CHAPTER 2 Modeling with Linear and Quadratic...

38 CHAPTER 2 MODELING WITH LINEAR AND QUADRATIC FUNCTIONS

CHAPTER 2 Modeling with Linear and Quadratic Functions Section 2-1

2. Answers will vary. 4. Answers will vary. 6. Answers will vary.

8. Using the points (-3, -3) and (1, 3),

Rise = 3 - (-3) = 6; Run = 1 - (-3) = 4; Slope =

6

4 =

3

2

y - 3 =

3

2(x - 1)

y - 3 =

3

2x -

3

2

y =

3

2x +

3

2

3x - 2y = -3

10. Using the points (0, -3) and (5, 3),

Rise = 3 - (-3) = 6; Run = 5 - 0 = 5; Slope =

6

5

y - 3 =

6

5(x - 5)

y - 3 =

6

5x - 6

y =

6

5x - 3

6x - 5y = 15

12. Using the points (-5, 3) and (-1, -2),

Rise = -2 - 3 = -5; Run = -1 - (-5) = 4; Slope = -

5

4

y - (-2) = -

5

4(x - (-1))

y + 2 = -

5

4x -

5

4

y = -

5

4x -

13

4

5x + 4y = -13

14. From the graph, x-intercept = 1, y-intercept = 1, slope = -1 y = -x + 1

16. From the graph, x-intercept = -1, y-intercept = -3, slope = -3 y = -3x - 3

18. From the graph, x-intercept = 4, y-intercept = -2, slope =

2

4 =

1

2

y =

1

2x - 2

20. y = 5 - 3x3; not linear because of the cubic term.

22. y =

3 ! x

2; linear--can be rewritten as y = -

1

2x +

3

2

24. y = -

1

5(2 - 3x) +

2

7(x + 8); linear--can be rewritten as y =

31

35x +

66

35.

26. y =

4

3(2 - x) +

2

3(x + 2); linear--can be rewritten as y = -

2

3x + 4.

28. y =

2

3 ! x; not linear because x is in the denominator.

SECTION 2-1 39

30. y = -

3

2x + 6

x intercept: 0 = -

3

2x + 6 y intercept: y = -

3

2(0) + 6

0 = -3x + 12 y = 6 3x = 12 x = 4

slope = m = -

3

2

10–10

10

–10

y

x

32. y =

2

3x - 3

x intercept: 0 =

2

3x - 3 y intercept: y =

2

3(0) - 3

0 = 2x - 9 y = -3 -2x = -9 x =

9

2

slope = m =

2

3

5–5

5

–5

y

x

34. 4x + 3y = 24 x intercept: 4x + 3(0) = 24 y intercept: 4(0) + 3y = 24

4x = 24 3y = 24 x = 6 y = 8

slope: 3y = -4x + 24 y = -

4

3x + 8

m = -

4

3

10–10

10

–10

y

x

36.

y

6 -

x

5 = 1

x intercept:

0

6 -

x

5 = 1 y intercept:

y

6 -

0

5 = 1

-

1

5x = 1

1

6y = 1

x = -5 y = 6

slope:

y

6=

x

5 + 1

y =

6

5x + 6

m =

6

5

10–10

10

–10

y

x

38. y = -2, horizontal line

x intercept: none y intercept: -2

slope = 0

5–5

5

–5

y

x

40. x = 2.5, vertical line

x intercept: 2.5 y intercept: none

undefined slope 5–5

5

–5

y

x


42. x = -7 44. y = 12 46. y = -2 48. x = -3

50. m = -1, b = 7: y = mx + b y = -x + 7 x + y = 7

52. m =

5

3, b = 6:

y = mx + b y =

5

3x + 6

3y = 5x + 18 -5x + 3y = 18

54. y – 0 = 2(x – 2) y = 2x – 4

56.

!

y " ("2) =1

2x " ("4)( )

!

y + 2 =1

2x + 4( )

y =1

2x

58.

!

m =1 " 4

6 " ("3)="3

9= "

1

3

!

y " 1 = "1

3x " 6( )

y " 1 = "1

3x + 2

y = "1

3x + 3

60.

!

m =5 " ("1)

10 " 2=6

8=3

4

!

y " 5 =3

4x " 10( )

y " 5 =3

4x "

15

2

y =3

4x "

5

2

62.

!

m =rise

run="5

4

!

y = mx + b

y = "5

4x " 5

64. m = -2 y – 0 = -2(x – (-4)) y = -2x - 8

66. m =

!

"3

2

!

y " ("4) = "3

2x " ("2)( )

y + 4 = "3

2x + 2( )

y = "3

2x " 7

68.

!

3x + 4y = 8

!

y = "3

4x + 2

!

m = "3

4

!

y " 5 = "3

4x " 3( )

y " 5 = "3

4x +

9

4

y = "3

4x +

29

4

70. 4x + 5y = 0

!

y = "4

5x; m = "

4

5

!

m" =5

4

!

y " 4 =5

4x " ("2)( )

y " 4 =5

4x +

5

2

y =5

4x +

13

2

72. y = -

1

2x + b

m = -

1

2

y-int = b Lines will be parallel.

74. (A) m =

5 ! (!3)

10 ! (!2) =

8

12 =

2

3

y - 5 =

2

3(x - 10)

y - 5 =

2

3x -

20

3

y =

2

3x -

5

3

(B) m =

10 ! (!2)

5 ! (!3) =

12

8 =

3

2

y - 10 =

3

2(x - 5)

y - 10 =

3

2x -

15

2

y =

3

2x +

5

2

(C)

x

y

10

10

–10

–10

The functions are inverses.

SECTION 2-1 41

76.

!

f(1) =1

2; f(3) =

9

2

!

m =9 2 " 1 2

3 " 1= 2

78.

!

f(1) =1

2; f

3

2

"

# $ %

& ' =

9

8

!

m =9 8 " 1 2

3 2 " 1=5 8

1 2=5

4

80. The estimated slope of the tangent line is 1.

82. (A) 3x + 4y = 12 4x - 3y = 12 4y = -3x + 12 -3y = -4x + 12 y = -

3

4x + 3 y =

4

3x - 4

–10

10

15.2–15.2

(B) 2x + 3y = 12 3x - 2y = 12 3y = -2x + 12 -2y = -3x + 12 y = -

2

3x + 4 y =

3

2x - 6

–10

10

15.2–15.2

(C) The lines are perpendicular to each other.

(D) Ax + By = C Bx - Ay = C By = -Ax + C -Ay = -Bx + C y = -

A

Bx +

C

B y =

B

Ax -

C

A

m1 = -

A

B m2 =

B

A

m1 · m2 = -

A

B ·

B

A = -1

84. The two points are (x1, mx1 + b) and (x2, mx2 + b)

slope =

!

mx2 + b " mx1 + b( )x2 " x1

=m x2 " x1( )x2 " x1

= m

86. (A) Rate of change of price = slope = 0.23 dollars per day; going up

(B) August 15: P(0) = $66.45; December 1: P(108) = $91.29 The price went up by $24.84 over that span, an average rate of $0.23 per day.

88. (A) Rate of change of temperature = slope = 0.0183 degrees per year; going up

(B) 1965: T(0) = 13.86 degrees Celsius; 2001: T(36) = 14.52 degrees Celsius The temperature went up by 0.66 degrees over that span, an average rate of 0.0183 degrees per day.

90. m = -235; b = 700; d = -235h + 700; d(2.5) = 112.5; the evacuation will be done before the tsunami reaches Japan.

92. (w, s): (5, 2), (0, 0), m =

2

5 = 0.4

(A) s - s1 = m(w - w1) (B) (w, 3.6) (C) slope = 0.4 s - 0 = 0.4(w - 0) 3.6 = 0.4w s = 0.4w w = 9 lbs


94. (C, R): (20, 33), (60, 93)

(A) (60, 93)

(20, 33)

R = retail

C = cost

m =

93 ! 33

60 ! 20 = 1.5

R - 33 = 1.5(C - 20) R - 33 = 1.5C - 30 R = 1.5C + 3, (C > 10)

(B) If R = 240: 240 = 1.5C + 3 (C) slope = 1.5; the retail price 237 = 1.5C rises $1.50 for each $1 in cost. C = $158

96. (A) Using the points (50, 4,174) and (60, 4,634),

m =

4,634 ! 4,174

60 ! 50 =

460

10 = 46

y - 4,174 = 46(x - 50) y - 4,174 = 46x - 2,300 C(x) = 46x + 1,874

(B) The daily fixed costs are $1,874 and the variable costs are $46 per racket.

98. (A) (t, N): (0, 4.76), (90, 2.5) where t = 0 at 1900.

(0, 4.76)

(90, 2.5)

N

t

slope =

4.76 ! 2.5

0 ! 90 = -

2.26

90 = -0.0251

N - 4.76 = -0.0251 (t - 0) N = -0.0251t + 4.76 (t ≥ 0)

(B) (115, N): N = -0.0251 (100) + 4.76 N ≈ 1.87 people per household

100. (A) T = 200 + 0.02(200)A where A = altitude in thousands of feet. T = 4A + 200 (A ≥ 0)

(B) T = 4(6.5) + 200 T = 226 mph

(C) slope = 4 which indicates that true air speed increases 4 mph for each one thousand foot increase in altitude.

Section 2-2

2. Answers will vary. 4. Answers will vary.


10. V(x) = 0 : {a, d} 12. u(x) - v(x) = 0 u(x) = v(x) : {b, e}

14. 4(x - 1) - 2(x + 2) = 2x + 7 4x - 4 - 2x - 4 = 2x + 7 2x - 8 = 2x + 7 Contradiction

16. 4(2 - x) + 2(x - 3) = 5x + 2 8 - 4x + 2x - 6 = 5x + 2 2 - 2x = 5x + 2 0 = 7x x = 0 Conditional Equation

SECTION 2-2 43

18. 2(x + 1) + 3(2 - x) = 8 - x 2x + 2 + 6 - 3x = 8 - x -x + 8 = 8 - x Identity

20. 11 + 3y = 5y – 5 -2y = -16 y = 8

22. 5x + 10(x - 2) = 40 5x + 10x - 20 = 40 15x = 60 x = 4

24. 5w - (7w - 4) - 2 = 5 - (3w + 2) 5w - 7w + 4 - 2 = 5 - 3w - 2 -2w + 2 = 3 - 3w w = 1

26.

