2. Copyright © 2004 Pearson Education, Inc. Chapter 3 Polynomial and Rational Functions

3. Copyright © 2004 Pearson Education, Inc. 3.1 Quadratic Functions and Models

4. Slide 3-4 Copyright © 2004 Pearson Education, Inc. Polynomial Function A polynomial function of degree n, where n is a nonnegative integer, is a function defined by an expression of the form where a n, a n  1, …a 1, and a 0 are real numbers, with a n  0. Examples

5. Slide 3-5 Copyright © 2004 Pearson Education, Inc. Quadratic Functions A function f is a quadratic function if f(x) = ax 2 + bx + c, where a, b, and c are real numbers with a  0. The simplest quadratic function is f(x) = x 2. This is the graph of a parabola. The line of symmetry for a parabola is called the axis of the parabola. The vertex is the point where the axis intersects the parabola.

6. Slide 3-6 Copyright © 2004 Pearson Education, Inc. Examples Graph each function. Give the domain and range. a) b) c)

7. Slide 3-7 Copyright © 2004 Pearson Education, Inc. Solutions a) 26 33 5 77 3 66 2 33 1 20 yx The domain is ( ,  ). The range is [  7,  ) The vertex is (3,  7)

8. Slide 3-8 Copyright © 2004 Pearson Education, Inc. Solutions continued b) The graph is broader than y = x 2 it is reflected across the x-axis. The vertex is (0, 0) The domain is ( ,  ) The range is ( , 0].

9. Slide 3-9 Copyright © 2004 Pearson Education, Inc. Solutions continued c) The graph is translated 2 units to the left and 3 units down. The vertex is (  2,  3). The domain is ( ,  ) The range is ( ,  3].

10. Slide 3-10 Copyright © 2004 Pearson Education, Inc. Completing the Square Graph f(x) = x 2  4x + 7 by completing the square and locating the vertex. Solution: The vertex is (2, 3) and the axis is the line x = 2.

11. Slide 3-11 Copyright © 2004 Pearson Education, Inc. Completing the Square continued Find additional ordered pairs that satisfy the equation. Plot the points and connect in a smooth curve. 74 43 32 41 70 yx

12. Slide 3-12 Copyright © 2004 Pearson Education, Inc. Example Graph f(x) =  3x 2 + 6x  1 by completing the square and locating the vertex.

13. Slide 3-13 Copyright © 2004 Pearson Education, Inc. Example continued The vertex is (1, 2). Find additional points by substituting x-values into the original equation. 11 2 21 11 0 yx

14. Slide 3-14 Copyright © 2004 Pearson Education, Inc. Graph of a Quadratic Function The quadratic function defined by f(x) = ax 2 + bx + c can be written in the form y = f(x) = a(x  h) 2 + k, a  0, where The graph of f has the following characteristics. 1.It is a parabola with vertex (h, k) and the vertical line x = h as axis.

15. Slide 3-15 Copyright © 2004 Pearson Education, Inc. Graph of a Quadratic Function continued 2.It opens up if a > 0 and down if a < 0. 3.It is broader than the graph of y = x 2 if |a| 1. 4.The y-intercept is f(0) = c. 5.If b 2  4ac  0, the x-intercepts are If b 2  4ac < 0, there are no x-intercepts.

16. Slide 3-16 Copyright © 2004 Pearson Education, Inc. Example Find the axis and vertex of the parabola having equation f(x) = 4x 2 + 8x + 3 using the formula. Solution: Here a = 4, b = 8, and c = 3. The axis of the parabola is the vertical line The vertex is (  1,  1).

17. Copyright © 2004 Pearson Education, Inc. 3.2 Synthetic Division

18. Slide 3-18 Copyright © 2004 Pearson Education, Inc. Division Algorithm Let f(x) and g(x) be polynomials with g(x) of lower degree than f(x) and g(x) of degree of one or more. There exists unique polynomials q(x) and r(x) such that f(x) = g(x)  q(x) + r(x) where either r(x) = 0 or the degree of r(x) is less than the degree of g(x).

