1 / 102

Inverse Variation

Inverse Variation. ALGEBRA 2 LESSON 9-1. (For help, go to Lesson 2-3.). In Exercises 1–3, y varies directly with x . 1. Given that x = 2 when y = 4, find y when x = 5. 2. Given that x = 1 when y = 5, find y when x = 3.

arnold
Download Presentation

Inverse Variation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Inverse Variation ALGEBRA 2 LESSON 9-1 (For help, go to Lesson 2-3.) In Exercises 1–3, y varies directly with x. 1. Given that x = 2 when y = 4, find y when x = 5. 2. Given that x = 1 when y = 5, find y when x = 3. 3. Given that x = 10 when y = 3, find y when x = 4. 9-1

  2. Inverse Variation ALGEBRA 2 LESSON 9-1 Solutions 1.y = kx with x = 2 and y = 4: 4 = k(2), so k = 2 y = 2x with x = 5: y = 2(5) = 10 2.y = kx with x = 1 and y = 5: 5 = k(1), so k = 5 y = 5x with x = 3: y = 5(3) = 15 3.y = kx with x = 10 and y = 3: 3 = k(10), so k = 0.3 y = 0.3x with x = 4: y = 0.3(4) = 1.2 9-1

  3. k x y = x and y vary inversely. k 7 4 = Substitute the given values of x and y. 28 x y = Use the value of k to write the function. Inverse Variation ALGEBRA 2 LESSON 9-1 Suppose that x and y vary inversely, and x = 7 when y = 4. Write the function that models the inverse variation. 28 = kFind k. 9-1

  4. = / a. x –2 4 6 y 5 –10 –15 x –2 –1.3 7 y 6 5 –4 x 2 4 14 y 0.7 0.35 0.1 So xy = 1.4 and the function is y = . 1.4 x b. Not all the products of x and y are the same (–2 • 6 –1.3 • 5). c. Inverse Variation ALGEBRA 2 LESSON 9-1 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. The product of each pair of x- and y-values is 1.4. 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. This is neither a direct variation nor an inverse variation. As x increases, y decreases. 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. 9-1

  5. Pressure (lb/in.2) Volume (in.3) 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 3 32 5 19.2 8 12 For each of the three states of gas in the table, PV 96. Inverse Variation ALGEBRA 2 LESSON 9-1 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 PV = 96 Substitute 96 for k. P(6) = 96 Substitute 6 for V. P = 16 A gas’s pressure that is contained in a volume of 6 in.3 is 16 lb/in.2. 9-1

  6. 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. 2k v2 mvaries directly as the kinetic energy k. mvaries inversely as the square of the velocity v. 2k v2 m = Inverse Variation ALGEBRA 2 LESSON 9-1 9-1

  7. 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. 3 3 = k(2)2Substitute the values for A and r. 3 3 4 3 3 4 = kSolve for k. A = r2Substitute the value for k. Inverse Variation ALGEBRA 2 LESSON 9-1 A = kr2A varies directly as the square of r. 9-1

  8. Inverse Variation ALGEBRA 2 LESSON 9-1 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 = 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. 1 x 11 x 100 x 1300 x 1 x 80 x 5 3x 1 6 56 x 3.6 x 250 x 42 x 0.3 x 9-1

  9. 3x2 y 300 r 2 Inverse Variation ALGEBRA 2 LESSON 9-1 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. 14,000 b. 226 29. 18 30. 3.6 31. 32. 6 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 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 20 9 5x y m d2 2 3 16 3 3 16 4 xy 1 9 40 3 300 d 1 4 9-1

  10. x1 x2 y2 y1 Inverse Variation ALGEBRA 2 LESSON 9-1 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 . 705 w h2 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 1 4 9-1

  11. e5 4 e2 8 Inverse Variation ALGEBRA 2 LESSON 9-1 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| 3 4 9-1

  12. 8 x y = 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 96 x inverse variation; y = 70x y2 z = , 3.5 Inverse Variation ALGEBRA 2 LESSON 9-1 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 = r2h. 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. 0.25 neither direct variation; y = 7x V varies jointly as the square of r and h. 9-1

  13. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 (For help, go to Lesson 2-6.) 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 9-2

  14. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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. 9-2

  15. 0.5 x Draw a graph of y = . 1 4 1 10 1 10 1 4 1 2 1 2 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 Graph the points. Connect them with a smooth curve. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 Make a table of values that includes positive and negative values of x. Notice that x cannot be 0. 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. 9-2