!

2x

3+1

2=3x

2" 2

#

$ %

&

' ( ) 6

4x + 3 = 9x – 12 -5x = -15 x = 3

28. 8x – 5(2 + x) = 2 + 3(x – 4) 8x - 10 – 5x = 2 + 3x – 12 3x – 10 = 3x – 10 Solution is all real numbers.

30.

x + 3

4 -

x ! 4

2 =

3

8

2(x + 3) - 4(x - 4) = 3 2x + 6 - 4x + 16 = 3 -2x = -19 x =

19

2

32.

9

w - 3 =

2

w

9 - 3w = 2 -3w = -7 w =

7

3

34.

2

3x

!

" # +

1

2 =

4

x +

4

3

!

" # 6x

4 + 3x = 24 + 8x -5x = 20 x = -4

36. (x - 2)(x - 4) = (x - 1)(x - 5) x2 - 6x + 8 = x2 - 6x + 5 No solution

38. (x - 2)(x + 4) = (x + 3)(x - 5) x2 + 2x - 8 = x2 - 2x - 15 4x = -7 x = -

7

4

40. (x + 3)2 = (x + 2)(x - 4) x2 + 6x + 9 = x2 - 2x - 8 8x = -17 x = -

17

8

42.

2x

x + 4 = 2 -

8

x + 4

2x = 2x + 8 - 8 2x = 2x Solution is all real numbers except -4.

44.

2x

x + 4 = 7 -

6

x + 4

2x = 7x + 28 - 6 -5x = 22 x = -

22

5

46.

2x

x + 4 = 7 -

8

x + 4

2x = 7x + 28 - 8 -5x = 20 x = -4 (undefined) No solution

48. D = L + 2W + 2H D – L – 2W = 2H

!

H =D " L " 2W

2=1

2D "

1

2L " W

50. F =

9

5C + 32

9

5C = F - 32

C =

5

9(F - 32)

52.

1

R =

1

R1

+

1

R2

for R1

RR1R2

1

R

! " # $

% & = RR1R2

1

R1

!

" # #

$

% & & + RR1R2

1

R2

!

" # #

$

% & &

R1R2 = RR2 + RR1 R1R2 - RR1 = RR2 R1(R2 - R) = RR2

R1 =

RR2

R2! R

1


54. A = 2ab + 2ac + 2bc 2ac + 2bc = A - 2ab c(2a + 2b) = A - 2ab c =

A ! 2ab

2a + 2b

56. x =

3y + 2

y ! 3

x(y - 3) = 3y + 2 xy - 3x = 3y + 2 xy - 3y = 3x + 2 y(x - 3) = 3x + 2 y =

3x + 2

x ! 3

58. y1 = |x|, y2 =

!

x2 Graph y1 and y2. The graphs are identical for all values of x. This tells us

that the equation

!

x2 = |x| is an identity.

60. w = 3l 62. l =

1

3w 64. w = l - 5 66. w = 10 + l

68. (A) Enter L1 = -1, 3; L2 = 0, 5. y = 1.25x + 1.25

(B) Enter L1 = 0, 5; L2 = -1, 3. y = 0.8x - 1 The graphs are symmetric with respect to the line y = x. The functions are inverses.

70.

x ! 1x

x + 1 ! 2x

= 1

x -

1

x = x + 1 -

2

x

-

1

x = 1 -

2

x

-1 = x - 2 1 = x This makes the original equation undefined, so no solution.

72.

x ! 1 ! 2x

1 ! 2x

= x

x - 1 -

2

x = x - 2

-1 -

2

x = -2

-x - 2 = -2x x = 2 This makes the original equation undefined, so no solution.

74. Write the four integers as x, x + 1, x + 2, and x + 3. x + (x + 1) + (x + 2) + (x + 3) = 182 4x + 6 = 182 4x = 176 x = 44 The four integers are 44, 45, 46, and 47.

76. Write the three even integers as x, x + 2, and x + 4. x + 2(x + 2) = 2(x + 4) 3x + 4 = 2x + 8 x = 4 The integers are 4, 6, and 8.

78. P = 2l + 2w

l =

1

2w

P = 2

1

2w

! " # $

% & + 2w = 60

3w = 60 w = 20 l = 10 10 inches by 20 inches

SECTION 2-2 45

80. C(x) = 1,200 + 45x 4,800 = 1,200 + 45x 3,600 = 45x 80 = x 80 tables can be produced.

82. (A) earnings = base salary + commission 3,170 = 1,175 + 0.05x 0.05x = 1,995 x = $39,900 in sales

(B) 2,150 + 0.08(x – 7,000) = 1,175 + 0.05x 2,150 + 0.08x – 560 = 1,175 + 0.05x 0.03x = -415

x = -13,833.-3 and since sales cannot be negative,

earnings will never be the same. Take the first payment method.

84. With current:

!

0.5 = 11.3 + x( ) " t1;

!

t1 =0.5

11.3 + x

Against current:

!

0.5 = 11.3 " x( ) # t2;

!

t2 =0.5

11.3 " x

t1 = 0.8t2;

!

0.5

11.3 + x = 0.8

!

0.5

11.3 " x

!

0.5

11.3 + x =

!

0.4

11.3 " x

5.65 – 0.5x = 4.52 + 0.4x -0.9x = -1.13 x ≈ 1.26 miles per hour

86. d = 8 · t1 for trip upstream (1) d = 12 · t2 for return trip (2)

from (1) t1 =

d

8 and from (2) t2 =

d

12

and since t1 + t2 = 5,

d

8 +

d

12 = 5

12d + 8d = 480 20d = 480 d = 24 miles from resort at turnaround

t1 =

24

8 = 3 hr which gives 10:00 a.m.

as turnaround time.

88. d = rt = 5,000t underwater d = 1,100(t + 39) above water 5,000t = 1,100(t + 39) 5,000t = 1,100t + 42,900 3,900t = 42,900 t = 11 d = 5,000(11) d = 55,000 ft.

90. 100%

30% 12 gallons

x

.03(12) + x = 0.4(x + 12) 3.6 + x = 0.4x + 4.8 0.6x = 1.2 x = 2 gallons

Adding 2 gallons of pure hydrochloric acid to a 30% mixture will produce a 40% solution.

92. 0.3%

0.9% 120,000 gallons

x

0.8(x + 120,000) = 0.9(120,000) + 0.3x 0.8x + 96,000 = 108,000 + 0.3x 0.5x = 12,000 x = 24,000 gallons

Adding 24,000 gallons of 0.3% sulfur solution to 120,000 gallons of 0.9% sulfur solution will produce a 0.8% solution.


94. (A) V = VS + (0.03A)VS = VS(1 + 0.03A) (B) V = 120(1 + 0.03(6.4))

V = 143.04 ≈ 143 mph (C) 125 = V(1 + 0.03(8.5))

V ≈ 99.6 mph

(D) 155 = 135(1 + 0.03(A)) 1 + 0.03A =

155

135

0.03A =

155

135 - 1

A =

155135 ! 1

0.03

A ≈ 4.94 The mountain resort has an altitude of 4,940 ft.

96. Enter t = 0, 1, …, 6 as

!

L1 and the “Private” column as

!

L2 and use the linear regression command.

!

y = 1,101.5x + 19,172.0 2008 is x = 9:

!

y(9) = 1,101.5(9) + 19,172.0 = $29,085.5 Finally, substitute 35,000 for y and solve for x:

!

35,000 = 1,101.5x + 19,172.0

!

15,828 = 1,101.5x

!

x = 14.4 Tuition will reach $35,000 in 2013.

98. Enter t = 0, 4, 8, 12, 16, 20, 24, 28, 32, and 36 as

!

L1, and the Men's 200m backstroke times in seconds as

!

L2. Use the linear regression command. Men: y = -0.286x + 125.334 Repeat with women's times: Women: y = -0.398x + 139.481 Set the regression equations equal: -0.286x + 125.334 = -0.398x + 139.481 .112x = 14.147 x = 126.3 In 2094, the times are predicted to be equal.

100. First enter the supply column as L1 and the first price column as L2. Use the linear regression command. y = 1.47x + 2.98 Repeat using the demand column as L1 and the second price column as L2. y = -2.39x + 11.1

Each of these equations gives price in terms of supply or demand. To find equilibrium price, we need to set supply = demand, so solve each regression equation for x, then set the results equal. y = 1.47x + 2.98 y = -2.39x + 11.1 y - 2.98 = 1.47x y - 11.1 = -2.39x 0.680y - 2.03 = x -0.418y + 4.64 = x 0.680y - 2.03 = -0.418y + 4.64 1.098y = 6.67 y = 6.07 The equilibrium price is $6.07 per bushel.

Section 2-3

2. Answers will vary. 4. Answers will vary. 6. Answers will vary.

SECTION 2-3 47

8. Vertex: (-2, -2); axis: x = -2 Hand graph:

10–10

10

–10

y

x

Graphing calculator graph:

-10 10

-10

10

10. Vertex:

!

11

2,3

"

# $

%

& ' ; axis: x =

!

11

2

Hand graph:

10–10

10

–10

y

x


-10 10

-10

10

12. Vertex: (-8, 12); axis: x = -8 Hand graph:

–8

12

y

x


-15 5

0

20

14. The graph of

!

y = x2 is shifted one unit left, reflected in the x axis and shifted two units down.

16. The graph of

!

y = x2 is shifted two units right.

18. The graph of

!

y = x2 is shifted one unit left, reflected in the x axis, and shifted four units up.

