Inverse Variation

ALGEBRA 2 LESSON 9-1ALGEBRA 2 LESSON 9-1

(For help, go to Lesson 2-3.)

Inverse VariationInverse Variation

In Exercises 1–3, y varies directly with x.

1. Given that x = 2 when y = 4, find y when x = 5.



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Solutions

1. y = kx with x = 2 and y = 4: 4 = k(2), so k = 2y = 2x with x = 5: y = 2(5) = 10

2. y = kx with x = 1 and y = 5: 5 = k(1), so k = 5y = 5x with x = 3: y = 5(3) = 15

3. y = kx with x = 10 and y = 3: 3 = k(10), so k = 0.3y = 0.3x with x = 4: y = 0.3(4) = 1.2

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Suppose that x and y vary inversely, and x = 7 when y = 4. Write the function that models the inverse variation.

28 = k Find k.

y = x and y vary inversely.kx

4 = Substitute the given values of x and y.k7

y = Use the value of k to write the function.28x

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This is neither a direct variation nor an inverse variation.

Since each y-value is –2.5 times the corresponding x-value, y varies directly with x and the constant of variation is –2.5, and the function is y = –2.5x.

The product of each pair of x- and y-values is 1.4.

Is the relationship between the variables in the table a direct

variation, an inverse variation, or neither? Write functions to model the

direct and inverse variations.



As x increases, y decreases.y varies inversely with x and

the constant of variation is 1.4.

As x increases, y decreases, but this is not an inverse variation.

As x increases, y decreases.

x 2 4 14

y 0.7 0.35 0.1

a.

x –2 –1.3 7

y 6 5 –4

b.

x –2 4 6

y 5 –10 –15

c.

So xy = 1.4 and the function is y = .1.4

x

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Not all the products of x and y are thesame (–2 • 6 –1.3 • 5). =/

The pressure P of a sample of gas at a constant temperature

varies inversely as the volume V. Use the data in the table to write a

function that models this inverse variation for the sample of gas. Use your

equation to estimate the pressure when the volume is 6 in.3



3 32

5 19.2

8 12

Volume(in.3)

Pressure(lb/in.2)

Relate: pressure times volume = a constant

Define: Let P = pressure of the gas (lb/in.2).Let V = volume of the gas (in.3).Let k = constant of variation (pressure times volume).

Write: P V = k

PV = 96 Substitute 96 for k.P(6) = 96 Substitute 6 for V.

P = 16A gas’s pressure that is contained in a volume of 6 in.3 is 16 lb/in.2.

For each of the three states of gas in the table, PV 96.

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The mass m of a moving object is related to its kinetic energy k

and its velocity v by the formula m = . Describe the relationship as a

combined variation.

2kv 2

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m varies directly as the kinetic energy k.m varies inversely as the square of the velocity v.

m = 2kv 2



The area of an equilateral triangle varies directly as the square

of the radius r of its circumscribed circle. The area of an equilateral triangle

for which r = 2 is 3 3 . Find the formula for the area A of an equilateral

triangle in terms of r.

A = kr2 A varies directly as the square of r.

3 3 = k(2)2 Substitute the values for A and r.

= k Solve for k.3 34

A = r 2 Substitute the value for k.3 34

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pages 481–483 Exercises

1. y =

2. y = –

3. y =

4. y = –

5. y =

6. y =

7. direct; y = 5x

8. inverse; y =

9. direct; y = 2x

10. inverse; y =

11x

1300x

1x

56x

3.6x

250x

42x

0.3x

11. inverse; y =

12. neither

13. y = ; 10

14. y = – ; –8

15. y = – ; –

16. A varies directly with the square of r.

17. A varies jointly with b and h.

18. h varies directly with A and inversely with b.

19. V varies jointly with B and h.

20. V varies jointly with h and the square of r.

21. h varies directly with V and inversely with the square of r.

1x

100x

80x

5 3x

16

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22. V varies jointly with , w, and h.

23. varies directly with V and inversely with the product of w and h.

24. z = ;

25. z = 10xy; 360

26. z = ;

27. z = ;

28. a. 14,000b. 226

29. 18

30. 3.6

31.

32. 6

5xy

209

3x2

y163

4 xy

19

14

33. 9

34. 16

35. 7200 rpm

36. F = k

37. 18

38. 10

39. 2

40. 5.4

41. 4.277

42. 3.64

43. 2.625

44. 2.5

md 2

45. 8

46. 15

47. 11.786

48. 1.857

49. 32

50.

51.

52. a. A =

b. 600 ft2; 300 ft2; 200 ft2

c. d =

53. 32

54. doubled; tripled

3 16403

300d

300 r 2

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23



55. quartered; divided by 16

56. Division by zero is undefined.

57. x1y1 = k and x2y2 = k (def. of inverse variation) x1y1 = x2y2

(transitivity) = (Divide both

sides by x2y1.)

58. Answers may vary. Sample: Quadruple the volume and leave the radius constant, halve the radius and leave the volume constant, multiply the volume by 16 and double the radius, and multiply the

volume and radius by .

x1

x2

y2

y1

14

59. BMI

60. B

61. F

62. A

63. [2] A varies directly with the square of r OR r varies directly with the square root of A.

[1] incomplete answer

64. [4] (24.4, 4.8) and (9.6, 12.2); in both pairs the product is 117.12, so k is the same for both. In the other 2 pairs k is not equal.

[3] appropriate methods, but one computational error

[2] incomplete explanation[1] answer only, with no explanation

705 wh2

9-1

4



65. 37.1

66. 3e4 163.79

67. 0.92

68. –90x x

69. 42x2 6

70. 10x2y3 2y

71. |x5|y50

72. –4ab2

73. 2m2|n| 4

74. |x|

e5

4

e2

8

9-1

3



1. Suppose x and y vary inversely, and x = 5 when y = 1.6.a. Write a function that models the inverse variation.b. Find y when x = 32.

Tell whether the relationship between the variables in each table is a direct variation, an inverse variation, or neither. Write functions to model the direct and inverse variations.

2. 3.

4.

5. Describe the combined variation modeled by the formula V = r 2h.

6. Suppose z varies directly as x and inversely as the square of y. When x = 35 and y = 7, the value of z is 50. Write the function that models the relationship and find z when x = 5 and y = 10.

x 0.1 1.5 4

y 3 2 1.6

x 48 –4 8

y 2 –24 12

x 3 7 –10

y 21 49 –70

0.25

neither

direct variation; y = 7x

V varies jointly as the square of r and h.

y = 8x

z = , 3.570xy 2

inverse variation; y =96x

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Graphing Inverse VariationsGraphing Inverse Variations

Each of the following equations is a translation of y = |x|. Describe each translation.

1. y = |x| + 2 2. y = |x + 2|

3. y = |x| – 3 4. y = |x – 3|

5. y = |x + 4| – 5 6. y = |x – 10| + 7

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Solutions



1. y = |x| + 2 is y = |x| translated 2 units up.

2. y = |x + 2| is y = |x| translated 2 units left.

3. y = |x| – 3 is y = |x| translated 3 units down.