19. Slide 3-19 Copyright © 2004 Pearson Education, Inc. Synthetic Division A shortcut method of performing long division with certain polynomials, called synthetic division, is used only when a polynomial is divided by a first-degree binomial of the form x  k, where the coefficient of x is 1. In the following example, the example on the right will show how the division process is simplified by omitting all variables and writing only coefficients, with 0 used to represent the coefficient of any missing terms. Since the coefficient of x in the divisor is always 1 in these divisions, it too can be omitted. These omissions simplify the problem.

20. Slide 3-20 Copyright © 2004 Pearson Education, Inc. Example 2x 3  4x 2 x 2 x 2  2x 2x  32 2x  4  28 2  4 1 1  2 2  32 2  4  28

21. Slide 3-21 Copyright © 2004 Pearson Education, Inc. Example continued The numbers that are repetitions of the numbers directly above them can also be omitted.  4 1  2 2  32  4  28 The entire problem can now be condensed vertically, and the top row of numbers can be omitted since it duplicates the bottom row if the 2 is brought down.  4  2  4 2 1 2  28 The rest of the bottom row is obtained by subtracting  4,  2, and  4 from the corresponding terms above them.

22. Slide 3-22 Copyright © 2004 Pearson Education, Inc. Example continued With synthetic division it is useful to change the sign of the divisor, so the  2 is changed to 2, which also changes the sign of the numbers in the second row. To compensate for this change, subtraction is changed to addition. Doing this gives the final result. 4 2 4 Signs changed 2 1 2  28 Quotient 2x 2 + x + 2  Remainder

23. Slide 3-23 Copyright © 2004 Pearson Education, Inc. Caution Note: To avoid errors, use 0 as coefficient for any missing terms, including a missing constant, when setting up the division.

24. Slide 3-24 Copyright © 2004 Pearson Education, Inc. Example Use synthetic division to divide 2x 3  3x 2  11x + 7 by x  3. Solution: Since x  3 in the form x  k use this and the coefficients of the polynomial to obtain Bring down the 2 and multiply (3)(2) = 6

25. Slide 3-25 Copyright © 2004 Pearson Education, Inc. Example continued Add  3 and 6 to obtain 3. Multiply: 3(3) = 9 6 9 2 3 Add  11 and 9, obtaining  2. Finally 3(  2) =  6 6 9  6 2 3  2

26. Slide 3-26 Copyright © 2004 Pearson Education, Inc. Example continued Add 7 and  6 to obtain 1. 6 9  6 2 3  2 1 Remainder Since the divisor x  k has degree 1, the degree of the quotient will always be one less than the degree of the polynomial to be divided. Thus

27. Slide 3-27 Copyright © 2004 Pearson Education, Inc. Remainder Theorem If a polynomial f(x) is divided by x  k, the remainder is f(k).  For example: In the synthetic division problem given previously, when f(x) = 2x 3  3x 2  11x + 7 was divided by x  3, the remainder was 1. Substituting 3 for x, in f(x) gives f(3) = 2(3) 3  3(3) 2  11(3) + 7 = 54  27  33 + 7 = 1

28. Slide 3-28 Copyright © 2004 Pearson Education, Inc. Applying the Remainder Theorem Example: Let f(x) = 3x 4  7x 3  4x + 5. Use the remainder theorem to find f(2). Solution: Use synthetic division with k = 2. 6  2  4  16 3  1  2  8  11 Remainder By this result, f(2) =  11

29. Slide 3-29 Copyright © 2004 Pearson Education, Inc. Testing Potential Zeros A zero of a potential function f is a number k such that f(k) = 0. The zeros are the x-intercepts of the graph of the function. The remainder theorem gives a quick way to decide if a number k is a zero of a polynomial function defined by f(x). Use synthetic division to find f(k); if the remainder is 0, then f(k) = 0 and k is a zero of f(x). A zero of f(x) is called a root or solution of the equation f(x) = 0.