20. f(x) = (x - 2)2 + 1 22. n(x) = -(x + 1)2 + 4 24. g(x) = -(x + 1)2 - 2

26.

!

x2 + 8x + 16 = x + 4( )2 28.

!

x2 " 3x +9

4= x "

3

2

#

$ %

&

' ( 2

30.

!

x2 +11

3x +

121

36= x +

11

6

"

# $

%

& ' 2

32.

!

g(x) = x2 " 6x + 9( ) + 1 " 9

!

f(x) = x " 3( )2" 8

Vertex: (3, -8); axis: x = 3


34.

!

k(x) = " x2 + 10x( ) + 3

!

k(x) = " x2 + 10x + 25( ) + 3 + 25

!

k(x) = " x + 5( )2

+ 28 Vertex: (-5, 28); axis: x = -5

36.

!

n(x) = 3 x2 + 2x( ) " 2

!

n(x) = 3 x2 + 2x + 1( ) " 2 " 3

!

n(x) = 3 x + 1( )2" 5

Vertex: (-1, -5); axis: x = -1

38.

!

g(x) = "3

2x2 " 6x( ) +

11

2

!

g(x) = "3

2x2 " 6x + 9( ) +

11

2+27

2

!

g(x) = "3

2x " 3( )

2+ 19

Vertex: (3, 19); axis: x = 3

40.

!

g(x) = 3 x2 + 8x( ) + 30

!

g(x) = 3 x2 + 8x + 16( ) + 30 " 48

!

g(x) = 3 x + 4( )2" 18

Vertex: (-4, -18); axis: x = -4

42.

!

x = "b

2a= "

10

2= "5; f("5) = ("5)2 + 10("5) + 10 = "15

The vertex is (-5, -15). The graph is symmetric about its axis, x = -5. The graph decreases on the interval (–∞, -5] and increases on [-5, ∞), reaching a minimum at (-5, -15). The range is [-15, ∞).

44.

!

x = "b

2a= "

"11

"2= "

11

2; f "

11

2

#

$ %

&

' ( = " "

11

2

#

$ %

&

' ( 2

" 11 "11

2

#

$ %

&

' ( + 1 =

125

4

The vertex is (-11/2, 125/4). The graph is symmetric about its axis, x = -11/2. The graph increases on the interval (–∞, -11/2] and decreases on [-11/2, ∞), reaching a maximum at (-11/2, 125/4). The range is (-∞, 125/4].

46.

!

x = "b

2a= "

30

10= "3; f "3( ) = 5 "3( )

2+ 30 "3( ) " 17 = "62

The vertex is (-3, -62). The graph is symmetric about its axis, x = -3. The graph decreases on the interval (–∞, -3] and increases on [-3, ∞), reaching a minimum at (-3, -62). The range is [-62, ∞).

48.

!

x = "b

2a= "

"24

"16= "

3

2; f "

3

2

#

$ %

&

' ( = "8 "

3

2

#

$ %

&

' ( 2

" 24 "3

2

#

$ %

&

' ( + 16 = 34

The vertex is (-3/2, 34). The graph is symmetric about its axis, x = -3/2. The graph increases on the interval (–∞, -3/2] and decreases on [-3/2, ∞), reaching a maximum at (-3/2, 34). The range is (-∞, 34].

50. y = a(x - h)2 + k Vertex at (-2, -12): y = a(x + 2)2 - 12 Through (-4, 0): 0 = a(-4 + 2)2 - 12 12 = a(-2)2 a = 3 y = 3(x + 2)2 - 12 y = 3(x2 + 4x + 4) - 12 y = 3x2 + 12x

52. y = a(x - h)2 + k Vertex at (5, 8): y = a(x - 5)2 + 8 Through (0, -2): -2 = a(0 - 5)2 + 8 -10 = 25a a = -

2

5 = -0.4

y = -0.4(x - 5)2 + 8 y = -0.4(x2 - 10x + 25) + 8 y = -0.4x2 + 4x - 10 + 8 y = -0.4x2 + 4x - 2

SECTION 2-3 49

54. y = a(x - h)2 + k Vertex at (6, -40): y = a(x - 6)2 - 40 Through (3, 50): 50 = a(3 - 6)2 - 40 90 = 9a a = 10 y = 10(x - 6)2 - 40 y = 10(x2 - 12x + 36) - 40 y = 10x2 - 120x + 360 - 40 y = 10x2 - 120x + 320

56. y = a(x - h)2 + k Vertex at (-2, -4): y = a(x + 2)2 - 4 Through (-1, -1): -1 = a(-1 + 2)2 - 4 3 = a(1)2 a = 3 y = 3(x + 2)2 - 4 y = 3(x2 + 4x + 4) - 4 y = 3x2 + 12x + 8

58. y = a(x - h)2 + k Vertex at (3, 3): y = a(x - 3)2 + 3 Through (0, 0): 0 = a(0 - 3)2 + 3 -3 = a(9) a = -

1

3

y = -

1

3(x - 3)2 + 3

y = -

1

3(x2 - 6x + 9) + 3

y = -

1

3x2 + 2x - 3 + 3

y = -

1

3x2 + 2x

60. y = a(x - h)2 + k The x-coordinate of the vertex is 2 (halfway between (-1, 0) and (5, 0) by symmetry). y = a(x - 2)2 + k Plug in x = -1, y = 0: 0 = a(-1 - 2)2 + k = 9a + k k = -9a y = a(x - 2)2 - 9a Plug in x = 0, y = 5: 5 = a(0 - 2)2 - 9a = -5a a = -1, k = 9 y = -1(x - 2)2 + 9 = -x2 + 4x - 4 + 9 y = -x2 + 4x + 5

62. y = a(x - h)2 + k The x-coordinate of the vertex is -3 (halfway between (-5, 0) and (-1, 0) by symmetry). y = a(x + 3)2 + k Plug in x = -1, y = 0: 0 = a(-1 + 3)2 + k = 4a + k k = -4a y = a(x + 3)2 - 4a Plug in x = 0, y = 2.5: 2.5 = a(0 + 3)2 - 4a 2.5 = 5a a =

1

2, k = -2

y =

1

2(x + 3)2 - 2

=

1

2x2 + 3x +

9

2 - 2

y =

1

2x2 + 3x +

5

2

64. (A)

!

f(x) = a x2+b

ax

"

# $

%

& ' + c

(B)

!

1

2"b

a

#

$ %

&

' ( 2

=b2

4a2

(C)

!

f(x) = a x2+b

ax +

b2

4a2

"

# $ $

%

& ' ' + c (

b2

4a

(D)

!

f(x) = a x2 +b

2a

"

# $

%

& ' 2

+ c (b2

4a; The x

coordinate of the vertex for any

such function is

!

"b

2a.


66. Complete the square: g(x) = x2 + kx + 1

g(x) = x +

k

2

! " # $

% & 2

+ 1 -

k2

4

This is a translation of y = x2.

68. f(x) = -(x - 2)2 + k, opens downward. If k = 0, the vertex is (2, 0), an x-intercept, so there is only one x-intercept.

If k < 0, the vertex is below the x-axis, so there are no x-intercepts. If k > 0, the vertex is above the x-axis, so there will be 2 x-intercepts. For f(x) = a(x - h)2 + k, a < 0, the results will be the same.

70. (A)

10–10

5

–5

x

y

x

(B) The graph fails the horizontal line test, and is not a function.

(C) Equations of this form are parabolas that open to the right or left, rather than up or down.

72. f(x) = 9 - x2; (-2, 5), (4, -7) m =

5 ! (!7)

!2 ! 4 =

12

!6 = -2

y - y1 = m(x - x1) y - 5 = -2(x + 2) y - 5 = -2x - 4 y = -2x + 1 10–10

10

–10

x

y

x

74. f(x) = x2 + 2x - 6; (2, f(2)), (2 + h, f(2 + h))

f(2) = 4 + 4 - 6 = 2 f(2 + h) = (2 + h)2 + 2(2 + h) - 6 = 4 + 4h + h2 + 4 + 2h - 6 = h2 + 6h + 2

(A) m =

f(2) ! f(2 + h)

2 ! (2 + h)

=

2 ! (h2 + 6h + 2)

!h

=

!h2 ! 6h

!h = h + 6

(B) h = 1: 1 + 6 = 7 h = 0.1: 0.1 + 6 = 6.1 h = 0.01: 0.01 + 6 = 6.01 h = 0.001: 0.001 + 6 = 6.001

The slope appears to be approaching 6.

76. x + y = 60 y = 60 – x Product: xy = x(60 - x) = -x2 + 60x = -(x2 - 60x + 900) + 900 = -(x - 30)2 + 900 Vertex of parabola: (30, 900)

Maximum product is 900 when both numbers are 30. There is no minimum since the graph of the function opens downward.

SECTION 2-3 51

78. Find the vertex of the parabola:

!

t = ""80.5

2(6.8)= 5.9

Since t is in years, we round up to 6;

!

P(6) = 6.8(6)2 " 80.5(6) + 427.3 = 189.1

The lowest profit was $189,100 dollars.

80. (A) Find the vertex:

!

x = "b

2a= "

2.45

2("0.025)= 49; the owner should drive 49 miles per hour

(B)

!

m(49) = 30.025; m(65) = 23.625

!

30.025 migal

" 14 gal = 420.35 mi.;

!

23.625 migal

" 14 gal = 330.75 mi.

You could drive 89.6 miles further at 49 mi./hr.

82. Perimeter: x + 50 + x + 2y = 140 2y = 90 - 2x y = 45 - x

(A) Area = (x + 50)y A(x) = (x + 50)(45 - x) A(x) = -x2 - 5x + 2250 domain: 0 ≤ x ≤ 45

(B) A(x) = - x2

+ 5x +25

4

! " # $

% & + 2250 +

25

4

A(x) = - x +

5

2

! " # $

% & 2 +

9025

4

vertex: !5

2,9025

4

" # $ %

& '

Maximum area occurs at x = -2.5, which is not in the domain, so x = 0.

