4.1 Polynomial Functions and Models

Transcription

1 4.1 Polynomial Functions and Models Determine the behavior of the graph of a polynomial function using the leading-term test. Factor polynomial functions and find the zeros and their multiplicities. Use a graphing calculator to graph a polynomial function and find its real-number zeros.

2 Polynomial Function A polynomial function P is given by P( x) a x a x a x... a x a, n n 1 n 2 n n 1 n where the coefficients a n, a n - 1,, a 1, a 0 are real numbers and the exponents are whole numbers. Polynomial Function Constant Linear Quadratic Cubic Quartic Degree Example f(x) = 4 f(x) = 3x + 1 f(x) = 4x 2 x + 9 f(x) = x 3 +2x 2 x + 11 f(x) = x 4 3.2x x Slide 4.1-2

3 Quadratic Function Slide 4.1-3

4 Cubic Function Slide 4.1-4

5 Examples of Polynomial Functions Slide 4.1-5

6 Examples of Nonpolynomial Functions Slide 4.1-6

7 Polynomial Functions The graph of a polynomial function is continuous and smooth. The domain of a polynomial function is the set of all real numbers Slide 4.1-7

8 The Leading-Term Test Slide 4.1-8

9 Example Using the leading term-test, match each of the following functions with one of the graphs A D, which follow. 4 3 a) f ( x) 3x 2x 3 b) f x x x x 3 2 ( ) c) f ( x) x x 1 d) 4 f ( x) x x 4x Slide 4.1-9

10 Graphs Slide

11 Solution Leading Term Degree of Leading Term Sign of Leading Coeff. Graph a) 3x 4 Even Positive D b) 5x 3 Odd Negative B c) x 5 Odd Positive A d) x 6 Even Negative C Slide

12 Graphs f ( x) x x f x x x x 3 2 ( ) f ( x) x x 4x f x x x 4 3 ( ) Slide

13 Finding Zeros of Factored Polynomial Functions If c is a real zero of a function (that is, f(c) = 0), then (c, 0) is an x-intercept of the graph of the function. Example: Find the zeros of f ( x) 5( x 1)( x 1)( x 1)( x 2) 3 5( x 1) ( x 2). Slide

14 Finding Zeros of Factored Polynomial Functions continued Solution: To solve the equation f(x) = 0, we use the principle of zero products, solving x 1 = 0 and x + 2 = 0. The zeros of f(x) are 1 and 2. See graph on right. Slide

15 Even and Odd Multiplicity If (x c) k, k 1, is a factor of a polynomial function P(x) and (x c) k + 1 is not a factor and: k is odd, then the graph crosses the x-axis at (c, 0); k is even, then the graph is tangent to the x-axis at (c, 0). Slide

16 Example Find the zeros of f(x) = x 4 + 8x Solution We factor as follows: f(x) = x 4 + 8x 2 33 = (x )(x 2 3). Solve the equation f(x) = 0 to determine the zeros. We use the principle of zero products. 2 2 ( x 11)( x 3) 0 x 11 0 or x x 11 or x x x i 2 11 or x 3 11 Slide

17 Finding Real Zeros on a Calculator Find the zeros of f(x) = 0.2x 3 1.5x 2 0.3x + 2. Approximate the zeros to three decimal places. Solution Use a graphing calculator to create a graph. Look for points where the graph crosses the x-axis. We use the ZERO feature to find them The zeros are approximately 1.164, 1,142, and Slide

18 4.2 Graphing Polynomial Functions Graph polynomial functions. Use the intermediate value theorem to determine whether a function has a real zero between two given real numbers.

19 Graphing Polynomial Functions If P(x) is a polynomial function of degree n, the graph of the function has: at most n real zeros, and thus at most n x- intercepts; at most n 1 turning points. (Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.) Slide

20 Steps to Graph a Polynomial Function 1. Use the leading-term test to determine the end behavior. 2. Find the zeros of the function by solving f(x) = 0. Any real zeros are the first coordinates of the x-intercepts. 3. Use the x-intercepts (zeros) to divide the x-axis into intervals and choose a test point in each interval to determine the sign of all function values in that interval. 4. Find f(0). This gives the y-intercept of the function. 5. If necessary, find additional function values to determine the general shape of the graph and then draw the graph. 6. As a partial check, use the facts that the graph has at most n x-intercepts and at most n 1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x-axis. We can also check the graph with a graphing calculator. Slide

