

solve for y:

( ) ( )2

3.9 1.77 6.9 1.77 3.4 6.45 y = − + − − = − . So the

vertex is ( )1.77, 6.45− − .


22/34


171

41. First, change the equation to standard form:2

2

2

2

3.6 2.63 8.3 7.1

3.6 8.3 2.63 7.1

8.3 2.63 7.1

3.6

2.31 7.31 1.97

y x x

y x x

x x

y

y x x

− = −

= + −

+ −=

= + −


the function:

( ) ( )2

2.31 0 7.31 0 1.97 1.97 y = + − = −

The y-intercept is ( )0, 1.97− . Next find the

symmetric point to ( )0, 1.97− . Substitute -1.97 for

y in the function and solve for x:

( )

2

2

1.97 2.31 7.31 1.97

0 2.31 7.31

0 2.31 7.31

0 or 2.31 7.31 0

0 or 2.31 7.31

7.310 or 3.16

2.31

x x

x x

x x

x x

x x

x x

− = + −

= +

= +

= + =

= = −

= = − ≈ −

Therefore the symmetric points are ( )0, 1.97− and

( )3.16, 1.97− −

. Since

( )0 3.16

1.582

+ −

= −

, the x-

coordinate of the vertex is 1.58− . To find the y-

coordinate of the vertex, substitute 1.58− for x and

solve for y:

( ) ( )2

2.31 1.58 7.31 1.58 1.97 7.75 y = − + − − = − .

So the vertex is ( )1.58, 7.75− − .

43. Since the x-intercepts are symmetric points and

2 64

2

+= , the x-coordinate of the vertex is 4.

45. Since the x-intercepts are symmetric points and

9 4 5

2 2

− += − , the x-coordinate of the vertex is

5

2− .

47. To find the x-intercepts, let 0 y = and solve for x:

( )

20 5 10

0 5 2

5 0 or 2 0

0 or 2

x x

x x

x x

x x

= −

= −

= − =

= =

The x-intercepts are ( )0,0 and ( )2,0 . The y-

intercept is, therefore, ( )0,0 . Since the x-intercepts

are symmetric points and0 2

12

+= , the x-

coordinate of the vertex is 1. Substitute 1 for x in

the function and solve for y:

( ) ( )2

5 1 10 1 5 y = − = − . So, the vertex is ( )1, 5− .


( )

20 4

0 4

0 or 4 0

0 or 4

x x

x x

x x

x x

= −

= −

= − =

= =




22

+= , the x-



23/34


172

the function and solve for y:

( ) ( )2

5 2 10 2 4 y = − = − . So, the vertex is ( )2, 4− .


( )( )

20 10 24

0 6 4

6 0 or 4 0

6 or 4

x x

x x

x x

x x

= − +

= − −

− = − =

= =

The x-intercepts are ( )6,0 and ( )4,0 . To find the

y-intercept, let x = 0 and solve for y:

( ) ( )2

0 10 0 24 24 y = − + = . The y-intercept is

( )0,24 . Since the x-intercepts are symmetric

points and6 4

52

+= , the x-coordinate of the vertex

is 5. Substitute 5 for x in the function and solve for

y: ( ) ( )2

5 10 5 24 1 y = − + = − . So, the vertex is

( )5, 1− .


( )( )

20 8 7

0 7 1

7 0 or 1 0

7 or 1

x x

x x

x x

x x

= − +

= − −

− = − =

= =

The x-intercepts are ( )7,0 and ( )1,0 . To find the

y-intercept, let x = 0 and solve for y:

( ) ( )2

0 8 0 7 7 y = − + = . The y-intercept is ( )0,7 .

Since the x-intercepts are symmetric points and

7 14

2

+= , the x-coordinate of the vertex is 4.

Substitute 4 for x in the function and solve for y:

( ) ( )2

4 8 4 7 9 y = − + = − . So, the vertex is ( )4, 9− .