30. Slide 3-30 Copyright © 2004 Pearson Education, Inc. Example Decide whether the given number k is a zero of f(x). f(x) = 2x 3  2x 2  34x  30 k =  1 Solution: Since the remainder is 0, f(  1) = 0, and  1 is a zero of the polynomial function defined by f(x) = 2x 3  2x 2  34x  30 f(x) = 2x 3 + 4x 2  x + 5 k =  2 Solution: Since the remainder is 7, not 0,  2 is not a zero of f(x) = 2x 3 + 4x 2  x + 5. In fact, f(  2 ) = 7.

31. Copyright © 2004 Pearson Education, Inc. 3.3 Zeros of Polynomial Functions

32. Slide 3-32 Copyright © 2004 Pearson Education, Inc. Factor Theorem By the remainder theorem, if f(k) = 0, then the remainder when f(x) is divided by x  k is 0. This means that x  k is a factor of f(x). Conversely, if x  k is a factor of f(x), then f(k) must equal 0. This is summarized in the following factor theorem. Factor Theorem The polynomial x  k is a factor of the polynomial f(x) if and only if f(k) = 0.

33. Slide 3-33 Copyright © 2004 Pearson Education, Inc. Example Determine whether x  3 is a factor of f(x) for f(x) = x 3  7x 2 + 11x + 3. Solution: By the factor theorem, x  3 will be the factor of f(x) only if f(3) = 0. Use synthetic division and the remainder theorem to decide. 3  12  3 1  4  1 0

34. Slide 3-34 Copyright © 2004 Pearson Education, Inc. Example continued Because the remainder is 0, x  3 is a factor. Additionally, we can determine from the coefficients in the bottom row that the other factor is x 2  4x  1, and f(x) = (x  3) (x 2  4x  1).

35. Slide 3-35 Copyright © 2004 Pearson Education, Inc. Rational Zeros Theorem The rational zeros theorem gives a method to determine all possible candidates for rational zeros of a polynomial function with integers. Rational Zeros Theorem If is a rational number written in lowest terms, and if is a zero of f, a polynomial function with integer coefficients, then p is a factor of the constant term and q is a factor of the leading coefficient.

36. Slide 3-36 Copyright © 2004 Pearson Education, Inc. Using the Rational Zeros Theorem Example: Do each of the following for the polynomial function defined by f(x) 6x 3  5x 2  7x + 4. List all possible rational zeros. Find all rational zeros and factors f(x) into linear factors.

37. Slide 3-37 Copyright © 2004 Pearson Education, Inc. Using the Rational Zeros Theorem continued Solution: For a rational number to be a zero, p must be a factor of a 0 = 4 and q must be a factor of a 4 = 6. Thus, p can be  1,  2,  4, and q can be  1,  2,  3, or  6. The possible rational zero, are Use the remainder theorem to show that  1 is a zero.  6 11  4 6  11 4 0

38. Slide 3-38 Copyright © 2004 Pearson Education, Inc. Using the Rational Zeros Theorem continued The new quotient for the polynomial is 6x 2  11x + 4. This factors to (3x  4)(2x  1). Setting 3x  4 = 0 and 2x  1= 0 yields zeros and Thus the rational zeros are and the linear factors of f(x) are x + 1, 2x  1, and 3x  4. Therefore, f(x) = (x + 1)(2x  1) (3x  4) = 6x 3  5x 2  7x + 4.

39. Slide 3-39 Copyright © 2004 Pearson Education, Inc. Number of Zeros Fundamentals of Algebra Every function defined by a polynomial of degree 1 or more has at least one complex zero. Number of Zeros Theorem A function defined by a polynomial of degree n has at most n distinct zeros  Example: f(x) = x 3 + 3x 2 + 3x + 1 = (x + 1) 3 is of degree 3, but has only one zero,  1. Actually, the zero  1 occurs three times, since there are three factors of x + 1; this zero is called a zero of multiplicity 3.