4. y = |x – 3| is y = |x| translated 3 units right.

5. y = |x + 4| – 5 is y = |x| translated 4 units left and 5 units down.

6. y = |x – 10| + 7 is y = |x| translated 10 units right and 7 units up.

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Draw a graph of y = .0.5x

Make a table of values that includes positive and negative values of x. Notice that x cannot be 0.

Graph the points.

The graph has two parts. Each part is called a branch.

The x-axis is the horizontal asymptote.The y-axis is the vertical asymptote.

x –10 –5 –2 – – – 2 5 10

y –0.05 –0.1 –0.25 –1 –2 –5 5 2 1 0.25 0.1 0.05

12

14

1 10

1 10

14

12

Connect them with a smooth curve.

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Compare the graphs of y = and y = shown below.0.25

x1x

The axes are the asymptotes for both graphs.

Both graphs are symmetric with respect to y = x and y = –x.

What points on the graphs are closest to the origin?

The points, (1, 1), (–1, –1), (0.5, 0.5), and (–0.5, –0.5) are closest to the origin.

The branches of y = are closer to the axis than are the branches of .0.25x

1x

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Compare the graphs of y = and y = – shown below.2x

2x

Both graphs are symmetric with respect to y = x and y = –x.

Each is a 90° rotation of the other about the origin.

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Graph the functions f = and f = 440.343w

Use the Intersection feature.



The wavelength is about 0.78 m/s.

The frequency f in hertz of a sound wave varies inversely with

its wavelength w. The function f = models the relationship

between f and w for a wave with a velocity of 343 m/s. Find the

wavelength of a sound wave with a frequency of 440 Hz.

343w

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Sketch the graph of y = + 2. 1 x + 2

Translate these points 2 units to the left and 2 units up to (–1, 3) and (–3, 1). Draw the branches through these points.

Step 2: Translate y = .

The graph y = includes (1, 1) and (–1, –1).

1x

1x

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Step 1: Draw the asymptotes.

For y = + 2, b = –2 and c = 2.

The vertical asymptote is x = –2. The horizontal asymptote is y = 2.

1 x + 2



Write an equation for the translation of y = – that has

asymptotes at x = 8 and y = –4.

7x

9-2

y = + c Use the general form of a translation. – 7 x – b

= + 4 Substitute 8 for b and 4 for c. – 7 x – 8

= – 4 Simplify. – 7 x – 8




1.

2.

3.

4. The graph of y = is closer to the x- and

y-axes than the graph of y = .

5. The graph of y = is closer to the axes.

6. The graph of y = is closer to the axes.

3x

5x

1x0.2x

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7. The branches of y = are in Quadrants

I and III. The branches of y = – are in

Quadrants II and IV. Each graph is a 90° rotation about the origin of the other graph.

8. The graphs of both equations are in Quadrants II and IV. The graph of y = –is closer to the axes.

9. The branches of y = are in Quadrants

I and III.The branches of y = – are in

Quadrants II and IV. Each graph is a 90° rotation about the origin of the other graph.

2x

8x

8x

12x

12x

10. 18.4 ft

11. 7.67 ft

12. 3.83 ft

13. 1.84 ft

14.

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15.

16.

17.

18.

19.

20.

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21.

22. y = + 4

23. y = + 3

24. y = – 8

2x

2 x + 2 2 x – 4

25. a. c =

a = 0, c = 0 b. Answers may vary. Sample: If the number of awards is large, the amount of money

available for each award approaches 0.

750a

25. b. (continued)If there are a small number of awards, then the amount of money available for each award gets larger.

26. Check students’ work.

27. y =

28. y =

29. y =

30. y =

31. y =

32. y =

0.5x

0.75x

–0.01x

4x–1.4

x

9-2

–8.3x



33.

34.

35.

36.

37.

38.

39. Answers may vary. Sample: The graph of the translation looks similar to the graph of

y = , so knowing the

asymptotes helps to position the translation; check students’ work.

kx

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40. a.

b. Sahara Desert: 26.74 in., Kalahari Desert: 23.93 in., Mt. Kilimanjaro: 11.59 in., Vinson Massif: 12.58 in.

c. No; p = 0 is an asymptote.

41.

(3, 6)

42.

(2.92, 6.2)

43.

(–1.75, –4)

44.

(–0.45, –10) and (0.45, –10)

45.

(3.76, 4.2)

46.

(–0.76, 9) and (0.76, 9)

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47. a. m =

b. m =

c. 25 mi/gal, 28.57 mi/gal

10,000g

10,000g – 50

48. The branches of y = are in Quadrants I

and III. The branches of y = are in

Quadrants I and II. The graphs intersect at

all points on y = in Quadrant I.

49. The branches of y = are in Quadrants I

and II. The branches of y = are in

Quadrants I and III. The graphs intersect at

(1, 1). The graph of y = is closer to the

x-axis for x > 1, and the graph of y = is

closer to the y-axis for 0 < x < 1.

1x

1x

1x

1 x2

1 x2

1x

1x

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50. The branches of both graphs are in Quadrants I and II. They intersect at (1, 1) and (–1, 1). The graph of

y = is closer to the x-axis for x > 1

and x < –1. The graph of y = is

closer to the y-axis for –1 < x < 0 and 0 < x < 1.

51. y = , y = –

52. a. y =

1 x2

1x

16x

16x

b. y =

c. y = – 2

d. y =

–0.25 x – 0.5

1x

1 x – 1

9-2

0.6 x – 2



53. D

54. H

55. B

56. F

57. A

58. [2] When x – 2 = 0, – is undefined,

so x = 2 is an asymptote. As x

becomes larger, the value of –

approaches 0, so y approaches 11,

and y = 11 is an asymptote.

[1] answer only, with no explanation

3 x – 2

3 x – 2

59. [4] The general form of a translation

of y = is y = + c. For

asymptotes at x = –5 and

y = –13, b = –5 and c = –13.

Substituting, y = – 13,

or y = – 13.

[3] minor error, such as a sign error

[2] several errors OR major error,

such as writing y = + (–5)


60. V varies jointly with the square of s and h.

–3 x – b

–3 x – (–5) –3

x + 5

–3 x – 13

9-2

–3x



61. h varies directly with V and inversely with the square of s.

62. B varies directly with V and inversely with h.

63. w varies directly with V and inversely with the product of and h.

64. b varies directly with A and inversely with h.

65. growth, 4

66. growth, 2

67. decay, 0.8

68. decay, 0.5

69. 79 – 20 3

70. –2

71. 50 + 35 2

72. 6

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5. Write an equation for the translation

of y = that has asymptotes at

x = 5 and y = –8.

1. Draw a graph of y = .



2. Compare the graphs of y = and y = .6x

12x

3. A group of college students rents a large two-story house. The amount of rent in dollars that each student pays per month is inversely proportional to the number of students. The rent r per month for one student is related to the number of students n by the equation r = . Find the monthly rent each student pays if there are 8 students in the house.