(C) Dimensions: 50 ft × 45 ft

84. h(t) = -16t2 + 144 0 = -16t2 + 144 16t2 = 144 t2 = 9 It hits the ground after 3 seconds. t = 3, -3

86. h(t) = -16t2 + h0, where h0 is the height of the tower. h(1.5) = 0, so -16(1.5)2 + h0 = 0 -16(2.25) + h0 = 0 h0 = 36 The tower is 36 feet high.

88. Maximum at (4.5, 324) through (0, 0). (A) d(t) = a(t - h)2 + k

d(t) = a(t - 4.5)2 + 324 (0, 0): 0 = a(0 - 4.5)2 + 324 0 = 20.25a + 324 -20.25a = 324 a = -16

d(t) = -16(t - 4.5)2 + 324 d(t) = -16(t2 - 9t + 20.25) + 324 d(t) = -16t2 + 144t - 324 + 324 d(t) = -16t2 + 144t; 0 ≤ t ≤ 9

(B) Find (x, 250): Graph y1 = -16t2 + 1.44t; y2 = 250.

The intersection is at t ≈ 2.35 sec, 6.65 sec.


90. (A) d(x) = a(x - h)2 + k Let the center of the bridge be the origin of the coordinate system. Then the vertex is at (0, 10). d(x) = ax2 + 10

(100, 60): 60 = a(10,000) + 10 50 = 10,000a a = 0.005 d(x) = 0.005x2 + 10; 10 ≤ d ≤ 60

(B) The center cable is 10 feet long. The next 2 cables occur at x = ±25: d(25) = 0.005(25)2 + 10 = 13.125 The next 2 cables occur at x = ±50; d(50) = 0.005(50)2 + 10 = 22.5 The next 2 cables occur at x = ±75: d(75) = 0.005(75)2 + 10 = 38.125 The sum of the 7 cables is 157.5 feet: 10 + 2(13.125) + 2(22.5) + 2(38.125)

92. (A) Enter the demand column as L1 and the price column as L2, then use the linear regression command. (Note that demand is in thousands, so enter 47,800 as 47.8, etc.) p = -0.168x + 12.5

(B) R(x) = p · x = -0.168x2 + 12.5x Find the vertex: x = -

b

2a=

!12.5

2(!0.168) = 37.2

p(37.2) = -0.168(37.2) + 12.5 = 6.25 A price of $6.25 will maximize revenue.

Section 2-4

2. Natural numbers, integers, rational numbers, real numbers, complex numbers


8. real, complex 10. imaginary, complex

12. imaginary, pure imaginary, complex 14.

!

i3 = i2 " i = #1 " i = #i

16.

!

11i( )2

= 112" i2 = 121 " #1( ) = #121

18.

!

"i 17( )2

= "1( )2# i2 # 17

2= 1 # "1( ) # 17 = "17 20.

!

3i( ) 8i( ) = 24i2 = 24("1) = "24

22. (3 + i) + (4 + 2i) = 3 + i + 4 + 2i = 3 + 4 + i + 2i = 7 + 3i

24. (6 - 2i) + (8 - 3i) = 6 - 2i + 8 - 3i = 6 + 8 - 2i - 3i = 14 - 5i

26. (9 + 8i) - (5 + 6i) = 9 + 8i - 5 - 6i = 9 - 5 + 8i - 6i = 4 + 2i

28. (8 - 4i) - (11 - 2i) = 8 - 4i - 11 + 2i = 8 - 11 - 4i + 2i = -3 - 2i

SECTION 2-4 53

30. 6 + (3 - 4i) = 6 + 3 - 4i = 9 - 4i

32. -2i(5 - 3i) = -10i + 6i2 = -10i - 6 = -6 - 10i

34. (-2 - 3i)(3 - 5i) = -6 + 10i - 9i + 15i2 = -6 + i - 15 = -21 + i

36. (3 + 2i)(2 - i) = 6 - 3i + 4i - 2i2 = 6 + i + 2 = 8 + i

38. (5 + 3i)(5 - 3i) = 52 - (3i)2 = 25 - 9i2 = 25 + 9 = 34 or 34 + 0i

40.

!

"2 + 8i( )2

= 4 " 32i + 64i2 = "60 " 32i

42.

1

3 ! i =

1

3 ! i ·

3 + i

3 + i

=

3 + i

9 ! i2

=

3 + i

9 + 1

=

3

10 +

1

10i

44.

2 ! i

3 + 2i =

2 ! i

3 + 2i ·

3 ! 2i

3 ! 2i

=

6 ! 7i + 2i2

9 ! 4i2

=

6 ! 7i ! 2

9 + 4

=

4 ! 7i

13

=

4

13 -

7

13i

46.

15 ! 3i

2 ! 3i =

15 ! 3i

2 ! 3i·

2 + 3i

2 + 3i

=

30 + 39i ! 9i2

4 ! 9i2

=

30 + 39i + 9

4 + 9

=

39

13 +

39

13i

= 3 + 3i

48.

!

3

!

12 =

!

36 = 6

50.

!

"3

!

12 = i

!

3·

!

12 = i

!

36 = 6i 52.

!

3

!

"12 =

!

3·i

!

12 = i

!

36 = 6i

54.

!

"3·

!

"12 = i

!

3·i

!

12 = i2

!

36 = -6

56. (3 -

!

"4 ) + (-8 +

!

"25) = (3 - i

!

4 ) + (-8 + i

!

25) = 3 - 2i - 8 + 5i = -5 + 3i

58. (-2 -

!

"36) - (4 +

!

"49) = -2 - 6i - 4 - 7i = -6 - 13i

60. (2 -

!

"1)(5 +

!

"9) = (2 - i)(5 + 3i) = 10 + i - 3i2 = 10 + i + 3 = 13 + i

62.

!

6 " "64

2 =

6 ! 8i

2

= 3 - 4i

64.

!

1

3 " "16 =

1

3 ! 4i ·

3 + 4i

3 + 4i

=

3 + 4i

9 + 16

=

3

25 +

4

25i

66.

1

3i =

1

3i ·

i

i

=

i

3i2

= -

1

3i or 0 -

1

3i


68.

2 ! i

3i =

2 ! i

3i·

i

i

=

2i ! i2

3i2

=

2i + 1

!3

= -

1

3 -

2

3i

70. (2 - i)2 + 3(2 - i) - 5 = 4 - 4i + i2 + 6 - 3i - 5 = 5 - 7i - 1 = 4 - 7i

72. g(x) = -x2 + 4x - 5 (A) If x = 2 + i then g(x) = -(2 + i)2 + 4(2 + i) - 5

= -(4 + 4i + i2) + 8 + 4i - 5 = -(3 + 4i) + 3 + 4i = 0

(B) No real zeros, no x-intercepts. Vertex at (1, -2) and parabola opens downward.

If x = 2 - i then g(x) = -(2 - i)2 + 4(2 - i) - 5 = -(4 - 4i + i2) + 8 - 4i - 5 = -(3 - 4i) + 3 - 4i = 0

74. i21 = i20·i i43 = i40·i3 i52 = (i4)13 = (i4)5·i = (i4)10·(-i) = (1)13 = (1)5·i = (1)10·(-i) = 1 = i = -i

76. 3x + (y - 2)i = (5 - 2x) + (3y - 8)i equate real and imaginary parts: 3x = 5 - 2x y - 2 = 3y -8 5x = 5 2y = 6 x = 1 y = 3

78.

(2 + x) + (y + 3)i

1 ! i = -3 + i

Multiply both sides by (1 - i): (2 + x) + (y + 3)i = (-3 + i)(1 - i) (2 + x) + (y + 3)i = -3 + 3i + i - i2 (2 + x) + (y + 3)i = -2 + 4i 2 + x = -2 y + 3 = 4 x = -4 y = 1

80. (3 - i)z + 2 = i (3 - i)z = -2 + i z =

!2 + i

3 ! i ·

3 + i

3 + i

=

!6 + i + i2

9 ! i2

=

!7 + i

10

= -0.7 + 0.1i

82. (2 - i)z + (1 - 4i) = (-1 + 3i)z + 4 + 2i 2z - iz + 1 - 4i = -z + 3iz + 4 + 2i 3z - 4iz = 3 + 6i z(3 - 4i) = 3 + 6i z =

3 + 6i

3 ! 4i ·

3 + 4i

3 + 4i

=

9 + 30i + 24i2

9 ! 16i2

=

!15 + 30i

25

= -

15

25 +

30

25i

or -0.6 + 1.2i

SECTION 2-5 55

84. Show (-3 + 2i) is a square root of 5 - 12i: (-3 + 2i)2 = 9 - 12i + 4i2 = 5 - 12i Show (3 - 2i) is a square root of 5 - 12i: (3 - 2i)2 = 9 - 12i + 4i2 = 5 - 12i

86.

1

i =

!

1

"1 =

!

1

"1 =

!

1

"1 =

!

"1 = i

The property for radicals,

!

a

b =

!

a

b, is true for real numbers, but is not true

for complex numbers; so

!

1

"1 ≠

!

1

"1.

Correctly worked:

1

i ·

i

i =

i

i2 =

i

!1 = -i

88. The sum of a complex number and its conjugate is always a real number since the imaginary parts cancel. The difference is always a pure imaginary number since the real parts cancel. The quotient is only a real number if either the real or imaginary part of the original number is zero.