( )( )

20 9

0 3 3

3 0 or 3 0

3 or 3

x

x x

x x

x x

= −

= − +

− = + =

= = −

The x-intercepts are ( )3,0 and ( )3,0− . To find the

y-intercept, let x = 0 and solve for y:( )

20 9 9 y = − = − . The y-intercept is ( )0, 9− .

Since the x-intercepts are symmetric points and

( )3 30

2

+ −

= , the x-coordinate of the vertex is 0.

So the vertex is ( )0, 9− .


24/34


173


( )( )

20 2 11 21

0 2 3 7

2 3 0 or 7 0

2 3 or 7

3= 1.5 or 7

2

x x

x x

x x

x x

x x

= − −

= + −

+ = − =

= =

= − =

The x-intercepts are ( )1.5,0− and ( )7,0 . To find

the y-intercept, let x = 0 and solve for y:

( ) ( )2

2 0 11 0 21 21 y = − − = − . The y-intercept is

( )0, 21− . Since the x-intercepts are symmetric

points and( )7 1.5

2.75

2

+ −

= , the x-coordinate of

the vertex is 2.75. Substitute 2.75 for x in the

function and solve for y:

( ) ( )2

2 2.75 11 2.75 21 36.125 y = − − = − . So, the

vertex is ( )2.75, 36.125− .

59. a. Begin by finding the h-coordinate of the h-

intercept:

( ) ( ) ( )2

0 16 0 140 0 3 3h = − + + =

This means the height of the ball is 3 feet

when the matter makes contact (t = 0).

b. Since the function is in the form2

( )h t at bt c= + + and a = -16 < 0, the vertexis the maximum point. Find the h(t )-coordinate

of the vertex. Since the h-intercept is (0, 3)

from part a, find the symmetric point by

substituting 3 for h(t ) in the function and solve

for t :

( )

2

2

3 16 140 3

0 16 140

0 4 4 35

4 0 or 4 35 0

0 or 4 35

350 or 8.75

4

t t

t t

t t

t t

t t

t t

= − + +

= − +

= − −

− = − =

= =

= = =

With the same h-coordinate as the h-intercept,

the symmetric point is (8.75, 3). Since the

average of the t -coordinates of the symmetric

points is approximately 4.37, the t -coordinate

of the vertex is approximately 4.37. Compute

h(4.37) to find the h-coordinate of the vertex:

( ) ( ) ( )2

4.37 16 4.37 140 4.37 3

309.25

h = − + +

≈

So the vertex is (4.37, 309.25), which means

that the maximum height of the ball is 309.25

feet and it is reached at 4.37 seconds.

c.

61. a. The function P is in the form2( )P t at bt c= + + and a = 0.99 > 0, so the

vertex is the minimum point where the profit

is the least. Find the vertex. Begin by finding

the P-intercept of the function: let t = 0,

( ) ( ) ( )

2

0 0.09 0 1.65 0 9.72 9.72 f = − + =

The P-intercept is (0, 9.72), Next, find the

symmetric point to (0, 9.72). Substitute 9.72

for P(t ) in the function and solve for t :


25/34


174

( )

2

2

9.72 0.09 1.65 9.72

0 0.09 1.65

0 0.09 1.65

0 or 0.09 1.65 0

0 or 0.09 1.65

1.650 or 18.3

0.09

x x

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =

= = ≈

With the same P-coordinate as the P-intercept,

the symmetric point is (18.3, 97.2). Since the

average of the t -coordinates is approximately

9.17, the t -coordinate of the vertex is 9.17.

Computer P(9.17) to find the P-coordinate of

the vertex:

( ) ( ) ( )2

9.17 0.09 9.17 1.65 9.17 9.72

2.16

P = − +

≈

So the vertex is (9.17, 2.16), which means that

in 1999 ( )9.17t ≈ , the profit was the least at

2.16 million dollars.

b.