1200n

4. Sketch the asymptotes and

the graph of y = – – . 0. 5 x + 2

12

$150

13x

12x

Answers may vary. Sample: both graphs lie in quadrants I and III. Both have the axes as asymptotes. Both are symmetric with respect to y = x and y = –x.The graph of

y = can be obtained by stretching the graph of y = by a factor of 2.

6x

12x

9-2

y = – 8 13 x – 5

(For help, go to Lessons 5-4 and 5-5.)


Rational Functions and Their GraphsRational Functions and Their Graphs

Factor.

1. x2 + 5x + 6 2. x2 – 6x + 8 3. x2 – 12x + 27

4. 2x2 + x – 28 5. 2x2 – 11x + 15 6. 2x2 – 19x + 24

Solve.

7. x2 + x – 12 = 0 8. x2 – 3x – 28 = 0 9. x2 – 9x + 18 = 0

9-3

Solutions



1. Factors of 6 with a sum of 5:3 and 2x2 + 5x + 6 = (x + 3)(x + 2)

3. Factors of 27 with a sum of –12:–3 and –9x2 – 12x + 27 = (x – 3)(x – 9)

5. 2x2 – 11x + 15 = (2x – 5)(x – 3)Check: (2x – 5)(x – 3) = 2x2 – 6x– 5x + 15 = 2x2 – 11x + 15

7. x2 + x – 12 = 0(x + 4)(x – 3) = 0

x + 4 = 0 or x – 3 = 0x = –4 or x = 3

9. x2 – 9x + 18 = 0(x – 3)(x – 6) = 0

x – 3 = 0 or x – 6 = 0x = 3 or x = 6

2. Factors of 8 with a sum of –6:–4 and –2x2 – 6x + 8 = (x – 4)(x – 2)

4. 2x2 + x – 28 = (2x – 7)(x + 4)Check: (2x – 7)(x + 4) = 2x2 + 8x– 7x – 28 = 2x2 + x – 28

6. 2x2 – 19x + 24 = (2x – 3)(x – 8)Check: (2x – 3)(x – 8) = 2x2 – 16x– 3x + 24 = 2x2 – 19x + 24

8. x2 – 3x – 28 = 0(x + 4)(x – 7) = 0

x + 4 = 0 or x – 7 = 0x = –4 or x = 7

9-3

For each rational function, find any points of discontinuity.



The function is undefined at values of x for which x2 – x – 12 = 0.

x2 – x – 12 = 0 Set the denominator equal to zero.

(x – 4)(x + 3) = 0 Solve by factoring or using the Quadratic Formula.

x – 4 = 0 or x + 3 = 0 Zero-Product Property

x = 4 or x = –3 Solve for x.

There are points of discontinuity at x = 4 and x = –3.

9-3

a. y = 3 x2 – x –12

b. y =

(continued)



2x 3x2 + 4

The function is undefined at values of 3x2 + 4 = 0.

3x2 + 4 = 0 Set the denominator equal to zero.

x2 = – Solve for x.43

x = = ± –4 3

± 2i 3

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Since is not a real number, there is no real value for x for

which the function y = is undefined. There is no point

of discontinuity.

± 2i 3

2x 3x2 + 4

Describe the vertical asymptotes and holes for the graph of

each rational function.



Since –1 and –5 are the zeros of the denominator and neither is a zero of the numerator, x = –1 and x = –5 are vertical asymptotes.

–3 is a zero of both the numerator and the denominator. The graph of this function is the same as the graph y = x, except it has a hole at x = –3.

a. y = x – 7 (x + 1)(x + 5)

b. y = (x + 3)x x + 3

c. y = (x – 6)(x + 9) (x + 9)(x + 9)(x – 6)

6 is a zero of both the numerator and the denominator.

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The graph of the function is the same as the graph y =

which has a vertical asymptote at x = –9, except it has a hole at x = 6.

1 (x + 9)

,

Find the horizontal asymptote of y =



–4x + 3 2x + 1

The horizontal asymptote is y = –2.

Divide the numerator by the denominator as shown at the right. –2

2x + 1 –4x + 3 –(–4x – 2)

5

9-3

.

The function y = –4x + 3 2x + 1

can be written as y = – 2. 5 2x + 1

Its graph is a translation of y = 5 2x + 1

.

Sketch the graph y = .



x + 1 (x – 3)(x + 2)

The degree of the denominator is greater than the degree of the numerator, so the x-axis is the horizontal asymptote.

When x > 3, y is positive. So as x increases, the graph approaches the y-axis from above.

When x < –2, y is negative. So as x decreases, the graph approaches the y-axis from below.

Since –1 is the zero of the numerator, the x-intercept is at –1.

Since 3 and –2 are the zeros of the denominator, the vertical asymptotes are at x = 3 and x = –2.

Calculate the values of y for values of x near the asymptotes. Plot those points and sketch the graph.

9-3

A vat contains 20 gallons of cleaning solution that is 15% bleach. A second vat has a solution that is 50% bleach. A few gallons from the second vat will be added to the first vat to get a solution that is more than 15% bleach.



a. Write a function for the percent bleach in the new solution if x gallons from the second vat are added to the first vat. Graph the function.

gallons of bleach in 1st vat + gallons of bleach in 2nd vat total gallons of cleaning solution

Relate: percent bleach =

Define: Let x = number of gallons added from the second vat.

Let (0.15)(20) = 3 = gallons of bleach in first vat.

Let 0.5x = gallons of bleach from second vat.

Let 20 + x = the total number of gallons in the final mixture.

Let y = the percent of bleach in the final mixture.

Write: y = • 1003 + 0.5x20 + x

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(continued)



Graph the function. Adjust the viewing window.

b. What percent bleach will the new solution be if 10 gallons from the second vat are used? If 17 gallons are used?

Use the CALC feature to evaluate the function at x = 10 and at x = 17.

If 10 gallons are added from the second vat, the percentage of bleach will be about 27%.

If 17 gallons are added from the second vat, the percentage of bleach will be about 31%.

9-3




1. x = 0, x = 2

2. none

3. x = 1, x = –1

4. x = 2, x = 3

5. x = –3

6. x = – , x = 1

7. x = 2

8. none

9. x = –2.77, x = 1.27

10. vertical asymptote at x = –2

11. hole at x = –5

72

12. vertical asymptotes at x = – and x = 1

13. vertical asymptote at x = –1, hole at x = 2

14. hole at x = –2

15. none

16. holes at x = ±3

17. none

18. vertical asymptote at x = –5, hole at x = –

19. y = 0

20. y = 0

21. y = 1

22. y =

23. y = 0

32

23

12

9-3



24. y =

25.

26.

34 27.

28.

29.

30.

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31. a. y =

b. $46.88; $14.68c. more than 21,916 discsd. x = 500, y = 0.19

32. vertical asymptotes at x = –3 and x = 3, horizontal asymptote at y = 0

33. vertical asymptote at x = –2

34. horizontal asymptote at y = 0

0.19x + 210,000x – 500

35.

36.

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37.

38.

39.

40.

41. Answers may vary. Sample: There is no value of x for which the denominator equals 0.

42. a.

b. 6 free throws

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43. a. y =

b. $65,000; $25,806.45

c. Answers may vary. Sample: No; the president’s salary throws off the average; the median or mode would be a better measure.