40. Slide 3-40 Copyright © 2004 Pearson Education, Inc. Example Find a function f defined by a polynomial of degree 3 that satisfies the given conditions. Zeros of  2, 3, and 4; f(1) = 3 Solution: These three zeros give x  (  2) = x + 2, x  3, x  4 as factors of f(x). Since f(x) is to be of degree 3, these are the only possible factors by the number of zeros theorem. Therefore, f(x) has the form f(x) = a(x + 2)(x  3)(x  4) for some real number a.

41. Slide 3-41 Copyright © 2004 Pearson Education, Inc. Example continued To find a, use the fact that f(1) = 3. f(1) = a(1 + 2)(1  3)(1  4) = 3 a(3)(  2)(  3) = 3 18a = 3 a = Thus, f(x) = (x + 2)(x  3)(x  4), or f(x) =

42. Slide 3-42 Copyright © 2004 Pearson Education, Inc. Conjugate Zeros Theorem The following Properties of complex conjugates are needed to prove the conjugate zeros theorem. Properties of Conjugates For any complex numbers c and d, Conjugate Zeros Theorem If f(x) is a polynomial having only real coefficients and if z = a + bi is a zero of f(x), where a and b are real numbers,then is also a zero of f(x).

43. Slide 3-43 Copyright © 2004 Pearson Education, Inc. Example Find a polynomial function of least degree having only real coefficients and zeros 3 and 5i. Solution: The complex number 5i also must be a zero, so the polynomial has at least three zeros, 3, 5i, and  5i. For the polynomial to be of least degree, these must be the only zeros. By the factor theorem there must be three factors, x  3, x  5i, and x + 5i, so f(x) = (x  3)(x  5i)(x + 5i) = (x  3)(x 2 + 25) = x 3  3x 2 + 25x  75. Any nonzero multiple also satisfies the given conditions.

44. Slide 3-44 Copyright © 2004 Pearson Education, Inc. Descartes’ Rule of Signs Let f(x) define a polynomial function with real coefficients and a nonzero constant term, with terms in descending powers of x. The number of positive real zeros of f either equals the number of variations in sign occurring in the coefficients of f(x), or is less than the number of variations by a positive even integer. The number of negative real zeros of f either equals the number of variations in sign occurring in the coefficients of f(  x), or is less than the number of variations by a positive even integer.

45. Slide 3-45 Copyright © 2004 Pearson Education, Inc. Example Determine the possible number of positive real number zeros and negative real number zeros of f(x) = x 6 + 7x 4  x 3  2x 2 + 6x  5. Solution: First consider the possible number of positive zeros by observing that f(x) has three variations in signs. + x 6 + 7x 4  x 3  2x 2 + 6x  5 1 2 3

46. Slide 3-46 Copyright © 2004 Pearson Education, Inc. Example continued For negative zeros, consider the variations in signs for f(  x): f(  x) = (  x) 6 + 7(  x) 4  (  x) 3  2(  x) 2 + 6(  x)  5 = x 6 + 7x 4 + x 3  2x 2  6x  5. Since there is only one variation in sign, f(x) has only 1 negative real zero.

47. Copyright © 2004 Pearson Education, Inc. 3.4 Polynomial Functions: Graphs, Applications, and Models

48. Slide 3-48 Copyright © 2004 Pearson Education, Inc. Graphing Functions of the Form f(x) = ax n Graph each function.  162 21 00 22 11 f(x)f(x)x  2.53  1.5  1/3 1 00 1/3 11 g(x)g(x)x

49. Slide 3-49 Copyright © 2004 Pearson Education, Inc. Horizontal and Vertical Translations Graph the function. a) f(x) = x 5 + 1 The graph of f(x) = x 5 + 1 will be the same as that of f(x) = x 5, but translated up 1 unit. b) g(x) = (x  1) 6 The graph is translated 1 unit to the right.