63. The function f is in the form 2( ) f t at bt c= + +

where a = 0.0038 > 0, so the vertex is the

minimum point of the parabola where the median

age is at a minimum. Find the vertex. Begin by

finding the f -intercept: let t = 0,

( ) ( ) ( )2

0 0.00248 0 0.29 0 31.48 31.48 f = − + = .

The f -intercept is (0, 31.48). Next, find the

symmetric point to (0, 31.48). Substitute 31.48 for

f (t ) in the function and solve for t :

( )

2

2

31.48 0.00248 0.29 31.48

0 0.00248 0.29

0 0.00248 0.29

0 or 0.00248 0.29 0

0 or 0.00248 0.29

0.290 or 116.94

0.00248

t t

t t

t t

t t

t t

t t

= − +

= −

= −

= − =

= =

= = ≈

With the same f -coordinate as the f -intercept, the

symmetric point is (116.94, 31.48). Since the


points is approximately 58.47, the t -coordinate of

the vertex is 58.47. Compute f (58.47) to find the f -

coordinate of the vertex:

( ) ( ) ( )2

58.47 0.00248 58.47 0.29 58.47 31.48

23.00

f = − +

≈

So the vertex is (58.47, 23) which means that the

minimum median age for men was 23 years old in

1958 (t • 58.47).

65. Since the function is in the form 2( ) f t at bt c= + +

and a = -5.07 < 0, the vertex is the maximum. Find

the vertex. Begin by finding the f -intercept: let t =

0,

( ) ( ) ( )2

0 5.07 0 102.93 0 460.40 460.40 f = − + − = − .

The f -intercept is (0, -460.4). Next, find the

symmetric point to (0, -460.4). Substitute -460.40

for f (t ) in the function and solve for t :

( )

2

2

460.4 5.07 102.93 460.40

0 5.07 102.93

0 5.07 102.93

0 or 5.07 102.93 0

0 or 5.07 102.93

102.930 or 20.3

5.07

t t

t t

t t

t t

t t

t t

− = − + −

= − +

= − −

− = − =

= =

= = ≈


symmetric point is (20.3, -460.4). Since theaverage of the t -coordinates of the symmetric


the vertex is approximately 10.15. Compute

f (10.15) to find the f -coordinate of the vertex:

( ) ( ) ( )2

10.15 5.07 10.15 102.93 10.15 460.4

62.02

f = − + −

≈


26/34

SSM: Intermediate Algebra Chapter 6 Review Exercises

175

So the vertex is (10.15, 62.02), which means that

the sales of premium sports cars was at a maximum

of 62.02 thousand cars in 2000 (t = 10.15).

67. a. First find the y-intercept:

2(0) 0 4(0) 12 12 f = + − = − .

The y-intercept is (0, -12). Next, find the

symmetric point to (0, -12) by substituting -12

for f ( x) in the function and solve for x:

( )

2

2

12 4 12

0 4

0 4

0 or 4 0

0 or 4

x x

x x

x x

x x

x x

− = + −

= +

= +

= + =

= = −

The symmetric points are (0, -12) and

( )4, 12− − . Since the average of the x-

coordinates is0 ( 4)

22

+ −= − , the x-coordinates

of the vertex is -2.

b. First find the x-intercepts:

( )( )

20 4 12

0 2 6

2 0 or 6 0

2 or 6

x x

x x

x x

x x

= + −

= − +

− = + =

= = −

The x-intercepts are (2, 0) and (-6, 0). These

points are symmetric and since the average of

the x-coordinates is2 ( 6)

22

+ −= − , the x-

coordinate of the vertex is -2.

c. The results for part a and b are the same.

d. Finding the x-coordinate of the vertex by

averaging the x-coordinates of the y-intercept

and its symmetric point is the easier method.

e. Finding the x-coordinate of the vertex by

averaging the x-coordinates of the y-intercept

and its symmetric point is the easier method

since this function is prime.

f. It is easier to find the x-coordinate of the

vertex for a prime function by averaging the x

coordinates of the y-intercept and its

symmetric point.