20,000x + 210,000x + 1

44. a. P(n) = 4n2

b. R(n) = 4n + 1

c. y = ; ; check students’ work.

45. a. The increase in production workers’ average hourly wage is greater.

b. rational

c. R(x) = d.

around the year 2106

4n2

4n + 1

M(x)A(x)

6417

9-3



46. C

47. [2] x + 3 5x + 7 5x + 15

– 8

So y = can be written as

y = + 5. As x becomes

larger, the fraction approaches 0 and y approaches the asymptote at 5.


48. B

49. A

50. C

5

5x + 7x + 3

–8 x + 3

51. vertical: x = 0, horizontal: y = 4

52. vertical: x = –3, horizontal: y = 0

53. vertical: x = –1, horizontal: y = 1

54. vertical: x = 7, horizontal: y = –3

55. vertical: x = 2, horizontal: y = 0

56. vertical: x = 5, horizontal: y = –6

57. 1 unit right

58. 3 units up

59. 1 unit left and 5 units down

9-3



1. Find any points of discontinuity for the rational function

y = .

2. Describe the vertical asymptotes and holes for the graph of

y = .

3. Find the horizontal asymptote of y = .

4. Sketch the graph of y = .

5. The cost in dollars of publishing x copies of a certain book is modeled by 30,000 + 10x + 0.0001x2.a. Write a function for the average cost per book.

Graph the function.

b. What is the average cost per book if 12,000 copies are published? if 15,000 copies are published?

x + 7 x2 + 5x – 14

x + 7 (x – 3)(x + 5)

x2 + 93x2 + 5

x2 – 9x(x2 + 2x – 15)

y = 2, x = –7

vertical asymptote at x = 3; hole at x = –5

y = 13

$13.70; $13.50

ƒ(x) = 30,000 + 10x + 0.0001x2

x

9-3

(For help, go to Lesson 5-4 and Skills Handbook page 843.)


Rational ExpressionsRational Expressions

Factor.

1. 2x2 – 3x + 1 2. 4x2 – 9 3. 5x2 + 6x + 1 4. 10x2 – 10

Multiply or divide.

5. • 6. • 7. • 8. •

9. ÷ 4 10. ÷ 11. ÷ 12. ÷

38

56

12

46

83

2 16

25

37

58

34

12

9 16

34

54

15 8

9-4



2. 4x2 – 9 = (2x)2 – 32 = (2x + 3)(2x – 3)

4. 10x2 – 10 = 10(x2 – 1) = 10(x2 – 12)= 10(x + 1)(x – 1)

Solutions

1. 2x2 – 3x + 1 = (2x – 1)(x – 1)Check: (2x – 1)(x – 1) = 2x2 – 2x – 1x + 1

= 2x2 – 3x + 1 3. 5x2 + 6x + 1 = (5x + 1)(x + 1)Check: (5x + 1)(x + 1) = 5x2 + 5x + 1x + 1

= 5x2 + 6x + 1

5. • = = =

7. • = = = 9. ÷ 4 = • = =

11. ÷ = • = = =

38

56

3 • 58 • 6

3 • 5 8 • 3 • 2

5 16/

/

83

2 16

8 • 2 3 • 16

1 • 16 3 • 16

13

58

58

14

5 • 18 • 4

5 32

9 16

34

9 16

43

9 • 4 16 • 3

3 • 3 • 44 • 4 • 3/

// 34

6. • = = =

8. • = =

10. ÷ = • = = = or 1

12. ÷ = • = = =

12

46

1 • 42 • 6

1 • 2 • 22 • 2 • 3/

/ //

13

25

37

2 • 35 • 7

6 35

54

15 8

54

8 15

5 • 8 4 • 15

5 • 2 • 44 • 5 • 3///

/ 23

34

12

34

21

3 • 24 • 1

3 • 2 2 • 2 • 1

32/

12

/

9-4

Simplify . State any restrictions on the variable.



x2 – 6x – 16x2 + 5x + 6

The restrictions on x are needed to prevent the denominator of the original expression from being zero.

= Factor the polynomials. Notice x = –3 or –2./

x2 – 6x – 16x2 + 5x + 6

(x – 8)(x + 2)(x + 3)(x + 2)

= Divide out common factors.(x – 8)(x + 2)(x + 3)(x + 2)

1

1

= x + 8x + 3

The simplified expression for x = –3 or –2.x + 8x + 3

/

9-4

Compare the ratio of the volume to surface area of a sphere

with radius r with the ratio of volume to surface area of a sphere with

radius 2r.



The ratio of volume to surface area is twice as great for a sphere whose radius is 2r than a sphere with a radius of r.

Use the formulas for volume and surface area of a sphere.

Volume (V) = • r 3

Surface Area (S.A.) = 4 r 2

43

= = Simplify.r3

2r3

9-4

4

3

(r ) 3

4 (r )2= = Substitute for r.4

3

(2r ) 3

4 (2r )2

Sphere with radius r Sphere with radius 2r

= = Write a ratio. V S.A.

V S.A.

4

3

r 3

4 r 2

4

3

r 3

4 r 2



Multiply and . State any restrictions on

the variable.

3x2 + 5x – 2x – 5

x2 – 253x2 – 7x + 2

• = • Factor.3x2 + 5x – 2

x – 5x2 – 25

3x2 – 7x + 2(3x – 1)(x + 2)

x – 5(x + 5)(x – 5)(3x – 1)(x – 2)

= • Divide outcommonfactors.

(3x – 1)(x + 2)x – 5

(x + 5)(x – 5)(3x – 1)(x – 2)

1

1 1

1

=(x + 2)(x + 5)

x – 2

=x2 + 7x + 10

x – 2

9-4

The product is for x 5, 2, or .x2 + 7x + 10x – 2

13

=/

Divide by . State any restrictions on

the variables.



3 – y(2x – 1)(x + 5)

6(y – 3)(2x – 1)(x – 7)

÷ = • Multiply by thereciprocal.

3 – y(2x – 1)(x + 5)

6(y – 3)(2x – 1)(x – 7)

3 – y(2x – 1)(x + 5)

(2x – 1)(x – 7)6(y – 3)

= • Divide out common factors.

3 – y(2x – 1)(x + 5)

(2x – 1)(x – 7)6(y – 3)

1

1 1

–1

= • –1 (x + 5)

(x – 7)6

= 7 – x 6(x + 5)

= Multiply. 7 – x 6x + 30

9-4

The quotient is for x = –5, , or 7, and y = 3. 7 – x 6x + 30 // 1

2




1. ; x = 0 or

2. 2c + 3; c = 0

3. b + 1; b = 1

4. z – 7; z = –7

5. ; x = –5

6. ; x = 6 or –4

7. ; x = 0, y = 0

8. ; x = 0, y = 0

9. ; y or 3

1 2x + 1

2 x + 5

x + 4x – 6 7 15x2

xy443

/ /

//

/

/

/

/

/

/ 12

10. ; x 3 or

11. ; x = 0, 1, –1, or –2

12. 1; x = –2, –1, 2, or 3

13. ; x = 0, y = 0

14. ; x = 0, y = 0

15. ; x = y

16. 1; y = –2 or 4

17. ; x = –1, 1, or 0

18. ; y = 2, –5, or 0

19. ; x = –3 or 10

4(x + 6) 3(3x + 8) x – 2 x(x – 1)

2 3x2y2

y2x2

5(x + y) 3

x(x – 1) 3(x + 1) 4(y – 3) y(y + 5) x – 8 x – 10

/

/

/

/

/

//

//

/

/

9-4

12=/

83=/



20. ; y = 2

21. ; x = 0, y = –4 or 3

22.