50. Slide 3-50 Copyright © 2004 Pearson Education, Inc. Odd Degree Typical graphs of polynomial functions of odd degree suggest that for every polynomial function f of odd degree there is at least one real value of x that make f(x) = 0. The zeros are the x-intercepts of the graph.

51. Slide 3-51 Copyright © 2004 Pearson Education, Inc. Even Degree A polynomial of even degree has a range of the form ( , k] or [k,  ) for some real number k.

52. Slide 3-52 Copyright © 2004 Pearson Education, Inc. Turning Points A polynomial function of degree n has at most n  1 turning points, with at least one turning point between each pair of successive zeros. The end behavior of a polynomial graph is determined by the dominating term, that is, the term of greatest degree.

53. Slide 3-53 Copyright © 2004 Pearson Education, Inc. End Behavior of Graphs of Polynomial Functions

54. Slide 3-54 Copyright © 2004 Pearson Education, Inc. End Behavior of Graphs of Polynomial Functions continued

55. Slide 3-55 Copyright © 2004 Pearson Education, Inc. Graphing Techniques A comprehensive graph of a polynomial function will show the following characteristics.  all x-intercepts (zeros)  the y-intercept  all turning points  enough of the domain to show the end behavior.

56. Slide 3-56 Copyright © 2004 Pearson Education, Inc. Graphing a Polynomial Function Let a n  0, be a polynomial of degree n. To sketch its graph, follow these steps. Step 1:Find the real zeros of f. Plot them as x-intercepts. Step 2:Find f(0) = a 0. Plot this as the y-intercept. Step 3:Use test points within the intervals formed by the x-intercepts to determine the sign of f(x) in the interval. This will determine whether the graph is above or below the x-axis in that interval. Use end behavior, whether the graph crosses or is tangent to the x-axis at the x-intercept, and selected points as necessary to complete the graph.

57. Slide 3-57 Copyright © 2004 Pearson Education, Inc. Example Graph f(x) = 3x 3 + 4x 2  5x  2. Solution: Step 1: The possible rational zeros are Use synthetic division to show that 1 is a zero.

58. Slide 3-58 Copyright © 2004 Pearson Education, Inc. Example continued Thus, f(x) = (x  1)(3x 2 + 7x + 2) = (x  1)(3x + 1)(x + 2) The three zeros of f are 1,  2, and  1/3. Step 2: f(0) =  2, so plot (0,  2) Step 3: The x-intercepts divide the x-axis into 4 intervals: ( ,  2), (  2,  1/3), (  1/3, 1) and (1,  )

59. Slide 3-59 Copyright © 2004 Pearson Education, Inc. Example continued Select test points AbovePositive282 (1,  ) BelowNegative 22 0 (  1/3, 1) AbovePositive4 11(  2,  1/3) BelowNegative  110 44( ,  2) Graph above or below x-axis Sign of f(x)Value of f(x)Test PointInterval

60. Slide 3-60 Copyright © 2004 Pearson Education, Inc. Example continued Plot the test points and join the x-intercepts, y-intercepts, and test points with a smooth curve to get the graph.

61. Slide 3-61 Copyright © 2004 Pearson Education, Inc. x-intercepts, Zeros, Solutions, and Factors If a is an x-intercept of the graph of y = f(x), then a is a zero of f, a is a solution of f(x) = 0, and x  a is a factor of f(x). Intermediate Value Theorem for Polynomials If f(x) defines a polynomial function with only real coefficients, and if for real numbers a and b, the values f(a) and f(b) are opposite in sign, then there exists at least one real zero between a and b.

62. Slide 3-62 Copyright © 2004 Pearson Education, Inc. Example: Locating a Zero Use synthetic division and a graph to show that f(x) = 2x 3  8x 2 + x + 15 has a zero between  1 and  2. Solution: Use synthetic division to find f(  1) and f(  2). Since f(  1) is positive and f(  2) is negative, by the theorem there must be a real zero between  1 and  2.