90. (a + bi) - (c + di) = a + bi - c - di = (a - c) + (b - d)i

92. (u - vi)(u + vi) = u2 + v2 or (u2 + v2) + 0i

94.

a + bi

c + di =

a + bi

c + di ·

c ! di

c ! di

=

ac + (bc ! ad)i ! bdi2

c2! d

2i2

=

(ac + bd) + (bc ! ad)i

c2+ d

2

=

ac + bd

c2+ d

2 +

(bc ! ad)

c2+ d

2 i

96. i4k+1 = i4k · i1 = (i4)k · i = (1)k · i = i

98. Tn = i2 + i4 + i6 + … + i2n, n ≥ 1. If n = 2: T2 = i2 + i4 = -1 + 1 = 0 If n = 3: T3 = i2 + i4 + i6 = -1 + 1 - 1 = -1 If n = 4: T4 = i2 + i4 + i6 + i8 = -1 + 1 - 1 + 1 = 0 If n is even, Tn = 0. If n is odd, Tn = -1.

100. Theorem: The complex numbers are commutative under multiplication. statement reason

1. (a + bi)(c + di) = (ac - bd) + (ad + bc)i 1. def. of multiplication 2. = (ca - db) + (da + cb)i 2. commutative (·) 3. = (c + di)(a + bi) 3. def. of multiplication

Section 2-5

2. Answers will vary. 4. Answers will vary. 6. Answers will vary. 8.

!

x + 4( ) 3x " 10( ) = 0

!

x + 4 = 0 or 3x " 10 = 0

!

x = "4

!

3x = 10

!

x =10

3

10.

!

x2+ 6x + 8 = 0

!

x + 4( ) x + 2( ) = 0

!

x + 4 = 0 or x + 2 = 0

!

x = "4

!

x = "2


12. 3A2 = -12A 3A2 + 12A = 0 3A(A + 4) = 0 3A = 0, A + 4 = 0 A = 0 A = -4

14. 16x2 + 8x = -1 16x2 + 8x + 1 = 0 (4x + 1)(4x + 1) = 0 4x + 1 = 0 4x = -1

x = -

1

4

16. 8 - 10x = 3x2 3x2 + 10x – 8 = 0 (3x - 2)(x + 4) = 0 3x - 2 = 0, x + 4 = 0 3x = 2 x = -4 x =

2

3

18. y2 - 10y - 3 = 0 y2 - 10y + 25 = 3 + 25 (y - 5)2 = 28 y - 5 = ±

!

28 = ±2

!

7 y = 5 ± 2

!

7

20. w2 - 6w + 25 = 0 w2 - 6w + 9 = -25 + 9 (w - 3)2 = -16 w - 3 = ±4i w = 3 ± 4i

22. n2 + 8n + 34 = 0 n2 + 8n + 16 = -34 + 16 (n + 4)2 = -18 n + 4 = ±

!

"18 = ±3i

!

2 n = -4 ± 3i

!

2

24. 2u2 + 7u + 3 = 0 u2 +

7

2u +

3

2 = 0

u2 +

7

2u +

49

16 = -

3

2 +

49

16

u +

7

4

! " # $

% & 2

=

25

16

u +

7

4 = ±

5

4

u = -

7

4 ±

5

4

u = -

1

2, -3

26. 9x2 - 12x + 5 = 0 x2 -

12

9x +

5

9 = 0

x2 -

4

3x = -

5

9

x2 -

4

3x +

4

9 = -

5

9 +

4

9

x !

2

3

"

# $ %

& ' 2

= -

1

9

x -

2

3 = ±

!

"1

9

x -

2

3 = ±

1

3i

x =

2

3 ±

1

3i

28. 5t2 + 2t + 5 = 0 t2 +

2

5t + 1 = 0

t2 +

2

5t +

1

25 = -1 +

1

25

t +

1

5

!

" # $

% & 2

= -

24

25

t +

1

5 = ±

!

"24

25 = ±

!

2i 6

5

t = -

1

5 ±

!

2i 6

5

30. x2 - 6x - 3 = 0

x =

!

"("6) ± ("6)2 " 4(1)("3)

2(1)

x =

!

6 ± 48

2 =

!

6 ± 4 3

2

x = 3 ± 2

!

3

32. y2 + 3 = 2y y2 - 2y + 3 = 0

y =

!

"("2) ± ("2)2 " 4(1)(3)

2(1)

y =

!

2 ± "8

2 =

!

2 ± 2i 2

2

y = 1 ± i

!

2

SECTION 2-5 57

34. 2m2 + 3 = 6m 2m2 - 6m + 3 = 0

m =

!

"("6) ± ("6)2 " 4(2)(3)

2(2)

m =

!

6 ± 12

4 =

!

6 ± 2 3

4

m =

!

3 ± 3

2

36. 7x2 + 6x + 4 = 0

x =

!

"6 ± 62 " 4(7)(4)

2(7)

x =

!

"6 ± "76

14 =

!

"6 ± 2i 19

14

x =

!

"3 ± i 19

7

x = -

3

7 ±

!

19

7i

38.

!

"2 5 " x2( ) + 2 = 0

!

2x2 " 8 = 0

!

x =0 ± 02 " 4(2)("8)

2(2)

!

x =± 64

4= ±2

40. a = 0.4, b = -3.2, c = 6.4 b2 - 4ac = (-3.2)2 - 4(0.4)(6.4) = 0 One real zero

42. a = 3.4, b = -2.5, x = -1.5 b2 - 4ac = (-2.5)2 - 4(3.4)(-1.5) = 26.65 Two real zeros

44. a = 1.7, b = 2.4, c = 1.4 b2 - 4ac = (2.4)2 - 4(1.7)(1.4) = -3.76 Two imaginary zeros

46.

Two real zeros

48.

Two imaginary zeros

50.

One real zero

52. y2 - 10y - 3 = 0 y2 - 10y + 25 = 3 + 25 (y - 5)2 = 28 y - 5 = ±

!

28 y = 5 ± 2

!

7

Graph y1 = x2 - 10x - 3 Zeros at ≈ -0.2915, 10.2915.

54. 2d2 - 4d + 1 = 0 d2 - 2d +

1

2 = 0

d2 - 2d + 1 = -

1

2 + 1

(d - 1)2 =

1

2

d - 1 = ±

!

12

d = 1 ±

!

2

2

=

!

2 ± 2

2

Graph y1 = 2x2 - 4x + 1 Zeros at ≈ 0.2929, 1.7071

56. 3x2 + 5x - 4 = 0 x2 +

5

3x -

4

3 = 0

x2 +

5

3x +

25

36 =

4

3 +

25

36

x +

5

6

!

" # $

% & 2

=

73

36

x +

5

6 = ±

!

7336

x = -

5

6 ±

!

73

6

=

!

"5 ± 73

6

Graph y1 = 3x2 + 5x - 4 Zeros at ≈ -2.257, 0.591


58. 9x2 + 9x = 4 9x2 + 9x - 4 = 0 (3x + 4)(3x - 1) = 0 3x + 4 = 0 3x - 1 = 0 3x = -4 3x = 1

x = -

4

3 x =

1

3

Graph y1 = 9x2 + 9x, y2 = 4

Intersection at x = -

4

3,

1

3

60. x2 + 2x = 2 x2 + 2x + 1 = 2 + 1 (x + 1)2 = 3 x + 1 = ±

!

3 x = -1 ±

!

3

Graph y1 = x2 + 2x, y2 = 2 Intersection at x ≈ -2.732, 0.732

62. a2 + b2 = c2 for a:

a2 = c2 - b2 a =

!

c2 " b2

64. A = P(1 + r)2 for r:

(1 + r)2 =

A

P

1 + r =

!

A

P

r =

!

A

P - 1

66. x2 +

!

11x + 3 = 0 a = 1, b =

!

11, c = 3

x =

!

" 11 ± 11 " 4(1)(3)

2(1)

x =

!

" 11 ± "1

2 =

!

" 11 ± i

2

68. x2 -

!

5x - 5 = 0 a = 1, b = -

!

5, c = -5

x =

!

5 ± 5 " 4(1)("5)

2(1)

x =

!

5 ± 25

2 =

!

5 ± 5

2

Graph y1 = x2 -

!

5x - 5. The zeros are 3.618 and -1.382.

70. x2 + 2

!

5x + 5 = 0 a = 1, b = 2

!

5, c = 5

x =

!

"2 5 ± 20 " 4(1)(5)

2(1)

x =

!

"2 5 ± 0

2 = -

!

5

Graph y1 = x2 + 2

!

5x + 5. The only zero is -2.236.

72. 1 +

25

x2 =

9

x

x2 + 25 = 9x x2 - 9x + 25 = 0 a = 1, b = -9, c = 25

x =

!

9 ± 81 " 4(1)(25)

2(1)

x =

!

9 ± "19

2 =

!

9 ± i 19

2

74. 1 +

25

x2 =

10

x

x2 + 25 = 10x x2 - 10x + 25 = 0 (x - 5)2 = 0 x = 5 Graph y1 = 1 +

25

x2 -

10

x. The only zero

is 5.

76. 1 +

25

x2 =

11

x

x2 + 25 = 11x x2 - 11x + 25 = 0 a = 1, b = -11, c = 25

x =

!

11 ± 121 " 4(1)(25)

2(1)

x =

!

11 ± 21

2

Graph y1 = 1 +

25

x2 =

11

x. The zeros

are 3.209 and 7.791.

SECTION 2-5 59

78. 5 +

6

x ! 2 =

4

x + 2 (Multiply by (x - 2)(x + 2))

5(x2 - 4) + 6(x + 2) = 4(x - 2) 5x2 - 20 + 6x + 12 - 4x + 8 = 0 5x2 + 2x = 0 x(5x + 2) = 0 x = 0, -

2

5

Graph y1 = 5 +

6

x ! 2 =

4

x + 2.

The zeros are 0 and -0.4

80.

6

x ! 3 =

4

x + 3 - 3 (Multiply by (x - 3)(x + 3))

6(x + 3) = 4(x - 3) - 3(x2 + 9) 6x + 18 - 4x + 12 + 3x2 - 27 = 0 3x2 + 2x + 3 = 0 a = 3, b = 2, c = 3

x =

!

"2 ± 4 " 4(3)(3)

2(3)

x =

!

"2 ± "32

6 =

!

"2 ± 4i 2

6

x =

!