23. The student is not correct; x = 2 will make the denominator of equal 0, so x = 2 is not a solution.


25. The numerator and the denominator have no common factors; check students’ work.

y + 6 y – 2 y(y + 3) 12(y + 4) 6(a + 1)

a – 3

26. a. =

b. =

c. The ratio for the cylindrical tank is larger.

d. The cylindrical tank will have a larger volume.

27. ; a = –4, –3, or 3

28. ; b = –5

29. ; x = 0, –5, 4, or 1

/ /

/ r 2

2 r 2 + r 2

23

r 2(r)2 r 2 + r 2

a + 3 (a – 3)(a – 3) 2(b – 5)

b + 54x

/

/

/

2r9r4

9-4

x x – 2



30. ; x –9, –3, or 3

31. ; x –3, , 2, or 4

32. ; x – , , 1, or –2

33. ; x –4, 0, 1

34. 2; x = –3, 1

35. ; y = 0

36.

37. a. 1.2 m/s2

b. 2.68 m/s2

38. a. 2xn + 1b. 2 is a factor of 2xn, so 2xn is even,

and 2xn + 1 is odd.

18x (x + 9)(x + 3) x + 1 x – 1

12

x + 1 x – 1

12

12

x(x – 1)3

(x + 4)

18x5

y 2

2(a + 8) 2a + 5

39. ; x = 0 or –1, y = 0

40. ; a = 0 or b, b = 0

41. ; m = 0 or n , n 0

42. D

43. G

44. D

45. H

46. [2] ÷ = • The restrictions

are all the values of x that make any denominator equal zero, so the restrictions are that g(x) = 0, k(x) = 0, and h(x) = 0.

[1] incomplete answer

4x3y3a2b2

4 15 4n2

ƒ(x)g(x)

h(x)k(x)

ƒ(x)g(x)

k(x)h(x)

/

/

/ /

/ /

/

/ //

9-4

=/

=/

=/

=/

=/23



47. [2] • =

= ÷ =

•

= • =

[1] answer only, with no work shown OR one mistake, such as in factoring

48. hole at x = 3

49. vertical asymptotes at x = – and x = –1

50. hole at x = 4, vertical asymptote at x = –3

51. 3

52. –5

x + 4 x + 5

x2 – 11x + 28x2 – 2x – 35

ƒ(x)g(x)

ƒ(x)g(x)

x + 4 x + 5

x2 – 11x + 28x2 – 2x – 35

x + 4 x + 5

x2 – 2x – 35x2 – 11x + 28

x + 4 x + 5

(x – 7)(x + 5) (x – 7)(x – 4)

x + 4 x – 4

53.

54.

55. x = 22

56. x = 6

23

3234

9-4



Simplify. State any restrictions on the variable.

1. 2.

3. Multiply. State any restrictions on the variable.

•

4. Divide. State any restrictions on the variable.

÷

x2 + x – 6x2 + 3x

4x2– 252x2 + 3x – 20

x2 + 6x – 7x2 + 5x

3x2 + 16x + 52x2 + 7x – 9

y 2 + 5y + 4y 2 – 49

2y 2 + 5y – 12y 2 + 9y + 14

; x = –3, 0/x – 2 x

; x = –5, – , 0, 1/3x2 + 22x + 72x2 + 9x

92

; y = –7, –4, –2, , 7/ 32

y 2 + 3y + 22y 2 – 17y + 21

9-4

; x = –4, / 2x + 5 x + 4

52


(For help, go to Skills Handbook page 843.)

Adding and Subtracting Rational ExpressionsAdding and Subtracting Rational Expressions

Find the least common multiple of the two numbers.

1. 7, 21 2. 6, 10 3. 11, 17 4. 30, 105

Add or subtract.

5. + 6. + 7. – 8. – 5 19

7 38

2 15

3 25

7 24

5 36

1112

7 45

9-5



Solutions

1. 7 = 7; 21 = 3 • 7; LCM = 3 • 7 = 21

2. 6 = 2 • 3; 10 = 2 • 5; LCM = 2 • 3 • 5 = 6 • 5 = 30

3. 11 = 11; 17 = 17; LCM = 11 • 17 = 187

4. 30 = 2 • 3 • 5; 105 = 3 • 5 • 7; LCM = 2 • 3 • 5 • 7 = 6 • 35 = 210

5. + = + = + = =

6. + = + = + = =

7. – = – = – = =

8. – = – = – = =

5 19

7 38

2 15

3 25

7 24

5 36

1112

7 45

5 • 2 19 • 2 2 • 5 15 • 5 7 • 3 24 • 311 • 1512 • 15

7 38 3 • 3 25 • 3 5 • 2 36 • 2 7 • 4 45 • 4

1038

7 38

10 75

9 75

2172

1072

165180

28 180

10 + 7 38 10 + 9

75 21 – 10

72 165 – 28

180

1738

19 751172

137180

9-5

An object is 24 cm from a camera lens. The object is in focus

on the film when the lens is 12 cm from the film. Find the focal length

of the lens.



= + Use the lens equation.1f

1di

1do

= + Substitute.1f

1 12

1 24

= + Write equivalent fractions with the LCD. 2 24

1 24

= = Add and simplify. 3 24

18

Since = , the focal length of the lens is 8 cm.1f

18

9-5

Find the least common multiple of 2x2 – 8x + 8 and 15x2 – 60.



Step 1: Find the prime factors of each expression.

2x2 – 8x + 8 = (2)(x2 – 4x + 4) = (2)(x – 2)(x – 2)

15x2 – 60 = (15)(x2 – 4) = (3)(5)(x – 2)(x + 2)

Step 2: Write each prime factor the greatest number of times it appears in either expression. Simplify where possible.

(2)(3)(5)(x – 2)(x – 2)(x + 2) = 30(x – 2)2(x + 2)

The least common multiple is 30(x + 2)(x – 2)2.

9-5

Simplify + .



13x2 + 21x +30

4x 3x + 15

+ = + Factor the denominators.

13x2 + 21x +30

4x 3x + 15

13(x + 2)(x + 5)

4x 3(x + 5)

= + • Identity for Multiplication.

13(x + 2)(x + 5)

4x 3(x + 5)

x + 2x + 2

= + Multiply.13(x + 2)(x + 5)

4x(x + 2)3(x + 2)(x + 5)

= Add.1 + 4x(x + 2)3(x + 2)(x + 5)

= Simplify thenumerator.