63. Slide 3-63 Copyright © 2004 Pearson Education, Inc. Example: Locating a Zero continued Graph of f(x) = 2x 3  8x 2 + x + 15.

64. Slide 3-64 Copyright © 2004 Pearson Education, Inc. Boundedness Theorem Let f(x) be a polynomial function of degree n  1 with real coefficients and with a positive leading coefficient. If f(x) is divided synthetically by x  c, and a) if c > 0 and all numbers in the bottom row of the synthetic division are nonnegative, then f(x) has no zero greater than c; b) if c < 0 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then f(x) has no zero less than c.

65. Slide 3-65 Copyright © 2004 Pearson Education, Inc. Example Show that f(x) = 2x 3  8x 2 + x + 15 has no real zero greater than 4. Solution: Since f(x) has real coefficients and the leading coefficient, 2, is positive, use the boundedness theorem. Divide f(x) synthetically by x  4. Since 4 > 0 and all numbers in the last row of the synthetic division are nonnegative, f(x) has no real zero greater than 4.

66. Copyright © 2004 Pearson Education, Inc. 3.5 Rational Functions: Graphs, Applications, and Models

67. Slide 3-67 Copyright © 2004 Pearson Education, Inc. Rational Function A function f of the form where p(x) and q(x) are polynomials, with q(x)  0, is called a rational function. Examples:

68. Slide 3-68 Copyright © 2004 Pearson Education, Inc. Reciprocal Function The simplest rational function with a variable denominator. Since x cannot equal 0, the graph will never intersect the vertical line x = 0. This line is called a vertical asymptote.

69. Slide 3-69 Copyright © 2004 Pearson Education, Inc. Reciprocal Function

70. Slide 3-70 Copyright © 2004 Pearson Education, Inc. Example Graph the function and give the domain and range. x and y axes are the horizontal and vertical asymptotes. The domain and range are ( , 0)  (0,  ).

71. Slide 3-71 Copyright © 2004 Pearson Education, Inc. Example Graph the function and give the domain and range. y = 0 is the horizontal asymptote. x =  2 is the vertical asymptote. Domain: ( ,  2)  (  2,  ) Range: ( , 0)  (0,  )

72. Slide 3-72 Copyright © 2004 Pearson Education, Inc. Rational Function

73. Slide 3-73 Copyright © 2004 Pearson Education, Inc. Example Graph the function and give the domain and range. The graph will be shifted 3 units to the left and 2 units down. Vertical asymptote x =  3 Horizontal asymptote: y =  2 Domain: ( ,  3)  (  3,  ) Range: (  2,  )

74. Slide 3-74 Copyright © 2004 Pearson Education, Inc. Asymptotes Let p(x) and q(x) define polynomials. For the rational function defined by written in lowest terms, for real numbers a and b: 1. If |f(x)|   as x  a, then the line x = a is a vertical asymptote. 2. If |f(x)|  b as |x|  , then the line y = b is a horizontal asymptote.

75. Slide 3-75 Copyright © 2004 Pearson Education, Inc. Determining Asymptotes To find the asymptotes of a rational function defined by a rational expression in lowest terms, use the following procedures. 1. Vertical Asymptotes Find any vertical asymptotes by setting the denominator equal to zero and solving for x. If a is a zero of the denominator, then the line x = a is a vertical asymptote.

76. Slide 3-76 Copyright © 2004 Pearson Education, Inc. Determining Asymptotes continued 2. Horizontal Asymptotes Determine any other asymptotes. Consider three possibilities: (a) If the numerator has a lower degree than the denominator, then there is a horizontal asymptote y = 0 (the x-axis). (b) If the numerator and denominator have the same degree, and the function is of the form then the horizontal asymptote has equation

77. Slide 3-77 Copyright © 2004 Pearson Education, Inc. Determining Asymptotes continued (c)If the numerator is of degree exactly one more than the denominator, then there will be an oblique (slanted) asymptote. To find it, divide the numerator by the denominator and disregard the remainder. Set the rest of the quotient equal to y to obtain the equation of the asymptote.