69. (3, 2) is the vertex for both f and k . The vertex of g

is approximately (2.7, 1.8). The vertex of h is

approximately (3.3, 1.7).

71. Answers may vary. See page 295 of the text and

the description of figure 54.

Chapter 6 Review Exercises

1.

2.

3.


27/34

Chapter 6 Review Exercises SSM : Intermediate Algebra

176

4.

5. Since the parabola has a maximum point, a < 0.

Since the vertex is in quadrant II, h < 0 and k > 0.

6. This is the graph of 2 y x= − , translated up 4 units

and to the right 4 units, so ( )2

4 4 y x= − − + . (The

vertex is (4, 4)).

7. 23 ( 2) ( 3 ) ( 3 )2 3 6 x x x x x x x− + = − + − = − −

8. 2( 4) ( 4)( 4) x x x+ = + +

2

2

4 4 16

8 16

x x x

x x

= + + +

= + +

9. 2 2( 2)( 5) 5 2 10 3 10 x x x x x x x+ − = − + − = − −

10. 23 (2 5) 3 (2 5)(2 5) x x x x x− = − −

2 2

2

3 2

3 [(2 ) 10 10 5 ]

3 (4 20 25)

12 60 75

x x x x

x x x

x x x

= − − +

= − +

= − +

11. 2(2 3)(5 1) (10 2 15 3) x x x x x− − + = − + − −

2

2

(10 13 3)

10 13 3

x x

x x

= − − −

= − + +

12. 22 (4 7)(4 7) 2 (16 28 28 49) x x x x x x x− − + = − − + −

2

3

2 (16 49)

32 98

x x

x x

= − −

= − +

13.

2 21 1 1 1 1 1

5 8 5 8 5 8 x x x

− + = −

21 1

25 64 x= −

14. 2 2 4 2 2( 3)( 7) 7 3 21 x x x x x− − = − − +

4 210 21 x x= − +

15.

2 3 2 2

( 2)( 5 4) 5 4 2 10 8 x x x x x x x x+ − + = − + + − + 3 23 6 8 x x x= − − +

16. 2 2(3 2 1)( 3 2) x x x x− + + −

4 3 2 3 2 2

4 3 2

3 9 6 2 6 4 3 2

3 7 11 7 2

x x x x x x x x

x x x x

= + − − − + + + −

= + − + −

17. The following functions factor down to

3( 1)( 4) y x x= − + :

2

3 9 12; 3( 4)( 1)3 ( 3) 12; ( 4)(3 3)

y x x y x x y x x y x x

= + − = + −= + − = + −

18. 2( ) 2( 1) 3 2( 1)( 1) 3 f x x x x= − + − = − + + −

2

2

2

2

2( 1) 3

2( 2 1) 3

2 4 2 3

2 4 5

x x x

x x

x x

x x

= − + + + −

= − + + −

= − − − −

= − − −

19. 24 12 ( 4 ) ( 4 )(3) 4 ( 3) x x x x x x x− + = − − − = − +

20. 2 2 24 ( 6)( 4) x x x x− − = − +

21. 2 2 2 27 63 7( 9) 7( 3 ) 7( 3)( 3) x x x x x− = − = − = − +

22.

2

2 21 1 1 1

4 2 2 2 x x x x

− = − = − +

23. 2 23 6 45 3( 2 15) 3( 5)( 2) x x x x x x− − = − − = − +

24. 22.4 7.9 (2.4 7.9) x x x x− = −

25. 2 22 16 32 2( 8 16) x x x x− + = − +

2

2( 4)( 4)

2( 4)

x x

x

= − −

= −

26. 28 14 15 (2 5)(4 3) x x x x+ − = + −


28/34


177

27. 2 2

6 33 36 3(2 11 12) x x x x − + = − +

3(2 3)( 4) x x = − −

28.

