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Inverse Variation

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Inverse Variation

## Inverse Variation

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1. Inverse Variation ALGEBRA 1 LESSON 12-1 (For help, go to Lesson 5-5.) Suppose y varies directly with x. Find each constant of variation. 1.y = 5x2.y = –7x3. 3y = x4. 0.25y = x Write an equation of the direct variation that includes the given point. 5. (2, 4) 6. (3, 1.5) 7. (–4, 1) 8. (–5, –2) 12-1

2. Inverse Variation ALGEBRA 1 LESSON 12-1 1.y = 5x; constant of variation = 5 2. y = –7x; constant of variation = –7 3. 3y = x • 3y = • x y = x; constant of variation = 4. 0.25y = x y = x 4 • y = 4 • x y = 4x; constant of variation = 4 Solutions 1 3 1 3 1 3 1 3 1 4 1 4 12-1

3. Inverse Variation ALGEBRA 1 LESSON 12-1 Solutions (continued) 5. Point (2, 4) in y = kx: 4 = k(2), so k = 2 and y = 2x. 6. Point (3, 1.5) in y = kx: 1.5 = k(3), so k = 0.5 and y = 0.5x. 7. Point (–4, 1) in y = kx: 1 = k(–4), so k = – and y = – x. 8. Point (–5, –2) in y = kx: –2 = k(–5), so k = and y = x. 1 4 1 4 2 5 2 5 12-1

4. 72 x The equation of the inverse variation is xy = 72 or y = . Inverse Variation ALGEBRA 1 LESSON 12-1 Suppose y varies inversely with x, and y = 9 when x = 85. Write an equation for the inverse variation. xy = kUse the general form for an inverse variation. (8)(9) = kSubstitute 8 for x and 9 for y. 72 = kMultiply to solve for k. xy = 72 Write an equation. Substitute 72 for k in xy = k. 12-1

5. Inverse Variation ALGEBRA 1 LESSON 12-1 The points (5, 6) and (3, y) are two points on the graph of an inverse variation. Find the missing value. x1 • y1 = x2 • y2Use the equation x1 • y1 = x2 • y2 since you know coordinates, but not the constant of variation. 5(6) = 3y2Substitute 5 for x1, 6 for y1, and 3 for x2. 30 = 3y2Simplify. 10 = y2Solve for y2. The missing value is 10. The point (3, 10) is on the graph of the inverse variation that includes the point (5, 6). 12-1

6. Relate:  A weight of 130 lb is 5 ft from the fulcrum.A weight of 93 lb is x ft from the fulcrum.Weight and distance vary inversely. Define:  Let weight1 = 130 lbLet weight2 = 93 lbLet distance1 = 5 ftLet distance2 = x ft Inverse Variation ALGEBRA 1 LESSON 12-1 Jeff weighs 130 pounds and is 5 ft from the lever’s fulcrum. If Tracy weighs 93 pounds, how far from the fulcrum should she sit in order to balance the lever? 12-1

7. Write: weight1 • distance1 = weight2 • distance2 130 • 5 = 93 • xSubstitute. 650 93 = xSolve for x. 6.99 = xSimplify. Inverse Variation ALGEBRA 1 LESSON 12-1 (continued) 650 = 93xSimplify. Tracy should sit 6.99, or 7 ft, from the fulcrum to balance the lever. 12-1

8. a. xy 3 10 5 6 10 3 Check each product xy. xy: 3(10) = 30    5(6) = 30    10(3) = 30 Inverse Variation ALGEBRA 1 LESSON 12-1 Decide if each data set represents a direct variation or an inverse variation. Then write an equation to model the data. The values of y seem to vary inversely with the values of x. The product of xy is the same for all pairs of data. So, this is an inverse variation, and k = 30. The equation is xy = 30. 12-1

9. y x b. xy 2 3 4 6 8 12 Check each ratio . 12 8 6 4 = 1.5 y x 3 2 = 1.5 = 1.5 y x The ratio is the same for all pairs of data. Inverse Variation ALGEBRA 1 LESSON 12-1 (continued) The values of y seem to vary directly with the values of x. So, this is a direct variation, and k = 1.5. The equation is y = 1.5x. 12-1

10. cost souvenirs Since the ratio is constant at \$10 each, this is a direct variation. Inverse Variation ALGEBRA 1 LESSON 12-1 Explain whether each situation represents a direct variation or an inverse variation. a. You buy several souvenirs for \$10 each. The cost per souvenir times the number of souvenirs equals the total cost of the souvenirs. b. The cost of a \$25 birthday present is split among several friends. The cost per person times the number of people equals the total cost of the gift. Since the total cost is a constant product of \$25, this is an inverse variation. 12-1