4x2 + 8x +13(x + 2)(x + 5)

= Simplify thedenominator.

4x2 + 8x +13x2 + 21x +30

9-5



Simplify + .2x

x2 – 2x – 3 3 4x + 4

– = – Factor the denominators.

2xx2 – 2x – 3

3 4x + 4

2x(x – 3)(x + 1)

3 4(x + 1)

= Simplify.4(2x) – (3)(x – 3)

4(x + 1)(x – 3)

= Simplify.5x + 94x2 – 8x – 12

9-5

= • – • Identity for Multiplication.

x – 3x – 3

2x(x – 3)(x + 1)

3 4(x + 1)

44

Simplify .



1x

1y

2y

1x

+

–

1x

1y

2y

1x

+

–

1x

1y

2y

1x

+

–

• xy

• xy= The LCD is xy. Multiply the numerator

and denominator by xy.

2 • xyy

1 • xyx

1 • xyx

1 • xyy+

–

= Use the Distributive Property.

y + x2x – y= Simplify.

Method 1: First find the LCD of all the rational expressions.

9-5

(continued)



= Write equivalent expressions withcommon denominators.

1x

1y

2y

1x

+

–

+

–

yxy

xxy

2xxy

xxy

2x – yxy

x + yxy

= Add.

Method 2: First simplify the numerator and denominator.

= ÷ Divide the numerator fraction by the denominator fraction.

x + yxy

2x – yxy

= • Multiply by the reciprocal.x + y

xyxy

2x – y

=x + y2x – y

9-5


1. 2.03 in.

2. 2.02 in.

3. For distances greater than 10 ft, di is nearly constant.

4. 9(x + 2)(2x – 1)

5. (x – 1)(x + 1)(x + 1)

6. 10(x – 2)(x + 3)2

7. 18(2x – 7)(x + 3)

8. 5(y + 4)(y – 4)

9. 2(x + 5)(x2 – 32x – 10)

10.

120 59



1x

11.

12.

13.

14.

15.

16. –

17.

18.

19.

20.

2(d – 2) 2d + 1

7x2 + 20x – 18(x – 3)(x + 3)(x + 4) –x + 6 (x – 3)(x – 3)

3x

xy + 8y + 4 2xy2

5x2 + 14x – 12(x – 3)(x + 2)2

–3(2y + 1) 2y – 1

y – 6 2(y + 2) x2 – 24

3x(x + 3) –5(y + 8) (y – 5)(y + 5)

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

–2x(x + 3) (x – 2)(x – 1)(x + 1) y 2x1528 2 3(x + y)b9 y x + y

3x 2 + xy25 3 x – 6

–3x 5 + xy

9-5

240119



31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

3x – 8 4x2

x2 + 4x – 3(x + 1)(x – 1)

2x3 – x2 + 1x2(x + 1)(x – 1)

3(4y – 21) y(y – 7)

4y 3 + 12y 2 – y – 2y(y + 3)

7x – 17 (x – 3)(x + 3)

4x – 1 2x(2x – 1)

x2 + 9x – 1(x – 1)(2x + 1)

5x2 + 6x + 12(x – 3)(x + 2)2

x(3x2 + x – 1)x2 – 2


42. Factoring is used to determine the least common multiple of the denominators; check students’ work.

43. Answers may vary. Sample: Substitute 0 for x in the three expressions,

and show that + = .

44.

45.

46.

47. x

48.

3x + 2y 7x – 5y 2x – 5y 2(3x + 2y)2(x + 2) 4x + 3

2(x + 5) x + 7

4 –9

73

25–9

/

9-5



49.


51. x –2, –3, –4; those values result in division by 0, which is undefined.

52. , ,

53. a. mi/h

b.

c. mi/h

d. mi/h

54. a.

b. 0.88 ohms

–5x + 13 2(x – 4)

23

35

23

55. a. ƒ =

b. = ƒ

56. B

57. H

58. C

59. F

60. [2] First factor both denominators: x2 – 5x – 6 = (x – 6)(x + 1), x2 – 12x + 36 = (x – 6)2. The least common denominator would have to include the factors (x – 6), (x + 1) and (x – 6)2, so the LCD is (x – 6)2(x + 1) = x3 – 11x2 + 24x + 36.

[1] answer only, no work shown

2x(x + a) 2x + a

247247

4009

R1R2R3

R1R2 + R1R3 + R2R3

x(2x + 1) 3x + 1

dido

di + do

9-5

=/



61. [4] = 9

Find a common denominator

for the numerator: = 9

Combine the numerators: = 9

Multiply the numerator by the reciprocal

of the denominator: • = 9

Simplify: = 9

Solve for x:2x – 15 = 27

2x = 42x = 21

2 – 15x

3x

– 15x

3x

2 – 15x3x

2x – 15x

x3

2x – 153

[3] one calculation error[2] one process error,

such as an error in simplifying the numerator

[1] answer only, no work shown

62. ; x 0, y 0

63. ; a 0, b 0, x 0, y 0

64. ; x 0, y 0

65. ; x –3, 2, or 3

66. ; x –2, 2, or 3

3x 4y2

1 4xy y 2

2x2

12x x + 33(x – 2)4(x – 3)

9-5

=/ =/

=/ =/ =/ =/

=/ =/

=/

=/

2x



67. ; x 1, –1, or –3

68. log3 t4y

69. log10 p7q

70. log5

71. 30

72. 82

73.

74. 101

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

15 4

x

y5

9-5

=/



1. The lens equation for a camera is = + , where f is the focal length

of the lens, di is the distance between the lens and the film, and do is the

distance between the lens and the object. For a certain camera, an object

that is 18 cm from the lens is in focus when the lens is 9 cm from the film.

What is the focal length of the lens?

2. Find the least common multiple of x(x2 – 5x + 6) and x3(x2 + 4x – 21).

Simplify.

3. + 4. –

5.

1f

1di

1do

x x2 – 4

7 3x + 6

m m + 3

6m m2 – 9

2y 2y + 1 – 1

2y 2y – 11 –

6 cm

x3(x – 2)(x – 3)(x + 7)

10x – 143x2 – 12

m2 – 9m m2 – 9

2y – 12y + 1

9-5

(For help, go to Lessons 5-8 and 9-1.)


Solving Rational EquationsSolving Rational Equations

Find the LCD of each pair of fractions.

1. , 2. ,

3. , 4. ,

5. , 6. ,

13t

1 5t2

4 3h2

2hh3

z 2z + 1

1z

x2

3x8

4 y + 2

3 y – 1

1 k + 2

3k k2 – 4

9-6



Solutions

1. 3t = 3 • t; 5t 2 = 5 • t • t; LCD = 3 • 5 • t • t = 15t 2

2. 2 = 2; 8 = 2 • 2 • 2; LCD = 2 • 2 • 2 = 8

3. 3h2 = 3 • h • h; h3 = h • h • h; LCD = 3 • h • h • h = 3h3

4. y + 2 = y + 2; y – 1 = y – 1; LCD = (y + 2)(y – 1) or y2 + y – 2

5. 2z + 1 = 2z + 1; z = z; LCD = z(2z + 1) or 2z2 + z

6. k + 2 = k + 2; k2 – 4 = (k + 2)(k – 2); LCD = (k + 2)(k – 2) or k2 – 4

9-6

Solve = .