3 2

5 45 5 ( 9) x x x x

− = − 2 2

5 ( 3 )

5 ( 3)( 3)

x x

x x x

= −= + −

29. 3 2 2

3 27 42 3 ( 9 14) x x x x x x − + = − +

3 ( 2)( 7) x x x = − −

30. 3 2 2

4 8 9 18 4 ( 2) 9( 2) x x x x x x + − − = + − +

2

2 2

( 2)(4 9)

( 2)[(2 ) 3 ]

( 2)(2 3)(2 3)

x x

x x

x x x

= + −

= + −= + − +

31. 2

2 24 0 x x − − =

( 6)( 4) 0

6 0 or 4 0

6 or 4

x x

x x

x x

− + =− = + == = −

32. 2

25 0 x − =

2 25 0

( 5)( 5) 0

5 0 or 5 0

5 or 5

x

x x

x x

x x

− =− + =

− = + == = −

33. 2 1

081

x − =

2

2 10

9

1 10

9 9

1 10 or 0

9 9

1 1 or

9 9

x

x x

x x

x x

− =

− + =

− = + =

= = −

34. 2

3 7 2 0 x x − + =

(3 1)( 2) 0

3 1 0 or 2 0

3 1 or 21

or 23

x x

x x

x x

x x

− − =− = − =

= == =

35. 2

10 100 x x =

210 100 0

10 ( 10) 0

10 0 or 10 0

0 or 10

x x

x x

x x

x x

− =− =

= − == =

36. 2 2

3 10 5 5 x x x x − + = − + +

22 9 5 0

(2 1)( 5) 0

2 1 0 or 5 0

2 1 or 5

1 or 5

2

x x

x x

x x

x x

x x

+ − =− + =

− = + == = −

= = −

37. 2 3

3 30 6 x x x + =

3 2

2

6 3 30 0

3 (2 10) 0

3 (2 5)( 2) 0

3 0 or 2 5 0 or 2 0

0 or 2 5 or 2

50 or or 2

2

x x x

x x x

x x x

x x x

x x x

x x x

− − =

− − =− + =

= − = + == = = −

= = = −

38. 3 2

4 12 3 x x x − = −

3 2

2

2

2 2

3 4 12 0

( 3) 4( 3) 0

( 3)( 4) 0

( 3)( 2 ) 0

( 3)( 2)( 2) 0

3 0 or 2 0 or 2 0

3 or 2 or 2

x x x

x x x

x x

x x

x x x

x x x

x x x

+ − + =

+ − + =

+ − =+ − =+ − + =

+ = − = + == − = = −


29/34


178

39. 23 ( 7) 13 2 7 x x x− + = −

2 2

2

3 21 13 2 7

21 20 0

( 20)( 1) 0

20 0 or 1 0

20 or 1

x x x

x x

x x

x x

x x

− + = −

− + =

− − =− = − == =

40. 2 220 24 4 9 x x x− = −

216 24 9 0

(4 3)(4 3) 0

4 3 0

4 3

3

4

x x

x x

x

x

x

− + =− − =

− ==

=

41. To find the x-intercepts, of the function f substitute

0 for f ( x) and solve for x:

2

2

( ) 6 7 5

0 6 7 5

0 (3 5)(2 1)

3 5 0 or 2 1 0

3 5 or 2 1

5 1 or

3 2

f x x x

x x

x x

x x

x x

x x

= − −

= − −= − +− = + == = −

= = −

So, the x-intercepts are5

,03

and1

,02

−

.

42. To find the x-intercepts, of the function f substitute

0 for f ( x) and solve for x:

2

2 2

( ) 64 49

0 (8 ) 7

0 (8 7)(8 7)

8 7 0 or 8 7 0

8 7 or 8 7

7 7 or

8 8

f x x

x

x x

x x

x x

x x

= −

= −= − +− = + =

= = −

= = −

So, the x-intercepts are7

,08

and7

,08

−

.