21 Example Graph the polynomial function f(x) = 2x 3 + x 2 8x 4. Solution: 1. The leading term is 2x 3. The degree, 3, is odd, the coefficient, 2, is positive. Thus the end behavior of the graph will appear as: 2. To find the zero, we solve f(x) = 0. Here we can use factoring by grouping. Slide

22 Example continued Factor: 3 2 2x x 8x x x x (2 1) 4(2 1) 0 2 (2x 1)( x 4) 0 (2x 1)( x 2)( x 2) 0 The zeros are 1/2, 2, and 2. The x-intercepts are ( 2, 0), ( 1/2, 0), and (2, 0). 3. The zeros divide the x-axis into four intervals: (, 2), ( 2, 1/2), ( 1/2, 2), and (2, ). We choose a test value for x from each interval and find f(x). Slide

23 Example continued Interval Test Value, x Function value, f(x) Sign of f(x) Location of points on graph (, 2) 3 25 Below x-axis ( 2, 1/2) ( 1/2, 2) (2, ) Above x-axis Below x-axis Above x-axis 4. To determine the y-intercept, we find f(0): f The y-intercept is (0, 4). f x x x x 3 2 ( ) ( ) 2( ) 8( ) Slide

24 Example continued 5. We find a few additional points and complete the graph. 6. The degree of f is 3. The graph of f can have at most 3 x- intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at 2, 1/2, and 2. The graph has the end behavior described in step (1). The graph appears to be correct. x f(x) Slide

25 Intermediate Value Theorem For any polynomial function P(x) with real coefficients, suppose that for a b, P(a) and P(b) are of opposite signs. Then the function has a real zero between a and b. Example: Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b. a) f(x) = x 3 + x 2 8x; a = 4 b = 1 b) f(x) = x 3 + x 2 8x; a = 1 b = 3 Slide

26 Solution We find f(a) and f(b) and determine where they differ in sign. The graph of f(x) provides a visual check. f( 4) = ( 4) 3 + ( 4) 2 8( 4) = 16 f( 1) = ( 1) 3 + ( 1) 2 8( 1) = 8 zero y = x 3 + x 2 8x By the intermediate value theorem, since f( 4) and f( 1) have opposite signs, then f(x) has a zero between 4 and 1. Slide

27 Solution f(1) = (1) 3 + (1) 2 8(1) = 6 f(3) = (3) 3 + (3) 2 8(3) = 12 By the intermediate value theorem, since f(1) and f(3) have opposite signs, then f(x) has a zero between 1 and 3. y = x 3 + x 2 8x zero Slide

28 4.3 Polynomial Division; The Remainder and Factor Theorems Perform long division with polynomials and determine whether one polynomial is a factor of another. Use synthetic division to divide a polynomial by x c. Use the remainder theorem to find a function value f(c). Use the factor theorem to determine whether x c is a factor of f(x).

29 Division and Factors When we divide one polynomial by another, we obtain a quotient and a remainder. If the remainder is 0, then the divisor is a factor of the dividend. Example: Divide to determine whether x + 3 and 3 2 x 1 are factors of x 2x 5x 4. Slide

30 Division and Factors continued Divide: x 2 5x 20 x x x x x 3x x 5 x 2 5x 15 x 20x 4 20x remainder Since the remainder is 64, we know that x + 3 is not a factor. Slide

31 Division and Factors continued Divide: x 2 x 4 x x x x x x 3 2 x x 2 2 5x x 4x 4 4x 4 0 remainder Since the remainder is 0, we know that x 1 is a factor. Slide

32 Division of Polynomials When dividing a polynomial P(x) by a divisor d(x), a polynomial Q(x) is the quotient and a polynomial R(x) is the remainder. The quotient must have degree less than that of the dividend, P(x). The remainder must be either 0 or have degree less than that of the divisor. P(x) = d(x) Q(x) + R(x) Dividend Divisor Quotient Remainder Slide

33 The Remainder Theorem If a number c is substituted for x in a polynomial f(x), then the result f(c) is the remainder that would be obtained by dividing f(x) by x c. That is, if f(x) = (x c) Q(x) + R, then f(c) = R. Synthetic division is a collapsed version of long division; only the coefficients of the terms are written. Slide