"1 ± 2i 2

3

82. Solving x2 - 2x + c = 0 with the quadratic formula gives x = 1 ±

!

1 " c . For 1 - c > 0 ⇔ c < 1, two distinct real roots. For 1 - c = 0 ⇔ c = 1, one real root (a double root). For 1 - c < 0 ⇔ c > 1, two complex roots (conjugates).

84. x2 - 7ix - 10 = 0

x =

!

7i ± 49i2 " 4("10)

2

=

7i ± 3i

2 =

4i

2,

10i

2 = 2i, 5i (0 + 2i, 0 + 5i)

86. x2 = 2ix - 3 x2 - 2ix + 3 = 0 (x - 3i)(x + i) = 0 x - 3i = 0, x + i = 0 x = 3i x = -i (0 + 3i) (0 - i)

88. x4 - 1 = 0 (x2 - 1)(x2 + 1) = 0 (x - 1)(x + 1)(x - i)(x + i) = 0 x - 1 = 0, x + 1 = 0, x - i = 0, x + i = 0 x = 1 x = -1 x = i x = -i

90. No. Complex roots of quadratics with real coefficients occur in conjugate pairs.

92. ax2 + bx + c = 0, r1 and r2 are the 2 roots,

x =

!

"b ± b2 " 4ac

2a

r1 + r2 =

!

"b + b2 " 4ac

2a +

!

"b " b2 " 4ac

2a

=

!

"b + b2 " 4ac " b " b2 " 4ac

2a

=

!2b

2a = -

b

a


94. (A)

!

i 4ac " b2

(B)

!

x ="b ± i 4ac " b2

2a

(C)

!

x ="b

2a+

4ac " b2

2ai, x =

"b

2a"

4ac " b2

2ai

The two solutions are conjugates.

96. Let x be the number:

x + x = x2 x2 = 2x x2 - 2x = 0 x(x - 2) = 0 x = 0, x - 2 = 0 x = 2

98. Let x be the first number; then x + 1 is the next consecutive integer.

x(x + 1) = 600 x2 + x - 600 = 0 (x - 24)(x + 25) = 0 x = 24, x = -25 (discard, not positive) x + 1 = 25 The two consecutive integers are 24 and 25.

100. Upstream: 24 = (10 - r)t ⇒ t =

24

10 ! r (1)

Downstream: 24 = (10 + r)(t - 1) ⇒ t - 1 =

24

10 + r (2)

Substituting (1) → (2):

24

10 ! r - 1 =

24

10 + r

24(10 + r) - 1(10 - r)(10 + r) = 24(10 - r) 240 + 24r - 100 + r2 = 240 - 24r r2 + 48r - 100 = 0 (r + 50)(r - 2) = 0 r = -50, (reject) r = 2 Rate of current is 2 mph.

102. No, the walkway in problem 101 requires (30 × 20) - 400 = 600 - 400 = 200 ft2.

(30 × 20) - (30 - 2x)(20 - 2x) = 160 600 - (600 - 60x - 40x + 4x2) = 160 -4x2 + 100x = 160 -4x2 + 100x - 160 = 0 x2 - 25x + 40 = 0 x = 23.28, 1.72; discard 23.28 since it is too large.

The width will be 1.72 feet.

104. If x is the width and y the length, then 2x + 2y = 1,600, and y = 800 – x. Area = xy = x(800 – x) = 140,000

!

"x2 + 800x " 140,000 = 0

!

x ="800 ± 8002 " 4("1)("140,000)

"2

!

x ="800 ± 80,000

"2= 258.6,541.4

If x = 258.6, y = 800 – 258.6 = 541.4, so the only solution is 258.6 ft. × 541.4 ft.

SECTION 2-5 61

106. A =

1

2bh =

1

2(5)(4) = 10

The small isosceles triangle at the top has base w and altitude 4 - h. It is similar to the large triangle. Using similar triangles:

4 ! h

4 =

w

5

5(4 - h) = 4w 20 - 5h = 4w -5h = 4w - 20 h = -0.8w + 4

(A) Adoor = wh A(w) = w(-0.8w + 4) = -0.8w2 + 4w, 0 ≤ w ≤ 5

(B) -0.8w2 + 4w ≥ 4.2 Graph y1 = -0.8x2 + 4x, y2 = 4.2 Intersection at x = 1.5, 3.5 The inequality is satisfied for 1.5 ≤ w ≤ 3.5.

(C) h = -0.8w + 4 ≥ 2 -0.8w + 4 ≥ 2 -0.8w ≥ -2 w ≤ 2.5

This restricts w to 0 ≤ w ≤ 2.5.

108. let x = length of straightaways and y = radius of semicircles, then (1) 2x + 2πy =

1

4 mile = 1320 ft

x + πy = 660 and (2) πy2 + 2yx = 100,000

Solving (1) for x gives x = 660 - πy and substitution of this result into (2) yields πy2 + 2y(660 - πy) = 100,000 πy2 + 1320y - 2πy2 = 100,000 -πy2 + 1320y = 100,000 πy2 - 1320y + 100,000 = 0

which may be solved using the quadratic formula to give y = 321.0102612 and y = 99.15878857 y = 321 is rejected since it gives a negative value for x y = 99.158… gives x = 660 - πy = 348.4834783

summary: straightaways: 348 ft diameter = 2y = 198 ft

110. (A) Enter 0, 5, 10, 15, 20, 25, 30, 35, and 40 as L1, and the wine column as L2. Then use the quadratic regression command. y = -0.000214x2 + 0.011x + 0.204

(B) Plug in y = 0.22, solve for x: 0.22 = -0.000214x2 + 0.011x + 0.204 0 = -0.000214x2 + 0.011x – 0.016

x =

!

".011 ± 0.0112 " 4("0.000214)("0.016)

2("0.000214)# 1.5,50 = 0, 44.8

Consumption will return to 1960 levels in 2010.

(C) Plug in x = 45: y(45) = 0.27 gallons


112. (A) Enter 0, 5, 10, …, 50 as L1 and the consumption column as L2, then use the quadratic regression command. y = -2.02x2 + 69.1x + 3,530

(B) Plug in 500 for y, solve for x: 500 = -2.02x2 + 69.1x + 3,530 0 = -2.02x2 + 69.1x + 3,030

x =

!

"69.1 ± 69.12 " 4("2.02)(3,030)

2("2.02)= "25.2,59.4 = -22.8, 57.3

Consumption will fall to 500 in 2009.

(C) Plug in x = 55: y(55) = 1,220

114. (A) Enter the speed column as L1 and the Auto B column as L2, then use the quadratic regression command. y = 0.0562x2 - 0.490x + 15.9

(B) Plug in 165 for y, solve for x: 165 = 0.0562x2 - 0.490x + 15.9 0 = 0.0562x2 - 0.490x - 149.1

x =

!

0.490 ± ("0.490)2 " 4(0.0562)("149.1)

2(0.0562) = 56.1, -0.149

To the nearest mile, the auto was traveling 56 mph.

116. (A) Enter the speed column for Boat B as L1 and the MPG column as L2, then use the quadratic regression command. y = -0.00148x2 + 0.0907x + 1.01

(B) Cost = 15t + 2.50g where t = hours and g = gallons of gas used. t =

200

speed =

200

x (letting x = average speed)

g =

200

MPG =

200

!0.00148x2+ 0.0907x + 1.01

C(x) = 15

200

x

! " # $

% & + 2.50

200

!0.00148x2 + 0.0907x + 1.01

"

# $

%

& '

C(x) =

!

3,000

x +

!

500

"0.00148x2+ 0.0907x + 1.01

Graph C(x) and use the MINIMUM command: The optimal speed is 38.1 mph. y(38.1) = 2.32 miles per gallon

t(38.1) =

!

200

38.1 = 5.25 hours

g(38.1) =

!

200

2.32 = 86.2 gallons

C(38.1) = $295

Section 2-6

2.

!

25 = ±5. False:

!

25 = 5 since it is the principal square root.

4. False;

!

2x " 1( )2

= 2x " 1( ) 2x " 1( ) = 4x2 " 4x + 1

6. (

!

x " 1)2 + 1 = x. True: (

!

x " 1)2 + 1 = x - 1 + 1 = x.

8. If x1/3 = 2, then x = 8. True, since 81/3 = (23)1/3 = 2.

SECTION 2-6 63

10.

4

7 -

3

x +

6

x2 = 0

Let u =

1

x

6u2 - 3u +

4

7 = 0

12. 7x-1 + 3x-1/2 + 2 = 0 Let u = x-1/2 7u2 + 3u + 2 = 0

14. 3x3/2 - 5x1/2 + 12 = 0 Not of quadratic type--first term's exponent is 3 times the middle term's exponent.

16. No. Explanations will vary. 18. Answers will vary.

20.

!

4 " x = 4

!

4 " x = 16

!

"x = 12

!

x = "12

22.

!

10x + 1 + 8 = 0

!

10x + 1 = "8

Left side is a positive root: no solution

24.

!

x " 34 = 2 x - 3 = 16 x = 19

26. 3 +

!

2x " 1 = 0

!

2x " 1 = -3 no solution,

!

2x " 1 ≥ 0

28. x4 - 7x2 - 18 = 0 (x2 - 9)(x2 + 2) = 0 (x - 3)(x + 3)(x2 + 2) = 0 x - 3 = 0, x + 3 = 0, x2 + 2 = 0 x = 3 x = -3 x2 = -2 x = ±i

!

2

30. x =

!

5x2 + 9 x2 = 5x2 + 9 -4x2 = 9 x2 = -

9

4

x = ±

3

2i

32.

!

2x + 3 = x2 " 12 2x + 3 = x2 – 12 0 = x2 – 2x – 15 0 = (x – 5)(x + 3) x = 5, -3

x = -3 results in imaginary outputs and cannot be checked graphically.

34. m - 13 =

!

m + 7 m2 - 26m + 169 = m + 7 m2 - 27m + 162 = 0 (m - 18)(m - 9) = 0 m = 18 m = 9, reject, does not check

, m = 18 36.