43. 2(0) 5(0) (0) 4 4 f = − − = −

44.

21 1 1

5 45 5 5

f = − −

1 15 4

25 51 1

45 5

4

= − −

= − −

= −

45. ( ) ( ) ( )2

2 5 2 2 4 f a a a+ = + − + −

( )( ) ( )

( )2

2

2

5 2 2 2 4

5 4 4 2 4

5 20 20 6

5 19 14

a a a

a a a

a a a

a a

= + + − + −

= + + − − −

= + + − −

= + +

46. ( ) 25 4 f x x x= − −

25 4 0

(5 4)( 1) 0

5 4 0 or 1 0

5 4 or 1

4 or 1

5

x x

x x

x x

x x

x x

− − =+ − =

+ = − == − =

= − =

47. ( ) 25 4 f x x x= − −

2

2

5 4 2

5 6 0

(5 6)( 1) 0

5 6 0 or 1 0

5 6 or 1

6 or 1

5

x x

x x

x x

x x

x x

x x

− − =

− − =− + =

− = + == = −

= = −

48. ( ) 25 4 f x x x= − −

2

2

5 4 4

5 0

(5 1) 0

0 or 5 1 0

0 or 5 1

10 or

5

x x

x x

x x

x x

x x

x x

− − = −

− =− =

= − == =

= =


30/34


179

49. Since2 9 7

2 2

− += the x-coordinate of the vertex

must be7

2.


the function:

( )22(0) 8 0 5 5 y = − + + =



and solve for x:

( )

2

2

5 2 8 5

0 2 8

0 2 4

2 0 or 4 0

0 or 4

x x

x x

x x

x x

x x

= − + +

= − +

= − −

− = − =

= =


( )4,5 . Since0 4

22




( )

22(2) 8 2 5 13 y = − + + = . So the vertex is

( )2,13 .

51. The equation is in vertex form. To sketch the graph

of

2 29 9 y x x= − = − + , which has a vertex of

( )0,9 , translate the graph of 2 y x= − up 9 units.


( )

20 3 6

0 3 2

3 0 or 2 0

0 or 2

x x

x x

x x

x x

= −

= −

= − =

= =




12

+= , the x-


the function and solve for y: ( ) ( )2

3 1 6 1 3 y = − = − .

So, the vertex is ( )1, 3− .

53. First, change the equation to standard form:2 2

2

1.7 2.6 6.7 10 2.1

4.1 11.7 2.1

x x y x x

y x x

+ + = − +

= − +


the function:

( ) ( )2

4.1 0 11.7 0 2.1 2.1 y = − + =

The y-intercept is ( )0,2.1 . Next find the symmetric

point to ( )0,2.1 . Substitute 2.1 for y in the

function and solve for x:


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180

( )

2

2

2.1 4.1 11.7 2.1

0 4.1 11.7

0 4.1 11.7

0 or 4.1 11.7 0

0 or 4.1 11.7

11.70 or 2.85

4.1

x x

x x

x x

x x

x x

x x

= − +

= −

= −

= − =

= =

= = ≈

Therefore the symmetric points are ( )0,2.1 and

( )2.85,2.1 . Since0 2.85

1.432

+≈ , the x-coordinate

of the vertex is 1.43 . To find the y-coordinate of

the vertex, substitute 1.43 for x and solve for y:

( ) ( )2

4.1 1.43 11.7 1.43 2.1 6.25 y = − + ≈ − . So the

vertex is ( )1.43, 6.25− .

54. Answers may vary. Example: 2( 5) y x= −


2( 5)( 6) 8 12 y x x x x= − − = − +

56. a. Since the function is in the form2( )h t at bt c= + + and a = -16 < 0, the vertex

is the maximum point. Find the h(t )-coordinate

of the vertex.