1 x – 3

6x x 2 – 9

x2 – 9 = 6x(x – 3) Write the cross products.

x2 – 9 = 6x2 – 18xDistributive Property

–5x2 + 18x – 9 = 0 Write in standard form.

5x2 – 18x + 9 = 0 Multiply each side by –1.

(5x – 3)(x – 3) = 0 Factor.

5x – 3 = 0 or x – 3 = 0

x = or x = 3 Zero-Product Property35

9-6

1 x – 3

6x x 2– 9

=



9-6

(continued)

Check: When x = 3, both denominators in the original equation are zero.The original equation is undefined at x = 3.So x = 3 is not a solution.

When is substituted for x in the original equation,

both sides equal – .

35 5

12

Solve – = .



3 5x

4 3x

13

3 5x

4 3x

13

– = .

9 – 20 = 5x Simplify.

15x – = 15x Multiply each side by the LCD, 15x. 3 5x

4 3x

13

– = Distributive Property45x5x

60x3x

15x3

– = x115

Since – makes the original equation true, the solution is x = – .115

115

9-6

Josefina can row 4 miles upstream in a river in the same time it takes her to row 6 miles downstream. Her rate of rowing in still water is 2 miles per hour. Find the speed of the river current.



Relate: speed with the current = speed in still water + speed of the current,speed against the current = speed in still water – speed of the current,time to row 4 miles upstream = time to row 6 miles downstream

9-6

Define:Distance (mi) Rate (mi/h) Time (h)

With current 6 2 + r

Against current 4 2 – r 4 (2 – r )

6 (2 + r )

Write: = 6 (2 + r )

4 (2 – r )

(continued)



(2 – r )(6) = (2 + r )(4) Simplify.

12 – 6r = 8 + 4r Distributive Property

4 = 10r Solve for r.

0.4 = r Simplify.

The speed of the river current is 0.4 mi/h.

= 6 (2 + r )

4 (2 – r )

9-6

(2 + r )(2 – r ) = (2 + r )(2 – r ) Multiply by the LCD(2 + r )(2 – r ).

6 (2 + r )

4 (2 – r )

Jim and Alberto have to paint 6000 square feet of hallway in an office building. Alberto works twice as fast as Jim. Working together, they can complete the job in 15 hours. How long would it take each of them working alone?



Relate: Jim’s work speed + Alberto’s work speed = combined work speed

Define:Time (hours) Rate (square feet per hour)

Jim 2x

Alberto x

Combined 15

60002x

6000x

600015

= 400

Write: + = 40060002x

6000x

9-6

(continued)



60002x

6000x

+ = 400

6000 + 12000 = 800x Simplify.

18000 = 800x Simplify.

22.5 = x Solve for x.

Alberto could paint the hallway in 22.5 hours.Jim could paint the hallway in 2(22.5) hours or 45 hours.

2x + = 2x(400) Multiply by the LCD, 2x.60002x

6000x

+ = 2x(400) Distributive Property2x(6000)2x

2x(6000)x

9-6

24. 2 h

25. 1 h

26. E =

27. E = mc2

28. F = ma

29. c = ± a2 – b2

30. T = ± 2

31. B = ±

32. 2 days

33. 4 test scores


1. 5

2. no solution

3. 10

4. 2 or –5

5.

6. – or 4

7.

8. 3

9. –1

10.

11. 10

g



43

52

73

29

12. 4

13. 2

14. –1 or 2

15. –1 or 12

16. –

17. about –1.45 or 1.65

18. 1

19. –3, –2

20. –9

21. 1

22. Carlos: 32 mi/h, Paul: 12 mi/h

23. passenger train: 112 mi/h, freight train: 92 mi/h

1 12

2357

mV2

2

2Vmr 2g

25

9-6



34. a. c(x) =

b. 14 students

35. a. L =

b. 32 in., about 28.24 in., about 25.26 in.

36. a. $1000

b. (1.60)

c. 1000 – (1.60)

d. 30 mi/gal


38. a. about 2037b. Check students’ work.

39. 3

5.50x + 60x

24(R – r)T

15,00024 + x

15,00024 + x

40. no solution

41. no solution

42. 30

43. 5

44.

45. no solution

46. –4

47. 21

48. –4

49. no solution

50. 6

51. 1, –

3821

23

9-6



52. –1

53. x 4.5 ft


55. a. t =

b. h

c. + x

d. + = 3.5; 90 mi/h

56. a–c. Check students’ work.

57. about 44.44 mi/h

58. 5 attendants

59. B

60. I

3518

ds

dt700360

700 360 + x

61. D

62. F

63. [2] + =

12x + = 12x

+ =

3x + 12 = 4x

12 = xIt would take the small plow 12 hours to clear the lot by itself.

[1] answer only, with no work shown

14

13

1x14

1x

13

12x4

12xx

12x3

9-6



64. [4] =

=x(x + 3) = 3(x + 3)

(x + 2)x = 3(x + 2)x = 3x + 6

–2x = 6x = –3

However, x = –3 makes the denominators in the original equation equal 0. So, the equation has no solution.

[3] one computational error[2] gets x = –3, but does not state that

the equation has no solution[1] answer only, with no work shown

x 3x + 9

x + 2x + 3

x 3(x + 3)

x + 2x + 3

65.

66.

67.

68. –3

69. –1

70. –0.999

71. –6

72. y = ; yes

73. y = ± x – 1; no

74. y = x + 4; yes

–y – 134(y + 1)5xy – 122y(y + 2)

x2 + 32(x – 1)(x + 3)

5 – x2

3

9-6



Solve each equation. Check each solution.

1. = 2. 1 + =

3. + x =

4. The speed of the current in a river is 5 miles per hour. A boat leaves a dock on the bank of the river, travels upstream 25 miles, and returns to the dock in 12 hours. What is the speed of the boat in still water?

5x x2 – 25

2 x – 5

x x – 1

4x – 3x – 1

2x

3x2

103

–3, 1

3

7.5 mi/h

9-6



Probability of Multiple EventsProbability of Multiple Events

A bag contains 24 green marbles, 22 blue marbles, 14 yellow marbles, and 12 red marbles. Suppose you pick one marble at random. Find each probability.

1. P(yellow) 2. P(not blue) 3. P(green or red)

4. Of 300 senior students at Howe High, 150 have taken physics, 192 have taken chemistry, and 30 have taken neither physics nor chemistry. How many students have taken both physics and chemistry?