!

3w " 2 -

!

w = 2

!

3w " 2 =

!

w + 2 3w - 2 = w + 4

!

w + 4 2w - 6 = 4

!

w w - 3 = 2

!

w w2 - 6w + 9 = 4w w2 - 10w + 9 = 0 (w - 9)(w - 1) = 0 w = 9 w = 1, reject, does not check


38. x2/3 - 3x1/3 - 10 = 0 Let u = x1/3: u2 - 3u - 10 = 0 (u - 5)(u + 2) = 0 u = 5 u = -2 x1/3 = 5 x1/3 = -2 x = 53 = 125, x = -23 = -8

40. (x2 + 2x)2 - (x2 + 2x) = 6 Let u = x2 + 2x:

u2 - u = 6 u2 - u - 6 = 0 (u - 3)(u + 2) = 0 u = 3 u = -2 x2 + 2x = 3 x2 + 2x = -2 x2 + 2x - 3 = 0 x2 + 2x + 2 = 0

(x - 1)(x + 3) = 0 x =

!

"2 ± 22 " 4(1)(2)

2(1) =

!

"2 ± "4

2

x = 1 x = -3 x =

!2 ± 2i

2

x = -1 ± i

42.

!

3t + 4 +

!

t = -3, no solution.

!

3t + 4 ≥ 0 and

!

t ≥ 0 and two non-negatives can never add to give a negative.

44.

!

2x " 1 -

!

x " 4 = 2

!

2x " 1 =

!

x " 4 + 2 2x - 1 = x - 4 + 4

!

x " 4 + 4 x - 1 = 4

!

x " 4 x2 - 2x + 1 = 16(x - 4) x2 - 18x + 65 = 0 (x - 5)(x - 13) = 0 x = 5 x = 13

46.

!

3x + 6 -

!

x + 4 =

!

2

!

3x + 6 =

!

2 +

!

x + 4 3x + 6 = 2 + 2

!

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

!

2x + 8 x =

!

2x + 8 x2 = 2x + 8 x2 - 2x - 8 = 0 (x - 4)(x + 2) = 0 x = 4 x = -2, reject, does not check

SECTION 2-6 65

48. 6x -

!

4x2" 20x + 17 = 15

6x - 15 =

!

4x2 " 20x + 17 36x2 - 180x + 225 = 4x2 - 20x + 17 32x2 - 160x + 208 = 0 2x2 - 10x + 13 = 0

x =

!

10 ± 100 " 4(2)(13)

4

x =

10 ± 2i

4

x =

5

2 ±

1

2i

50. 6x-2 - 5x-1 - 6 = 0

Let u = x-1,or

1

x

! " # $

% & :

6u2 - 5u - 6 = 0 (3u + 2)(2u - 3) = 0 u = -

2

3 u =

3

2

x-1 = -

2

3 x-1 =

3

2

x = -

3

2 x =

2

3

52. 4x-4 - 17x-2 + 4 = 0

Let u = x-2,or

1

x2

!

" #

$

% & :

4u2 - 17u + 4 = 0 (4u - 1)(u - 4) = 0

u =

1

4 u = 4

1

x2 =

1

4

1

x2 = 4

x2 = 4 x2 =

1

4

x = ±2 x = ±

1

2

54. 4x-1 - 9x-1/2 + 2 = 0

Let u = x-1/2,or

!

1

x

"

# $

%

& ' :

4u2 - 9u + 2 = 0 (4u - 1)(u - 2) = 0

u =

1

4 u = 2

!

1

x =

1

4

!

1

x = 2

!

x = 4 2

!

x = 1

x = 16

!

x =

1

2

x =

1

4

56. (x - 3)4 + 3(x - 3)2 = 4 Let u = (x - 3)2

u2 + 3u - 4 = 0 (u + 4)(u - 1) = 0 u = -4 u = 1

(x - 3)2 = -4 (x - 3)2 = 1 x - 3 = ±2i x - 3 = ±1 x = 3 ± 2i, x = 4, 2


58. The left side was simplified incorrectly;

!

x2 " 16 is not equal to x – 4.

60.

!

2x + 3 -

!

x " 2 =

!

x + 1

2x + 3 - 2

!

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

2x = 2

!

2x2 " x " 6 4x2 = 8x2 - 4x - 24 x2 = 2x2 - x - 6 x2 - x - 6 = 0 (x - 3)(x + 2) = 0 x - 3 = 0, x + 2 = 0 x = 3 x = -2, reject, does not check

62. 4m-2 = 2 + m-4 m-4 - 4m-2 + 2 = 0

Let u = m-2,or

1

m2

!

" #

$

% & :

u2 - 4u + 2 = 0

u =

!

4 ± 8

2 =

!

4 ± 2 2

2 = 2 ±

!

2

1

m2 = 2 +

!

2

1

m2 = 2 -

!

2

m2 =

!

1

2 + 2 ·

!

2 " 2

2 " 2 m2 =

!

1

2 " 2 ·

!

2 + 2

2 + 2

m2 =

!

2 " 2

2 m2 =

!

2 + 2

2

m = ±

!

2 " 2

2 m = ±

!

2 + 2

2

64. y - 6 +

!

y = 0

!

y = 6 - y Squaring: y = 36 - 12y + y2 y2 - 13y + 36 = 0 (y - 9)(y - 4) = 0 y = 9 y = 4 9 is rejected since it does not check

Substitution: y - 6 +

!

y = 0 let u =

!

y , then u2 = y

u2 - 6 + u = 0 u2 + u - 6 = 0 (u - 2)(u + 3) = 0 u = 2 u = -3 y = u2 = 22 y = u2 = (-3)2 y = 4, as before y = 9, reject as before

66. x = 15 - 2

!

x Squaring: x - 15 = -2

!

x x2 - 30x + 225 = 4x x2 - 34x + 225 = 0 (x - 9)(x - 25) = 0 x = 9 x = 25, reject, does not check

SECTION 2-6 67

Substitution: x = 15 - 2

!

x ; let u =

!

x , u2 = x u2 = 15 - 2u u2 + 2u - 15 = 0 (u + 5)(u - 3) = 0 u = -5 u = 3; u2 = 9 u2 = 25 x = 9 as before x = 25, reject as before

68. 3

!

x " 1 = 0.05x + 2.9 Algebraically: 9(x - 1) = 0.0025x2 + 0.29x + 8.41

9x - 9 = 0.0025x2 + 0.29x + 8.41 0 = 0.0025x2 - 8.71x + 17.41 Using the quadratic formula,

x =

!

8.71 ± 75.69

0.005= 2, 3,482

Graphically:

x = 2, 3482

70. x-2/5 - 3x-1/5 + 1 = 0

Algebraically: Let u = x-1/5, or

!

1

x5

"

# $

%

& '

u2 - 3u + 1 = 0

u =

!

3 ± 9 " 4

2 =

!

3 ± 5

2

!

1

x5 =

!

3 ± 5

2

Graphically:

x ≈ 0.008131, 122.991869

!

x5 =

!

2

3 ± 5

x =

!

2

3 ± 5

"

# $

%

& ' 5 ≈ 0.008131, 122.991869

72. 12x

!

144 " x2

A =

1

2bh =

1

2

!

144 " x2 · x = 24

1

2x

!

144 " x2 = 24

x

!

144 " x2 = 48 x2(144 - x2) = 2.304 x4 - 144x2 + 2.304 = 0 u = x2 u2 - 144u + 2.304 = 0

u =

!

144 ± 1442 " 4(1)(2.304)

2 = 125.7, 18.3

If u = 125.7, x2 = 125.7 and x = 11.2 in. If u = 18.3, x2 = 18.3 and x = 4.3 in. The dimensions are 4.3 in × 11.2 in.


74. t1 = time for stone to reach the water x = 16t

12

t1 =

!

x

16 =

!

x

4

t2 = time for sound of splash to reach surface x = 1,100t2 (see Example 6)

t2 =

x

1,100

t1 + t2 = 2

!

x

4 +

x

1,100 = 2 u =

!

x

u2

1,100 +

u

4 - 2 = 0

u2 + 275u - 2,200 = 0

u =

!

"275 ± 2752 " 4(1)("2,200)

2(1) = 7.78, -283 (reject)

!

x = 7.78, so x = 61 feet

76. (A) Let the dimensions of one box be w (width) and l (length), then use the Pythagorean Theorem noting that the box diagonal is a radius of the circle (6 in). w2 + l2 = 36

l =

!

36 " w2

A = wl = w

!

36 " w2 , 0 < w < 6

(B) w

!

36 " w2 = 15 w2(36 - w2) = 225 w4 - 36w2 + 225 = 0 Let u = w2 u2 - 36u + 225 = 0

u =

!

36 ± 362 " 4(1)(225)

2(1) = 27.9, 8.05

If u = 27.9, w2 = 27.9, and w = 5.3 in; l = 2.8 in. If u = 8.05, w2 = 8.05, and w = 2.8 in; l = 5.3 in. The dimensions are 2.8 in by 5.3 in.

(C) Graph y1 = x

!

36 " x2 and find the maximum on 0 < x < 6. The maximum is (4.2, 18), so the maximum area is 18 in2 when the width is 4.2 in.

78. πr

!

r2+ h2 = S

πr

!

r2 + 102 = 125

π2r2(r2 + 100) = 15625 π2r4 + 100π2r2 - 15625 = 0 which may be solved using the quadratic formula r2 = 13.8994796 from which r = 3.73 cm to two decimal places.