78. Slide 3-78 Copyright © 2004 Pearson Education, Inc. Example For the rational function f, find all asymptotes. Vertical: Set the denominator equal to 0 and solve. The equation of the vertical asymptote is x = 2.

79. Slide 3-79 Copyright © 2004 Pearson Education, Inc. Example continued Horizontal: Divide each term by the largest power of x in the expression. The line y = 3 is therefore the horizontal asymptote.

80. Slide 3-80 Copyright © 2004 Pearson Education, Inc. Steps for Graphing Rational Functions A comprehensive graph of a rational function exhibits these features:  all x- and y-intercepts;  all asymptotes: vertical, horizontal, and/or oblique;  the point at which the graph intersects its nonvertical asymptote (if there is any such point);  enough of the graph to exhibit the correct end behavior.

81. Slide 3-81 Copyright © 2004 Pearson Education, Inc. Graphing a Rational Function Let define a function where p(x) and q(x) are polynomials and the rational expression is written in lowest terms. To sketch its graph, follow these steps: 1.Find any vertical asymptotes. 2.Find any horizontal or oblique asymptotes. 3.Find the y-intercept by evaluating f(0). 4.Find the x-intercepts, if any, by solving f(x) = 0. These will be the zeros of the numerator, p(x).

82. Slide 3-82 Copyright © 2004 Pearson Education, Inc. Graphing a Rational Function continued 5. Determine whether the graph will intersect its nonvertical asymptote y = b or y = mx + b by solving f(x) = b or f(x) = mx + b. 6. Plot selected points, as necessary. Choose an x-value in each domain interval determined by the vertical asymptotes and x-intercepts. 7. Complete the sketch.

83. Slide 3-83 Copyright © 2004 Pearson Education, Inc. Example Graph Step 1: Set the denominator equal to zero.  The vertical asymptotes are x = 3 and x =  1. Step 2: The horizontal asymptote is the x-axis.

84. Slide 3-84 Copyright © 2004 Pearson Education, Inc. Example continued Step 3: The y-intercept is 0, since Step 4: The x-intercept is found by solving f (x) = 0.

85. Slide 3-85 Copyright © 2004 Pearson Education, Inc. Example continued Step 5 Solve f(x) = 0, this solution was found in step (4). The graph intersects the horizontal asymptote at (0, 0). Step 6 Plot a point in each interval determined the x-intercepts and vertical asymptotes to get an idea of how the graph behaves in each interval.

86. Slide 3-86 Copyright © 2004 Pearson Education, Inc. Example continued Test Points AbovePositive16/54 (3,  ) BelowNegative 11 1(0, 3) AbovePositive1.143  1/2(  1, 0) BelowNegative  8/5 22( ,  1) Graph above or below x-axis Sign of f(x)Value of f(x)Test PointInterval

87. Slide 3-87 Copyright © 2004 Pearson Education, Inc. Example continued Step 7 Complete the graph.

88. Slide 3-88 Copyright © 2004 Pearson Education, Inc. Behavior of Graphs of Rational Functions Near Vertical Asymptotes

89. Slide 3-89 Copyright © 2004 Pearson Education, Inc. Example: Oblique Graph Vertical Asymptote x =  1 Oblique Asymptote y = x  1 x-intercept (0, 0) y-intercept (0, 0)

90. Copyright © 2004 Pearson Education, Inc. 3.6 Variation

91. Slide 3-91 Copyright © 2004 Pearson Education, Inc. Direct Variation y varies directly as x, or y is directly proportional to x, if a nonzero real number k, called the constant of variation, such that y = kx. The phrase “directly proportional” is sometimes abbreviated to just “proportional”.

92. Slide 3-92 Copyright © 2004 Pearson Education, Inc. Solving Variation Problems Step 1 Write the general relationship among the variables as an equation. Use the constant k. Step 2 Substitute given values of the variables and find the value of k. Step 3 Substitute this value of k into the equation from Step 1, obtaining a specific formula. Step 4 Substitute the remaining values and solve for the required unknown.