34 Example Use synthetic division to find the quotient and remainder. 4x x 6x 2x 50 ( x 2) The quotient is 4x 4 7x 3 8x 2 14x 28 and the remainder is Note: We must write a 0 for the missing term. Slide

35 Example Determine whether 4 is a zero of f(x), where f(x) = x 3 6x x 6. We use synthetic division and the remainder theorem to find f(4) f (4) Since f(4) 0, the number 4 is not a zero of f(x). Slide

36 The Factor Theorem For a polynomial f(x), if f(c) = 0, then x c is a factor of f(x). Example: Let f(x) = x 3 7x + 6. Factor f(x) and solve the equation f(x) = 0. Solution: We look for linear factors of the form x c. Let s try x 1: Slide

37 Example continued Since f(1) = 0, we know that x 1 is one factor and the quotient x 2 + x 6 is another. So, f(x) = (x 1)(x + 3)(x 2). For f(x) = 0, we have x = 3, 1, or 2. Slide

38 4.4 Theorems about Zeros of Polynomial Functions Find a polynomial with specified zeros. For a polynomial function with integer coefficients, find the rational zeros and the other zeros, if possible. Use Descartes rule of signs to find information about the number of real zeros of a polynomial function with real coefficients.

39 The Fundamental Theorem of Algebra Every polynomial function of degree n, with n 1, has at least one zero in the system of complex numbers. Slide

40 The Fundamental Theorem of Algebra Example: Find a polynomial function of degree 4 having zeros 1, 2, 4i, and 4i. Solution: Such a polynomial has factors (x 1),(x 2), (x 4i), and (x + 4i), so we have: f ( x) a ( x 1)( x 2)( x 4 i)( x 4 i) n Let a n = 1: f ( x) ( x 1)( x 2)( x 4 i)( x 4 i) 2 2 ( x 3x 2)( x 16) x x x x x x x x x Slide

41 Zeros of Polynomial Functions with Real Coefficients Nonreal Zeros: If a complex number a + bi, b 0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a bi, is also a zero. (Nonreal zeros occur in conjugate pairs.) Irrational Zeros: If a c b, where a, b, and c are rational and b is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then its conjugate is also a zero. a c b, Slide

42 Example Suppose that a polynomial function of degree 6 with rational coefficients has 3 + 2i, 6i, and 1 2 as three of its zeros. Find the other zeros. Solution: The other zeros are the conjugates of the given zeros, 3 2i, 6i, and 1 2. There are no other zeros because the polynomial of degree 6 can have at most 6 zeros. Slide

43 Rational Zeros Theorem Let n P( x) a x a x... a x a x a, n n 1 2 n where all the coefficients are integers. Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides 1 and 1). If p/q is a zero of P(x), then p is a factor of a 0 and q is a factor of a n. Slide

44 Example Given f(x) = 2x 3 3x 2 11x + 6: a) Find the rational zeros and then the other zeros. b) Factor f(x) into linear factors. Solution: a) Because the degree of f(x) is 3, there are at most 3 distinct zeros. The possibilities for p/q are: Possibilities for p 1, 2, 3, 6 : Possibilities for q 1, 2 Possibilities for p / q : 1, 1,2, 2,3, 3,6, 6,,,, Slide

45 Example continued Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions. We try 1: We try 1: Since f(1) = 6, 1 is not a zero. Since f( 1) = 12, 1 is not a zero. Slide

46 Example continued We try 3: Since f(3) = 0, 3 is a zero. Thus x 3 is a factor. Using the results of the division above, we can express f(x) as f x x x x 2 ( ) ( 3)(2 3. 2) We can further factor 2x 2 + 3x 2 as (2x 1)(x + 2). Slide

47 Example continued The rational zeros are 2, 3 and 1. 2 The complete factorization of f(x) is: f ( x) (2x 1)( x 3)( x 2) Slide

48 Descartes Rule of Signs Let P(x) be a polynomial function with real coefficients and a nonzero constant term. The number of positive real zeros of P(x) is either: 1. The same as the number of variations of sign in P(x), or 2. Less than the number of variations of sign in P(x) by a positive even integer. The number of negative real zeros of P(x) is either: 3. The same as the number of variations of sign in P( x), or 4. Less than the number of variations of sign in P( x) by a positive even integer. A zero of multiplicity m must be counted m times. Slide