( ) ( ) ( )2

0 16 0 100 0 3 3h = − + + =

So, the h-intercept is (0, 3). Next, find thesymmetric point by substituting 3 for h(t ) in

the function and solve for t :

( )

2

2

3 16 100 3

0 16 100

0 4 4 25

4 0 or 4 25 0

0 or 4 25

250 or 6.25

4

t t

t t

t t

t t

t t

t t

= − + +

= − +

= − −

− = − =

= =

= = =

The symmetric points are (0, 3) and (6.25, 3).

Since the average of the t -coordinates is

0 6.253.125

2

+= , the t -coordinate of the

vertex is 3.125. Substitute 3.125 for t in the

function to find the h-coordinate of the vertex:

( ) ( ) ( )2

3.125 16 3.125 100 3.125 3

159.25

h = − + +

=

So the vertex is (3.125, 159.25), which means

that the maximum height of the ball is 159.25

feet and it is reached in 3.125 seconds.

b. Solve for t when h(t ) = 3. From part a, we see

that when h(t ) = 3, t is either 0 or 6.25. In this

case, the fielder had 6.25 seconds to get into

position.

c.

61. The function f is in the form 2( ) f t at bt c= + +

where a = 0.0027 > 0, so the vertex is the minimumpoint of the parabola where the median age is at a

minimum. Find the vertex. Begin by finding the f -

intercept: let t = 0,

( ) ( ) ( )2

0 0.0027 0 0.31 0 29.29 29.29 f = − + = .


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SSM: Intermediate Algebra Chapter 6 Test

181

The f -intercept is (0, 29.29). Next, find the

symmetric point to (0, 29.29). Substitute 29.29 for

f (t ) in the function and solve for t :

( )

2

2

29.29 0.0027 0.31 29.29

0 0.0027 0.31

0 0.0027 0.31

0 or 0.0027 0.31 0

0 or 0.0027 0.31

0.310 or 114.81

0.0027

t t

t t

t t

t t

t t

t t

= − + =

= −

= −

= − =

= =

= = ≈


symmetric point is (114.81, 29.29). Since the



the vertex is 57.41. Compute f (57.41) to find the f -coordinate of the vertex:

( ) ( ) ( )

257.41 0.0027 57.41 0.31 57.41 29.29

20.39

f = − +

=

So the vertex is (57.41, 20.39) which means that

the median age for women at their first marriage

was at a minimum of 20.4 years old in 1957 (t •

57.41).

Chapter 6 Test

1.

2. Since the vertex lies on the x-axis when x > 0, h > 0

and k = 0. Since the parabola is turned upward (has

a minimum point), a > 0.

3. Answers may vary. Example: 2( 2) 7 y x= − +

4.22( 2)( 7) 2( 7 2 14) x x x x x− − + = − + − −

2

2

2( 5 14)

2 10 28

x x

x x

= − + −

= − − +

5. 2 2(3 7)(3 7) (3 ) 49 9 49 x x x x− + = − = −

6. 22 (5 8) 2 (5 8)(5 8) x x x x x+ = + +

2

2

3 2

2 [(5 ) 40 40 64]2 (25 80 64)

50 160 128

x x x x x x x

x x x

= + + +

= + +

= + +

7. 2 2(2 3)( 2 1) x x x x− + + −

4 3 2 3 2 2

4 3 2

2 4 2 2 3 6 3

2 3 7 3

x x x x x x x x

x x x x

= + − − − + + + −

= + − + −

8. 2( ) 2( 6) 11 2( 6)( 6) 11 f x x x x= − + + = − + + +

2

2

2

2

2( 6 6 36) 11

2( 12 36) 11

2 24 72 11

2 24 61

x x x

x x

x x

x x

= − + + + +

= − + + +

= − − − +

= − − −

9. 28 13 6 ( 2)(8 3) x x x x+ − = + −

10. 3 2 22 12 18 2 ( 6 9) x x x x x x− + = − +

2

2 ( 3)( 3)