11. Inverse Variation ALGEBRA 1 LESSON 12-1 pages 640–642  Exercises 1. xy = 18 2.xy = 2 3.xy = 56 4.xy = 1.5 5.xy = 24 6.xy = 7.7 7. xy = 2 8.xy = 0.5 9.xy = 0.06 10. 8 11. 15 12. 6 13. 7 14. 3 15. 130 16. 12 17. 96 18. 3125 19. 2 20. 21. 20 22. 3 h 23. 13.3 mi/h 24. direct variation; y = 0.5x 25. inverse variation; xy = 60 26. inverse variation; xy = 72 27. Direct variation; the ratio is constant at \$1.79. 28. Inverse variation; the total number of slices is constant at 8. 29. Inverse variation; the product of the length and width remains constant with an area of 24 square units. 30. 32; xy = 32 31. 1.1; rt = 1.1 32. 2.5; xy = 2.5 cost pound 1 6 12-1

12. 33. 1; ab = 1 34. 15.6; pq = 15.6 35. 375; xy = 375 36. Direct variation; the ratio of the perimeter to the side length is constant at 3. 37. Inverse variation; the product of the rate and the time is always 150. 38. Direct variation; the ratio of the circumference to the radius is constant at 2 . 39. 121 ft 40. 2.4 days Inverse Variation ALGEBRA 1 LESSON 12-1 41. direct variation; y = 0.4x; 8 42. direct variation; y = 70x; 0.9 43. inverse variation; xy = 48; 0.5 44. a. greater b. greater c. less 45. a. 16 h; 10 h; 8 h; 4 h b. hr worked, rate of pay c.rt = 80 46. Check students’ work. 47. 10.2 L 48.p: y = 0.5x; q: xy = 8 49.a.y is doubled. b.y is halved. 12-1

13. k d2 x4y z Inverse Variation ALGEBRA 1 LESSON 12-1 50. 4; s d = sd2 = k, so s = 4 . 51.a.x4y = k b. = k 52. C 53. F 54.[2] Direct variation: y = kx, 10 = 5k, k = 2. So when x = 8, y = 2 • 8 = 16. Inverse variation: xy = k, 5 • 10 = 50, So when x = 8, y = , or 6.25. [1] no work shown OR one computational error 1 2 1 4 2 55. [4] a. b. The variables speed and time are inversely related. c. 2 h [3] one computational error [2] one part missing [1] two parts missing 56. 57. 1 2 50 8 8 17 15 17 12-1

14. Inverse Variation ALGEBRA 1 LESSON 12-1 58. 59. 60. 61. 62. 5.1 63. 12.0 64. 12.0 65. 10.6 66. 2.2 67. 2.5 68. (3a – 1)(a + 4) 69. (5x + 2)(3x + 7) 70. (2y – 3)(y + 8) 15 17 8 17 8 15 15 8 12-1

15. xy 1 2 6 1 3 1 6 1 3 xy = 1 9 3 1 18 Inverse Variation ALGEBRA 1 LESSON 12-1 1. The points (5, 1) and (10, y) are on the graph of an inverse variation. Find y. 2. Find the constant of variation k for the inverse variation where a = 2.5 when b = 7. 0.5 17.5 3. Write an equation to model the data and complete the table. 4. Tell whether each situation represents a direct variation or an inverse variation. a. You buy several notebooks for \$3 each. b. The \$45 cost of a dinner at a restaurant is split among several people. direct variation Inverse variation 12-1

16. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 (For help, go to Lessons 5-3, 8-7, and 10-1.) Evaluate each function for x = –2, 0, 3. 1. ƒ(x) = x – 8 2.g(x) = x2 + 4 3.y = 3x Graph each function. 4. ƒ(x) = 2x + 1 5.g(x) = –x26.y = 2x 12-2

17. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 Solutions 1. ƒ(x) = x – 8 for x = –2, 0, 3: ƒ(–2) = –2 – 8 = –10 ƒ(0) = 0 – 8 = –8 ƒ(3) = 3 – 8 = –5 2.g(x) = x2 + 4 for x = –2, 0, 3: g(–2) = (–2)2 + 4 = 4 + 4 = 8 g(0) = 02 + 4 = 0 + 4 = 4 g(3) = 32 + 4 = 9 + 4 = 13 3.y = 3x for x = –2, 0, 3: y = 3–2 = = = y = 30 = 1 y = 33 = 3 • 3 • 3 = 9 • 3 = 27 1 9 1 32 1 3 • 3 12-2

18. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 Solutions (continued) 4. ƒ(x) = 2x + 1 5. g(x) = – x2 6.y = 2x 12-2

19. rt 15 4.67 20 3.5 30 2.33 40 1.75 60 1.17 Step 1: Make a table of values. Step 2:Plot the points. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 The function t = models the time it will take you to travel 70 miles at different rates of speed. Graph this function. 70 r 12-2

20. 4 x – 3 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 Identify the vertical asymptote of y = . Then graph the function. Step 1:Find the vertical asymptote. x – 3 = 0 The numerator and denominator have no common factors. Find any value(s) where the denominator equals zero. x = 3 This is the equation of the vertical asymptote. 12-2

21. Step 3: Graph the function. xy 2 –4 1 –2 0 –1 –1 4 4 5 2 6 4 3 – 4 3 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 (continued) Step 2:Make a table of values. Use values of x near 3, the asymptote. 12-2

22. xy –10 2 –8 2 –6 1 –5 –1 –3 7 –2 5 –1 4 2 3 1 3 Step 2:  Make a table of values using values of x near –4. 1 3 2 3 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 4 x + 4 Identify the asymptotes of y = + 3. Then graph the function. Step 1: From the form of the function, you can see that there is a vertical asymptote at x = –4 and a horizontal asymptote at y = 3. Sketch the asymptotes. 12-2

23. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 (continued) Step 3:Graph the function. 12-2

24. x 6 a.y = 1 6 The graph is a line with slope and y-intercept 0. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 Describe the graph of each function. b.y = 6x The graph is of exponential growth. c.y = 6x2 The graph is a parabola with axis of symmetry at x = 0. d.y = | x – 6| The graph is an absolute value function with a vertex at (6, 0). 12-2

25. 6 x e.y = f.y = x – 6 The graph is the radical function y = x shifted right 6 units. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 (continued) The graph is a rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 0. g.y = 6x The graph is a line with slope 6 and y-intercept 0. 12-2

26. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 pages 648–650  Exercises 1. 2. 5. 6. 0 7. 2 8. –2 9. 2 10. x = 2, y = 0 11. x = –1, y = 0 12. x = 1, y = –1 13. x = 0, y = 2 3. 4. 12-2

27. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 14.x = 0; 15.x = 0; 16.x = –1; 17.x = 5; 18.x = –4; 19.x = –4; 12-2

28. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 20.x = 0, y = –5; 21.x = 0, y = 5; 22.x = 0, y = –6; 23. x = –1, y = 4; 24.x = 3, y = –5; 25.x = 1, y = –2; 12-2

29. 36. moves graph 3 units to the right 37. lowers graph 15 units 38. moves graph 12 units left 39. moves the graph up 12 units 40. moves the graph left 3 units 41. moves the graph down 2 units 42. moves the graph 3 units left and 2 units down 26. line with slope 4, y-int. 1 27. absolute value function with vertex (4, 0) 28. exponential decay 29. line with slope , y-int. 0 30. rational function, with asymptotes x = 0, y = 1 31. radical function; y = x shifted right 4, up 1 32. parabola with axis of symmetry x = 0 33. rational function with asymptotes x = –4, y = –1 34. parabola with axis of symmetry x = – 35. moves graph 1 unit to the left 1 4 1 4 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 12-2

30. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 43.x = 0, y = 0; 44.x = 0, y = 0; 45.x = –4, y = 0; 46.x = 0, y = 1; 47.x = –1, y = 4; 48.x = –1, y = –3; 12-2

31. 49.x = 1, y = 3; 50.x = –5, y = 1; 51.x = 3, y = –2; 52. Answers may vary. Sample: ƒ(x) = + 3, g(x) = 53. 17.8 lumens; 1.97 lumens 54.a. b.x = 0, y = 0; x = 0, y = 0 c.y is any real number except 0; y > 0. 1 x 1 x Graphing Rational Functions ALGEBRA 1 LESSON 12-2 12-2

32. > – Graphing Rational Functions ALGEBRA 1 LESSON 12-2 55.a. d 40 b. 16; 1600; 160,000 c. The signal is extremely strong when you are in the immediate vicinity of a transmitter and it will interfere with the other station. 56. The graph of y = and y = – are both composed of two curves with asymptotes x = 0 and y = 0. The graph of y = – is a reflection of the graph of y = over the y-axis. 57. 58. 3 x 3 x 3 x 3 x 12-2