9-7

Probability of Multiple EventsProbability of Multiple EventsALGEBRA 2 LESSON 9-7ALGEBRA 2 LESSON 9-7

Solutions

1. total number of marbles = 24 + 22 + 14 + 12 = 72

P(yellow) = = =


P(not blue) = = = =


P(green or red) = = = =

1472

72 – 2272

2 • 7 2 • 36

24 + 1272

7 36

5072

3672

2 • 25 2 • 36

2536

1 • 36 2 • 36

12

//

//

9-7


Solutions (continued)

4. Let x = the number of students who have taken both physics and chemistry.Then (150 – x) is the number of students who have taken physics, but not chemistry. And (192 – x) is the number of students who have taken chemistry, but not physics. 30 students have taken neither, and there are 300 students altogether.

x + (150 – x) + (192 – x) + 30 = 300(150 + 192 + 30) + (1 – 1 – 1)x = 300

372 – x = 300372 – 300 = x

72 = x

So, 72 students have taken both physics and chemistry.

9-7

Classify each pair of events as dependent or independent.


a. Spin a spinner. Select a marble from a bag that contains marbles of different colors.

Since the two events do not affect each other, they are independent.

b. Select a marble from a bag that contains marbles of two colors. Put the marble aside, and select a second marble from the bag.

Picking the first marble affects the possible outcome of picking the second marble. So the events are dependent.

9-7

A box contains 20 red marbles and 30 blue marbles. A second box contains 10 white marbles and 47 black marbles. If you choose one marble from each box without looking, what is the probability that you get a blue marble and a black marble?


9-7

3050

4757

Relate: probability of both events is probability of first event

times probability of second event

Define: Event A = first marble is blue. Then P(A) = .

Event B = second marble is black. Then P(B) = .

Write: P(A and B) = P(A) • P(B)

(continued)


P(A and B) = •3050

4757

14102850

= Multiply.

= Simplify.4795

The probability that a blue and a black marble will be drawn is , or 49%.4795

9-7


Are the events mutually exclusive? Explain.

a. rolling an even number or a prime number on a number cube

By rolling a 2, you can roll an even number and a prime number at the same time.

b. rolling a prime number or a multiple of 6 on a number cube

Since 6 is the only multiple of 6 you can roll at a time and it is not a prime number, the events are mutually exclusive.

So the events are not mutually exclusive.

9-7


At a restaurant, customers get to choose one of four vegetables with any main course. About 33% of the customers choose green beans, and about 28% choose spinach.What is the probability that a customer will choose beans or spinach?

Since a customer cannot not choose both beans and spinach, the events are mutually exclusive.

P(beans or spinach) = P(beans) + P(spinach) Use the P(A or B)formula for mutually exclusive events.= 0.33 + 0.28

= 0.61

The probability that a customer will choose beans or spinach is about 0.61 or about 61%.

9-7


A spinner has twenty equal-size sections numbered from 1 to 20. If you spin the spinner, what is the probability that the number you spin will be a multiple of 2 or a multiple of 3?

P(multiple of 2 or 3) = P (multiple of 2) + P (multiple of 3) – P (multiple of 2 and 3)

= + – 1020

6 20

3 20

= 1320

The probability of spinning a multiple of 2 or 3 is .1320

9-7



1. independent

2. dependent

3. dependent

4. independent

5.

6.

7. 0.54

8.

9.

10. Not mutually exclusive since 2 is a prime number and less than 4.

16 9 34

2x 9 25

11. Mutually exclusive since if the numbers are equal, then the sum is even.

12. Not mutually exclusive since 6 • 4 = 24, which is greater than 20 and a multiple of 3.

13. 47%

14.

15.

16. 39%

17.

18.

19.

341415

26351212

9-7


20.

21.

22.

23.

24.

25. 1

26.

27.

28.

29.

5656235656

13 7 12253156

30. a. ; ; no; 20 is a

multiple of both 4 and 5.

b.

31. 13%

32. 42%

33. 98%

34. 75%

35. 58%

36.

37.

38.

15

14

4 15 4 15 8 15

1 20

39.

40.

41.

42.

43.

44.

45. a.

b.


1 15

7 15 1 11 5 12 7 15 5 2x

14 1 64

9-7

61.

62. 10.04

63. 201.71

64. ±e5 593.65

65. ±e2 47.39

66. e2 7.39

67. ln 12 2.48

68. 3 • ln(2) – 1 1.08

69. about ±1.048

70. ln 2 0.3466

16e3

2e6

2

12


52. (continued)

d.

53.

54.

55.

56.

57.

58.

59.

60. – or 2

x2 + 3x – 8(x + 2)(2x – 1)

101915191219 7 191719131937

32

9-7

47. F and G are mutually exclusive, so P(F or G) = P(F) + P(G) 0 and P(F and G) = 0. So P(F or G) > P(F and G).

48. 5%

49. 6%

50. 11.4%

51.

52. a.

b.

c.

49

2x + 2 x – 3 2x – 1 2(x – 3) (x + 2)(2x – 1)

>–


1. Classify each pair of events as dependent or independent.a. Flip a quarter. Then flip a penny.b. Choose a time for a Monday dental appointment. Then choose a time for

that same Monday to meet a friend.

2. M and N are independent events. If P(M) = and P(N) = , find P(M and N).

3. Two standard number cubes are tossed. State whether the events are mutually exclusive.a. The product is greater than 6; the sum is less than 7.b. The sum is odd; the product is odd.

4. A spinner has ten equal–size sections numbered from 1 to 10. Find the probability of each event when you spin the spinner.

a. P(even or multiple of 5)

b. P(multiple of 3 or multiple of 4)

16

3 10

independent

dependent

noyes

1 20

35

12

9-7

ALGEBRA 2 CHAPTER 9ALGEBRA 2 CHAPTER 9

Rational FunctionsRational Functions

1. y =

2. z = xy

3. z =

4. neither

5. neither

6. inverse variation; y =

7.

y = + 2

–16x

1032.7xr 2y

128x

7 x – 1

8.

y = + 2

9. vertical asymptote x = 1, horizontal asymptote y = 1

10. hole at x = –3

11. hole at x = 2, vertical asymptote x = –1, horizontal asymptote y = 0

12. hole at x = 0, vertical asymptote x = 4, horizontal asymptote y = 2

7 x + 3

9-A

Page 530



13. vertical asymptote x = –2, horizontal asymptote y = –3

14. vertical asymptote x = 2, horizontal asymptote y = 1

15. vertical asymptote x = 5

16. hole at x = –2, vertical asymptote x = 3; horizontal asymptote y = 0

17. ; x = –3 or 3

18. – ; x = –4, –3, – , or 0


20. 9x2 – 25

21. 10(x + 3)(x + 1)(x – 3)

x + 4x – 3

2x – 1 (x + 4)(2x + 1)

12

22.

23. –

24.

25.

26. 9

27. 1

28. –2

29. 0

30.

31. –6 or 3

32. 2

x2(x + 2)(x – 3)(x + 1)

(x + 1)(2x2 – 9x + 5)(x + 2)(3x – 1)(x + 3)

x3 + 6x2 + 9x – 1x2 – 4

2y x(y – 1)

43

9-A

/

/



33. 5.4 h

34. mutually exclusive;

35. not mutually exclusive;

36. a. No; 96 is a multiple of both 3 and 4.

b.

c.

5 18

49

1 1012

9-A

Documents

Inverse Variation