93. Slide 3-93 Copyright © 2004 Pearson Education, Inc. Example At a distance of 6 feet from a 60-cd (candlepower) light source, a photographer’s light meter registers an illuminance of 9 units. Find the illuminance of a 45-cd light source at the same distance. Solution: Step 1 Since the illuminance varies directly to the intensity at a given distance, E = kI, where I represent intensity and E represent illuminance, and k is a nonzero constant.

94. Slide 3-94 Copyright © 2004 Pearson Education, Inc. Example continued Step 2 Since E = 9 when I = 60, the equation E = kI becomes 9 = k(60) k = 0.15 Step 3 Using the value of k, we can express the relationship between the illuminance and intensity as E = 0.15I Step 4 To find the illuminance of the 45-cd source at this distance, we replace I with 45. E = 0.15(45) = 6.75 The illuminance is 6.75 units.

95. Slide 3-95 Copyright © 2004 Pearson Education, Inc. Direct Variation as nth Power Let n be a positive real number. Then y varies directly as the nth power of x, or y is directly proportional to the nth power of x if there exists a nonzero real number k such that y = kx n. For example, the area of a square of side x is given by the formula A = x 2, so the area varies directly as the square of the length of a side. Here k = 1.

96. Slide 3-96 Copyright © 2004 Pearson Education, Inc. Inverse Variation Let n be a positive real number. Then y varies inversely as the nth power of x, or y is inversely proportional to the nth power of x, if there exists a nonzero real number k such that If n = 1, then, and y varies inversely as x.

97. Slide 3-97 Copyright © 2004 Pearson Education, Inc. Example In a certain manufacturing process, the cost of producing a single item varies inversely as the square of the number of items produced. If 50 items are produced, each costs $3. Find the cost per item if 250 items are produced. Solution: Step 1 Let x represent the number of items produced and y represent the cost per item. Then for some nonzero constant k,

98. Slide 3-98 Copyright © 2004 Pearson Education, Inc. Example continued Step 2 Since y = 3 when x = 50, Step 3 The relationship between x and y is Step 4 When 250 items are produced, the cost per item is

99. Slide 3-99 Copyright © 2004 Pearson Education, Inc. Combined and Joint Variation One variable may depend on more than one other variable. Such variation is called combined variation. When a variable depends on the product of two or more other variables, it is referred to as joint variation. Joint Variation Let m and n be real numbers. Then y varies jointly as the nth power of x and the mth power of z if there exists a nonzero number k such that y = kx n z m.

100. Slide 3-100 Copyright © 2004 Pearson Education, Inc. Example The area of a triangle varies jointly as the lengths of the base and the height. A triangle with base 4 ft and a height 5 ft has area 10 ft 2. Find the area of a triangle with base 6 ft and height 2 ft. Solution: Step 1 Let A represent the area, b the base, and h the height of the triangle. Then for some number k, A = kbh. Step 2 Since A is 10 when b is 4 and h is 5, 10 = k(4)(5)

101. Slide 3-101 Copyright © 2004 Pearson Education, Inc. Example continued Step 3 The relationship among the variables is the familiar formula for the area of a triangle. Step 4 When b = 6ft. and h = 2 ft,

102. Slide 3-102 Copyright © 2004 Pearson Education, Inc. Combined Variation Example:The current (I) in an electric circuit varies directly with the voltage (E) and inversely with resistance (R). When the voltage is 16 volts and the resistance is 8 ohms, the current is 2 amps. Give the current when the voltage is 10 volts and the resistance is 20 ohms.

103. Slide 3-103 Copyright © 2004 Pearson Education, Inc. Combined Variation continued Solution: Using the formula and substituting the values for I, E, and R, we are able to find k. Now use the second set of values for E and R, to find I. The current is amps when the voltage is 10 volts and the resistance is 20 ohms.