  16. 1 x 0.25 x Compare the graphs of y = and y = shown below. 1 x 0.25 x The branches of y = are closer to the axis than are the branches of . Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 What points on the graphs are closest to the origin? The axes are the asymptotes for both graphs. Both graphs are symmetric with respect to y = x and y = –x. The points, (1, 1), (–1, –1), (0.5, 0.5), and (–0.5, –0.5) are closest to the origin. 9-2

  17. 2 x 2 x Compare the graphs of y = and y = – shown below. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 Both graphs are symmetric with respect to y = x and y = –x. Each is a 90° rotation of the other about the origin. 9-2

  18. 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. 343 w Use the Intersection feature. 343 w Graph the functions f = and f = 440. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 The wavelength is about 0.78 m/s. 9-2

  19. 1 x + 2 Sketch the graph of y = + 2. 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 1 x Step 2: Translate y = . The graph y = includes (1, 1) and (–1, –1). 1 x Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 Translate these points 2 units to the left and 2 units up to (–1, 3) and (–3, 1). Draw the branches through these points. 9-2

  20. 7 x Write an equation for the translation of y = – that has asymptotes at x = 8 and y = –4. – 7 x – b y = + cUse the general form of a translation. – 7 x – 8 = + 4Substitute 8 for b and 4 for c. – 7 x – 8 = – 4 Simplify. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 9-2

  21. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 pages 488–490  Exercises 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. 3 x 5 x 1 x 0.2 x 9-2

  22. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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. 10. 18.4 ft 11. 7.67 ft 12. 3.83 ft 13. 1.84 ft 14. 8 x 8 x 2 x 12 x 12 x 9-2

  23. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 15. 16. 17. 18. 19. 20. 9-2

  24. 21. 22.y = + 4 23.y = + 3 24.y = – 8 2 x 2 x + 2 2 x – 4 –8.3 x Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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. 750 a 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 = 0.5 x 0.75 x –0.01 x 4 x –1.4 x 9-2

  25. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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. k x 9-2

  26. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 44. (–0.45, –10) and (0.45, –10) 45. (3.76, 4.2) 46. (–0.76, 9) and (0.76, 9) 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) 9-2

  27. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 10,000 g 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. 1 x 47.a.m = b.m = c. 25 mi/gal, 28.57 mi/gal 1 x 1 x 1 x2 1 x 10,000 g – 50 1 x2 1 x 9-2

  28. 0.6 x – 2 Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 –0.25 x – 0.5 b.y = c.y = – 2 d.y = 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 1 x 1 x 16 x 16 x 1 x – 1 9-2

  29. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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 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) [1] answer only, with no explanation 60.V varies jointly with the square of s and h. –3 x –3 x –b –3 x – (–5) –3 x + 5 3 x – 2 3 x – 2 –3 x – 13 9-2

  30. Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 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 9-2

  31. 6 x 12 x 2. Compare the graphs of y = and y = . 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. 12 x 6 x 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. 1200 n 4. Sketch the asymptotes and the graph of y = – – . 0. 5 x + 2 1 2 13 x – 5 y = – 8 Graphing Inverse Variations ALGEBRA 2 LESSON 9-2 12 x 1. Draw a graph of y = . $150 5. Write an equation for the translation of y = that has asymptotes at x = 5 and y = –8. 13 x 9-2

  32. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 (For help, go to Lessons 5-4 and 5-5.) 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

  33. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 Solutions 1. Factors of 6 with a sum of 5: 3 and 2 x2 + 5x + 6 = (x + 3)(x + 2) 3. Factors of 27 with a sum of –12: –3 and –9 x2 – 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 = 0 x = –4 or x = 3 9.x2 – 9x + 18 = 0 (x – 3)(x – 6) = 0 x – 3 = 0 or x – 6 = 0 x = 3 or x = 6 2. Factors of 8 with a sum of –6: –4 and –2 x2 – 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 = 0 x = –4 or x = 7 9-3

  34. 3 x2 – x –12 a. y = Rational Functions and Their Graphs ALGEBRA 2 LESSON 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

  35. 4 3 x2 = – Solve for x. ± 2i 3 ± –4 3 x = = 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 Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 (continued) 2x 3x2 + 4 b.y = The function is undefined at values of 3x2 + 4 = 0. 3x2 + 4 = 0 Set the denominator equal to zero. 9-3