33. 1 x + 3 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 59. 60. 61.a.x = –3, y = –2 b.y = – 2 62. No; ƒ(x) = is equivalent to g(x) = x + 1 for all values except x = –2. 63. C 64. I 65. A 66. C (x + 2)(x + 1) x + 2 12-2

34. Graphing Rational Functions ALGEBRA 1 LESSON 12-2 67.[2] a. The graph of g(x) = + 5 is a translation of ƒ(x) = 5 units up and 1 unit right. b.x = 1 and y = 5 [1] one part answered correctly 68.xy = 21 69.xy = 16 70.xy = 22 71. xy = 21.08 72. 0 73. 1 74. 2 75. 3(d – 6)(d + 6) 76. 2(m – 12)(m + 5) 77. (t2 + 3)(t – 1) 4 x – 1 4 x 12-2

35. 3 x + 2 1. Identify the vertical asymptote of y = . Then graph the function. 1 x – 2 2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function. 3. Describe the graph of each function. a.y = x + 3 b.y = c.y = + 2 d.y = 3x2 x 3 3 x Graphing Rational Functions ALGEBRA 1 LESSON 12-2 12-2

36. 3 x + 2 1. Identify the vertical asymptote of y = . Then graph the function. 1 x – 2 2. Identify the vertical and horizontal asymptotes of y = – 3. Then graph the function. x = –2 x = 2; y = –3 Graphing Rational Functions ALGEBRA 1 LESSON 12-2 12-2

37. 1 3 line with slope and y-intercept 0 radical function y = x shifted left 3 units Graphing Rational Functions ALGEBRA 1 LESSON 12-2 3. Describe the graph of each function. a.y = x + 3 b.y = c.y = + 2 d.y = 3x2 x 3 3 x rational function with vertical asymptote at x = 0 and horizontal asymptote at y = 2 parabola with axis of symmetry at x = 0. 12-2

38. 15 24 – Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 (For help, go to Lessons 9-5 and 9-6.) Write each fraction in simplest form. 8 2 25 35 1. 2. 3. Factor each quadratic expression. 4.x2 + x – 12 5.x2 + 6x + 8 6.x2 – 2x – 15 7.x2 + 8x + 16 8.x2 – x – 12 9.x2 – 7x + 12 12-3

39. 5 • 5 5 • 7 3 • 5 3 • 8 25 35 5 7 3. = = 15 24 – Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 Solutions 8 2 5 8 1. = 8 ÷ 2 = 4 2. = – = – 4. Factors of –12 with a sum of 1: 4 and –3.x2 + x – 12 = (x + 4)(x – 3) 5. Factors of 8 with a sum of 6: 2 and 4.x2 + 6x + 8 = (x + 2)(x + 4) 6. Factors of –15 with a sum of –2: 3 and –5.x2 – 2x – 15 = (x + 3)(x – 5) 12-3

40. Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 Solutions (continued) 7. Factors of 16 with a sum of 8: 4 and 4.x2 + 8x + 16 = (x + 4)(x + 4) or (x + 4)2 8. Factors of –12 with a sum of –1: 3 and –4.x2 – x – 12 = (x + 3)(x – 4) 9. Factors of 12 with a sum of –7: –3 and –4.x2 – 7x + 12 = (x – 3)(x – 4) 12-3

41. Factor the numerator. The denominator cannot be factored. 3x + 9 x + 3 3(x + 3) x + 3 = = Divide out the common factor x + 3. 1 3(x + 3) x + 3 1 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 3x + 9 x + 3 Simplify . = 3 Simplify. 12-3

42. 4x – 20 x2 – 9x + 20 = Factor the numerator and the denominator. 4(x – 5) (x – 4) (x – 5) 4(x – 5) (x – 4) (x – 5) 4x – 20 x2 – 9x + 20 = Divide out the common factor x – 5. 1 1 = Simplify. 4 x – 4 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 Simplify . 12-3

43. 3x – 27 81 – x2 3(x – 9) (9 – x) (9 + x) = Factor the numerator and the denominator. 3(x – 9) – 1 (x – 9) (9 + x) = Factor –1 from 9 – x. = Divide out the common factor x – 9. 1 3(x – 9) – 1 (x – 9) (9 + x) 1 3 9 + x = – Simplify. Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 3x – 27 81 – x2 Simplify . 12-3

44. = Substitute 8 for r and 3 for h. 30rh r + h 720 11 = Simplify. 60 • volume surface area 30rh r + h 30 (3) (8) 3 + 8 Round to the nearest whole number. 65 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 The baking time for bread depends, in part, on its size and shape. A good approximation for the baking time, in minutes, of a cylindrical loaf is , or , where the radius r and the length h of the baked loaf are in inches. Find the baking time for a loaf that is 8 inches long and has a radius of 3 inches. Round your answer to the nearest minute. The baking time is approximately 65 minutes. 12-3