2 ( 3)

x x x

x x

= − −

= −

11. 2 3 10 0 x x− − =

( 5)( 2) 0

5 0 or 2 0

2 or 2

x x

x x

x x

− + =

− = + =

= = −

12. (2 7)( 3) 10 x x− − =

2

2

2

2 6 7 21 10

2 13 21 10

2 13 11 0

(2 11)( 1) 0

2 11 0 or 1 0

2 11 or 1

11 or 1

2

x x x

x x

x x

x x

x x

x x

x x

− − + =

− + =

− + =

− − =

− = − =

= =

= =

13. 225 16 0 x − =


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Chapter 6 Test SSM : Intermediate Algebra

182

( )

( )( )

22(5 ) 4 0

5 4 5 4 0

5 4 0 or 5 4 0

5 4 or 5 44 4

or5 5

x

x x

x x

x x

x x

− =

− + =

− = + =

= = −

= = −

14. 3 22 3 18 27 x x x+ = +

3 2

2

2

2 2

2 3 18 27 0

(2 3) 9(2 3) 0

(2 3)( 9) 0

(2 3)( 3 ) 0

(2 3)( 3)( 3) 0

2 3 0 or 3 0 or 3 02 3 or 3 or 3

3 or 3 or 3

2

x x x

x x x

x x

x x

x x x

x x x x x x

x x x

+ − − =

+ − + =

+ − =

+ − =

+ − + =

+ = − = + =

= − = = −

= − = = −

15. Since the vertex of the parabola is (4, 7), the

standard form of its equation is

2( 4) 7 y a x= − +

The point (1, 3) satisfies this equation so we can

substitute 1 for x and 3 for y to solve for a:

2

2

3 (1 4) 7

3 ( 3) 7

3 9 7

4 9

4

9

a

a

a

a

a

= − +

= − +

= +

− =

= −

So the equation of the parabola is

24 ( 4) 79

y x= − − +

Any point that satisfies this equation will lie on the

parabola. One such point is (7, 3), which is the

symmetric point to (1, 3).

16. a. To find the x-intercept, solve for x when

( ) 0 f x =

20 5 24

0 ( 8)( 3)

8 0 or 3 0

8 or 3

x x

x x

x x

x x

= − −

= − +

− = + =

= = −

b. Since the x-intercepts are symmetric points,

the average of the x-coordinates for these

points is the x-coordinate of the vertex, which

is8 ( 3)

2.52

+ −= . Substitute 2.5 for x in the

function to find the y-coordinate of the vertex:

2(2.5) 5(2.5) 24 30.25 y = − − = −

So, the vertex is (2.5, -30.25).

17. ( ) ( ) ( )2

0 3 0 10 0 8 8 f = + − = −

18. ( ) ( ) ( )2

3 3 3 10 3 8 11 f − = − + − − = −

19. 20 3 10 8 x x= + −

0 (3 2)( 4)

3 2 0 or 4 0

3 2 or 4

2 or 4

3

x x

x x

x x

x x

= − +

− = + =

= = −

= = −

20. 21 3 10 8 x x= + −

2

2

5 3 10 8

0 3 10 13

0 (3 13)( 1)

3 13 0 or 1 0

3 13 or 1

13 or 1

3

x x

x x

x x

x x

x x

x x

= + −

= + −

= + −

+ = − =

= =

= =

21. Since (0, 0.75) lies on the curve, (0) 0.75 f = .

22. Since (-3, 0) lies on the curve, ( 3) 0 f − = .

23. This is not true for any value of x.

24. Since (1, 1) lies on the curve, x = 1 when ( ) 1 f x = .

25. Since (-3, -3) and (5, -3) lie on the curve, x = 3 and

x = 5 when ( ) 3 f x = −


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SSM: Intermediate Algebra Chapter 6 Test

2

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Lehmann IA SSM Ch6