  36. x – 7 (x + 1)(x + 5) a.y = (x + 3)x x + 3 b.y = (x – 6)(x + 9) (x + 9)(x + 9)(x – 6) c.y = 1 (x + 9) , 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. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 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. 6 is a zero of both the numerator and the denominator. 9-3

  37. Divide the numerator by the denominator as shown at the right. –4x + 3 2x + 1 –2 2x + 1 –4x + 3 –(–4x – 2) 5 The function y = 5 2x + 1 can be written as y = – 2. 5 2x + 1 . Its graph is a translation of y = Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 –4x + 3 2x + 1 . Find the horizontal asymptote of y = The horizontal asymptote is y = –2. 9-3

  38. 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. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 x + 1 (x – 3)(x + 2) Sketch the graph y = . 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. 9-3

  39. 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 = • 100 3 + 0.5x 20 + x Rational Functions and Their Graphs ALGEBRA 2 LESSON 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. 9-3

  40. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 (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

  41. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 pages 495–498  Exercises 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 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 3 2 7 2 2 3 1 2 9-3

  42. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 24.y = 25. 26. 3 4 27. 28. 29. 30. 9-3

  43. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 0.19x + 210,000 x – 500 31.a.y = b. $46.88; $14.68 c. more than 21,916 discs d.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 35. 36. 9-3

  44. Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 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 9-3

  45. 4n2 4n + 1 M(x) A(x) Rational Functions and Their Graphs ALGEBRA 2 LESSON 9-3 20,000x + 210,000 x + 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 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. 64 17 9-3

  46. Rational Functions and Their Graphs ALGEBRA 2 LESSON 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. [1] answer only, with no explanation 48. B 49. A 50. C 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 5 5x + 7 x + 3 –8 x + 3 9-3

  47. 1 3 x2 + 9 3x2 + 5 y = x2 – 9 x(x2 + 2x – 15) 30,000 + 10x + 0.0001x2 x ƒ(x) = Rational Functions and Their Graphs ALGEBRA 2 LESSON 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 y = 2, x = –7 x + 7 (x – 3)(x + 5) vertical asymptote at x = 3; hole at x = –5 $13.70; $13.50 9-3

  48. 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.  ÷ 3 8 5 6 1 2 4 6 8 3 2 16 2 5 3 7 5 8 3 4 1 2 9 16 3 4 5 4 15 8 Rational Expressions ALGEBRA 2 LESSON 9-4 (For help, go to Lesson 5-4 and Skills Handbook page 843.) 9-4

  49. 6. • = = = 8. • = = 10. ÷ = • = = = or 1 12. ÷ = • = = = 8 3 2 16 8 • 2 3 • 16 1 • 16 3 • 16 1 3 2 5 3 7 2 • 3 5 • 7 6 35 / 1 2 / 5 32 5 8 5 8 1 4 5 • 1 8 • 4 / 3 • 5 8 • 3 • 2 3 8 5 6 3 • 5 8 • 6 5 16 5. • = = = 7. • = = = 9. ÷ 4 = • = = 11. ÷ = • = = = / 3 4 1 2 3 4 2 1 3 • 2 4 • 1 3 • 2 2 • 2 • 1 3 2 1 • 4 2 • 6 / / 1 2 4 6 1 • 2 • 2 2 • 2 • 3 1 3 / / 3 4 9 16 3 4 9 16 4 3 9 • 4 16 • 3 / / 3 • 3 • 4 4 • 4 • 3 / / 5 4 15 8 5 4 8 15 5 • 8 4 • 15 5 • 2 • 4 4 • 5 • 3 2 3 / / / Rational Expressions ALGEBRA 2 LESSON 9-4 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 2. 4x2 – 9 = (2x)2 – 32 = (2x + 3)(2x – 3) 4. 10x2 – 10 = 10(x2 – 1) = 10(x2 – 12) = 10(x + 1)(x – 1) 9-4

  50. x2 – 6x – 16 x2 + 5x + 6 x2 – 6x – 16 x2 + 5x + 6 (x – 8)(x + 2) (x + 3)(x + 2) = Factor the polynomials. Notice x = –3 or –2. / 1 (x – 8)(x + 2) (x + 3)(x + 2) = Divide out common factors. 1 x + 8 x + 3 = x + 8 x + 3 The simplified expression for x = –3 or –2. / Rational Expressions ALGEBRA 2 LESSON 9-4 Simplify . State any restrictions on the variable. The restrictions on x are needed to prevent the denominator of the original expression from being zero. 9-4

More Related