45. pages 654–656  Exercises 1. 2. 3. 4. 5. 3x 6. 7. 8. 9. 10. 11. 12. 13. 14.b + 3 15. 16. –1 17. 18. –2 19. – 20. – 21. – 22. 36 min 23. 13 min 24. 13 min 25. 26. 27. 28. 29. 1 v + 5 w w – 7 2a + 3 4 1 w – 4 a + 1 5 1 7x m + 3 m + 2 1 3 c – 4 c + 3 1 2 2r – 1 r + 5 1 m – 2 x + 2 x2 7z + 2 z – 1 2 3 –4 t + 1 5t – 4 3t – 1 2 b + 4 4a2 2a – 1 1 m – 7 1 2 3(z + 4) z3 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 12-3

46. 3y 4(y + 4) m – n m + 10n x2 – 9 x + 3 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 30. 31. – 32. 33. 34. Answers may vary. Sample: 35. a. i.  ii.  b. ; 36. The student canceled terms instead of factors. 37. –3 is not in the domain of . 38. 39. 40. 41. 42. 43. 44. 45. sometimes 46. sometimes 47. never 48. C 49. I 50. B 51. D 52. C 53.[2] The student put the 4 in the numerator rather than in the denominator. = = [1] no explanation OR incorrectly simplified expression 2s + 1 s2 5w 5w + 6 2a + 1 a + 3 1 4 4 + 3m m – 7 t + 3 3(t + 2) –c(3c + 5) 5c + 4 3 (x – 2)(x + 3) a – 3b a + 4b 2b + 4h bh 2h + 2r rh 6v – 7w 3v – 2w x – 5 4x – 20 4 9 4 9 x – 5 4(x – 5) 1 4 12-3

47. 56. vertical asymptote: x = 0; horizontal asymptote: y = –4; 57. 10 2 58.a2b3c4 b 59. 2 2 60. 61. y = x2, y = –2x2, y = 3x2 62.y = x2, y = x2, y = x2 63.y = 0.5x2, y = 2x2, y = –4x2 64.y = –x2, y = 2.3x2, y = –3.8x2 54. vertical asymptote: x = 0; horizontal asymptote: y = 2; 55. vertical asymptote: x = 4; horizontal asymptote: y = 0; 2 5m2 1 4 1 3 2 5 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 12-3

48. x2 + 8x x2 – 64 –1 x + 6 x x – 8 6x2 – x – 12 8x2 – 10x – 3 x – 6 36 – x2 4x2 + x x3 Simplifying Rational Expressions ALGEBRA 1 LESSON 12-3 Simplify each expression. 3. 1. 2. 6 – 2x x – 3 –2 4. 5. 4x + 1 x2 3x + 4 4x + 1 12-3

49. Multiplying and Dividing Rational Expressions ALGEBRA 1 LESSON 12-4 (For help, go to Lessons 8-3 and 9-6.) Simplify each expression. 1.r2 • r82.b3 • b43.c7 ÷ c2 4. 3x4 • 2x55. 5n2 • n26. 15a3 (–3a2) Factor each polynomial. 7. 2c2 + 15c + 7 8. 15t2 – 26t + 11 9. 2q2 + 11q + 5 12-4

50. Multiplying and Dividing Rational Expressions ALGEBRA 1 LESSON 12-4 Solutions 1.r2 • r8 – r(2 + 8) = r102.b3 • b4 = b(3+4) = b7 3.c7÷c2 = c(7 – 2) = c54. 3x4 • 2x5 = (3 • 2)(x4 • x5) = 6x(4 + 5) = 6x9 5. 5n2 • n2 = 5(n2 • n2) = 5n(2 + 2) = 5n4 6. 15a3(–3a2) = 15(–3)(a3 • a2) = –45a(3 + 2) = –45a5 7. 2c2 + 15c + 7 = (2c + 1)(c + 7)Check: (2c + 1)(c + 7) = 2c2 + 14c + 1c + 7= 2c2 + 15c + 7 8. 15t2– 26t + 11 = (15t– 11)(t– 1)Check: (15t– 11)(t– 1) = 15t2– 15t– 11t + 11= 15t2– 26t + 11 9. 2q2 + 11q + 5 =(2q + 1)(q + 5)Check: (2q + 1)(q + 5) = 2q2 + 10q + 1q + 5= 2q2 + 11q + 5 12-4