SECTION 2-7 69

Section 2-7


6. v(x) ≥ 0 (-∞, a] ∪ [d, ∞)

8. v(x) < 0 (a, d)

10. u(x) - v(x) ≥ 0 u(x) ≥ v(x) [b, e]

12. v(x) < u(x) (b, e)

14. |w - 2| > 4 16. |z - (-2)| < 8 |z + 2| < 8

18. |c - (-4)| ≥ 7 |c + 4| ≥ 7

20. |m - 1| ≤ 6

22. The distance from t to the origin is no more than 5.

24. The distance from r to the origin is greater than 5.

26. 4x + 8 ≥ x - 1 3x ≥ -9 x ≥ -3; [-3, ∞); The intersection is at x = -3.

y1 ≥ y2 for x ≥ -3.

28. -7n ≥ 21 n ≤ -3; (-∞, -3]

The intersection is at x = -3. y1 ≥ y2 for n ≤ -3.

30. 2(1 - u) ≥ 5u 2 - 2u ≥ 5u 2 ≥ 7u

2

7 ≥ u

u ≤

2

7; !",

2

7

# $ % &

' (

32. Zeros of the left side: -10, 15

!

IntervalTest

NumberResult

x < -10 "15 positive

-10 < x < 15 0 negative

x > 15 20 positive

Solution: -10 < x < 15, or (-10, 15)

34. Zeros of left side: 0, 5/3

!

IntervalTest

NumberResult

x < 0 "1 negative

0 < x < 5/3 1 positive

x > 5/3 2 negative

Solution: 0 ≤ x ≤ 5/3, or [0, 5/3]

36.

!

t " 3 < 4

!

"4 < t " 3 < 4

!

"1 < t < 7 or (-1, 7)

38.

!

t " 3 > 4

!

t " 3 < "4 or t " 3 > 4

!

t < "1 or t > 7, or

!

"#,"1( ) $ 7,#( )

40. 2 ≤ 3m - 7 < 14 9 ≤ 3m < 21 3 ≤ m < 7; [3, 7)

42. 24 ≤

2

3(x - 5) < 36

72 ≤ 2x - 10 < 108 82 ≤ 2x < 118 41 ≤ x < 59; [41, 59)

44.

p

3 -

!

p " 2

3 ≤

p

4 - 4

4p - 6p + 12 ≤ 3p - 48 -5p ≤ -60 p ≥ 12; [12, ∞)

46.

!

x2+ x " 12 < 0

!

x + 4( ) x " 3( ) < 0 Zeros of the left side: -4, 3

!

IntervalTest

NumberResult

x < -4 "5 positive

-4 < x < 3 0 negative

x > 3 5 positive

Solution: -4 < x < 3 or (-4, 3)


48. Zeros of the left side: -5, -2

!

IntervalTest

NumberResult

x < -5 "6 positive

-5 < x < -2 "3 negative

x > "2 0 positive

Solution: x < -5 or x > -2 or (-∞, -5) ∪ (-2, ∞)

50.

!

x2" 4x # 0

Zeros of the left side: 0, 4

!

IntervalTest

NumberResult

x < 0 "1 positive

0 < x < 4 2 negative

x > 4 5 positive

Solution: 0 ≤ x ≤ 4, or [0, 4]

52. x2 + 25 < 10x x2 - 10x + 25 < 0 (x - 5)2 < 0 No solution

54. Zeros of the left side: -8, -5

!

IntervalTest

NumberResult

x < -8 "10 positive

-8 < x < -5 "6 negative

x > "5 0 positive

Solution: -8 ≤ x ≤ -5, or [-8, -5]

56. Zeros of the left side: -4.7, 1.7

!

IntervalTest

NumberResult

x < -4.7 "6 positive

-4.7 < x < 1.7 0 negative

x > 1.7 2 positive

Solution: x < -4.7 or x > 1.7, or (-∞, -4.7) ∪ (1.7, ∞)

58. -3x2 – 10x – 1 ≥ 0 Zeros of the left side: -3.2, -0.10

!

IntervalTest

NumberResult

x < -3.2 "5 negative

-3.2 < x < -0.10 "1 positive

x > "0.10 0 negative

Solution: -3.2 ≤ x ≤ -0.10, or [-3.2, -0.10]

60. |5y + 2| ≥ 8 5y + 2 ≤ -8 or 5y + 2 ≥ 8 5y ≤ -10 5y ≥ 6 y ≤ -2 y ≥

6

5

y ≥ 1.2 y ≤ -2 or y ≥ 1.2; (-∞, -2]

!

" [1.2, ∞)

62. |10 + 4s| < 6 -6 < 10 + 4s < 6 -16 < 4s < -4 -4 < s < -1; (-4, -1)

64. |0.5v - 2.5| > 1.6 0.5v - 2.5 < -1.6 or 0.5v - 2.5 > 1.6 0.5v < 0.9 0.5v > 4.1 v < 1.8 v > 8.2 v < 1.8 or v > 8.2; (-∞, 1.8)

!

" (8.2, ∞)

66. If u - v = -2, u = v + (-2), u < v since the difference is less than 0.

68. If a > 0, b > 0, and

b

a > 1, then a < b since

b

a > 1.

70. The distance from x to 5 is between 0 and 0.01. Solution:

!

4.99,5( ) " 5,5.01( )

72. The distance from x to 2c is between 0 and c. Solution:

!

c,2c( ) " 2c,3c( )

SECTION 2-7 71

74. Zeros: -10/3, 1

!

IntervalTest

NumberResult

x < -10/3 "4 negative

-10/3 < x < 1 0 positive

x > 1 2 negative

g is positive on (-10/3, 1), and negative on (–∞, –10/3) and (1, ∞).

76. Zeros: -4/3, 1/2

!

IntervalTest

NumberResult

x < -4/3 "3 negative

-7/3 < x < 1/2 0 positive

x > 1 / 2 2 negative

k is positive on (-4/3, 1/2), and negative on (–∞, -4/3) and (1/2, ∞).

78. Answers may vary. One example is x2 < 0, whose solution set is the empty set.

80. |V - 6.94| < 0.02 -0.02 < V - 6.94 < 0.02 6.92 < V < 6.96 (6.92, 6.96)

82. C(x) = 550,000 + 120x ; R(x) = 140x (A) To make a profit, revenue > cost 140x > 550,000 + 120x 20x > 550,000 x > 27,500

(B) To break-even, revenue = cost 140x = 550,000 + 120x x = 27,500

(C) Answers may vary; but to make a profit, the number sold must be greater than the number to break-even.

84. C(x) = 660,000 + 120x; R(x) = 140x (A) Answers may vary; but the company could increase the price of each unit, or increase the number of units sold.

(B) 140x > 660,000 + 120x 20x > 660,000 x > 33,000

(C) Refer to #82, x = 27,500: 27,500p = 660,000 + 120(27,500) 27,500p = 3,960,000 p = $144, new price Raise the wholesale price from $140 to $144 ($4).

86. P(x) = R(x) - C(x) = (10x - 0.05x2) - (200 + 2.25x) = -0.05x2 + 7.75x - 200

P(x) ≥ 60: -0.05x2 + 7.75x - 200 ≥ 60

Graph y1 = P(x), y2 = 60. Intersection at ≈ 49.105, 105.89 Production levels: 50 ≤ x ≤ 106

88. C =

5

9(F - 32), 20 < C < 30

20 <

5

9(F - 32) < 30

36 < F - 32 < 54 68°F < F < 86°F

90. (A) The vertex is (7,784), so d(t) = a(t - 7)2 + 784. d(0) = 0, so d(0) = a(0 - 7)2 + 784 = 49a + 784 = 0 49a = -784 a = -16

d(t) = -16(t - 7)2 + 784 = -16(t2 - 14t + 49) + 784 = -16t2 + 224t - 784 + 784 d(t) = -16t2 + 224t


(B) -16t2 + 224t < 640 -16t2 + 224t - 640 < 0 t2 - 14t + 40 > 0 (t - 10)(t - 4) > 0

Zeros of the left side: 4, 10

!

IntervalTest

NumberResult

t < 4 0 positive

4 < t < 10 6 negative

t > 10 11 positive

Solution to inequality: t < 4 or t > 10.

Note that d(0) = d(14) = 0, so the object is below 640 feet when 0 ≤ t < 4 or 10 < t ≤ 14.

92. x = depth, T = temp (1, 35) and (2, 71) are on the graph since temperature increases 36° over ten 100m intervals.

m =

71 ! 35

2 ! 1 = 36

T - 35 = 36(x - 1) T = 36x - 1 100 < 36x - 1 < 150 101 < 36x < 151 2.806 < x < 4.194 Between 2.806 and 4.194 kilometers

94. Enter the HGT-Minneapolis column as L1 and the temperature column as L2, then use the linear regression command. y = -0.00713x + 21.9 -20 < -0.00713x + 21.9 < -40 -41.9 < -0.00713x < -61.9 5,880 > x > 8,680 Altitudes between 5,880 and 8,680 feet

96. Enter the HGT-Minneapolis column as L1 and the air pressure column as L2, then use the linear regression command. y = -0.0615x + 876 350 < -0.0615x + 876 < 650 -526 < -0.0615x < -226 8,550 > x > 3,675 Altitudes between 3,675 and 8,550 feet

98. (A) Enter the Grapefruit Juice Demand column as L1 and the price column as L2, then use the linear regression command. p(x) = -0.000352x + 3.00 price can't be negative, so -0.000352x + 3.00 > 0 x < 8,520 The domain is 0 ≤ x ≤ 8,520.

(B) R(x) = p(x) · x R(x) = -0.000352x2 + 3.00x 0 ≤ x ≤ 8,250 C(x) = 3,000 + 0.40x x ≥ 0 Fixed cost + $0.40 per gallon

SECTION 2-7 73

(C) Profit = P(x) = R(x) - C(x) = -0.000352x2 + 3.00x - (3,000 + 0.40x) P(x) = -0.000352x2 + 2.6x - 3,000

The company breaks even if sales are 1,430 or 5,960 gallons. Any sales between

these two levels result in a profit, and sales less than 1,430 gallons or greater than 5,960 gallons result in a loss.

(D) The maximum profit is $1,800 when sales are 3,690 gallons.

CHAPTER 2 Modeling with Linear and Quadratic...

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