49 Example What does Descartes rule of signs tell us about the number of positive real zeros and the number of negative real zeros? P( x) 3x 6x x 7x x 6x x 7x 2 There are two variations of sign, so there are either two or zero positive real zeros to the equation. Slide

50 Example continued P( x) 3( x) 6( x) ( x) 7( x) x x x x There are two variations of sign, so there are either two or zero negative real zeros to the equation. Total Number of Zeros = 4: Positive Negative Nonreal Slide

51 4.5 Rational Functions For a rational function, find the domain and graph the function, identifying all of the asymptotes. Solve applied problems involving rational functions.

52 Rational Function A rational function is a function f that is a quotient of two polynomials, that is, ( ) f( x) px, qx ( ) where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial. The domain of f consists of all inputs x for which q(x) 0. Slide

53 Example Consider. f( x) Find the domain and graph f. 1 x 4 The graph of the function is the graph of y = 1/x translated to the left 4 units. Solution: When the denominator x + 4 = 0, we have x = 4, so the only input that results in a denominator of 0 is 4. Thus the domain is {x x 4} or (, 4) ( 4, ). Slide

54 Vertical Asymptotes The vertical asymptotes of a rational function f(x) = p(x)/q(x) are found by determining the zeros of q(x) that are not also zeros of p(x). If p(x) and q(x) are polynomials with no common factors other than constants, we need to determine only the zeros of the denominator q(x). If a is a zero of the denominator but not the numerator, then the line x = a is a vertical asymptote for the graph of the function. Slide

55 Example Determine the vertical asymptotes of the function. f( x) 2x 3 x 2 4 Factor to find the zeros of the denominator: x 2 4 = (x + 2)(x 2) Thus the vertical asymptotes are the lines x = 2 and x = 2. Slide

56 Horizontal Asymptote The line y = b is a horizontal asymptote for the graph of f if either or both of the following are true: f ( x) b as x or f ( x) b as x. When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively. Slide

57 Horizontal Asymptote Example: Find the horizontal asymptote: 4 2 6x 3x 1 4 9x 3x 2 f( x). The numerator and denominator have the same degree. The ratio of the leading coefficients is 6/9, so the line y = 2/3 is the horizontal asymptote. Slide

58 Determining a Horizontal Asymptote When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively. When the degree of the numerator of a rational function is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote. When the degree of the numerator of a rational function is greater than the degree of the denominator, there is no horizontal asymptote. Slide

59 True Statements The graph of a rational function never crosses a vertical asymptote. The graph of a rational function might cross a horizontal asymptote but does not necessarily do so. Slide

60 Example Graph 2x hx ( ). x 3 Vertical asymptotes: x + 3 = 0, so x = 3. The degree of the numerator and denominator is the same. Thus y = 2, is the horizontal asymptote. 1. Draw the asymptotes with dashed lines. 2. Compute and plot some ordered pairs and draw the curve. Slide

61 Example continued x h(x) /5 Slide

62 Oblique or Slant Asymptote 2 2x 3x 5 Find all the asymptotes of hx ( ). x 2 The line x = 2 is a vertical asymptote. There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator. Note that 2 2x 3x 5 3 h( x) (2x 1). x 2 x 2 Slide

63 Oblique or Slant Asymptote continued Divide to find an equivalent expression. The line y = 2x 1 is an oblique asymptote. 2 2x 3x 5 3 h( x) (2x 1) x 2 x 2 2x 1 2 x 2 2x 3x 5 2 2x 4 x x 5 x 2 3 Slide

64 Occurrence of Lines as Asymptotes of Rational Functions For a rational function f(x) = p(x)/q(x), where p(x) and q(x) have no common factors other than constants: Vertical asymptotes occur at any x-values that make the denominator 0. The x-axis is the horizontal asymptote when the degree of the numerator is less than the degree of the denominator. A horizontal asymptote other than the x-axis occurs when the numerator and the denominator have the same degree. Slide

65 Occurrence of Lines as Asymptotes of Rational Functions continued An oblique asymptote occurs when the degree of the numerator is 1 greater than the degree of the denominator. There can be only one horizontal asymptote or one oblique asymptote and never both. An asymptote is not part of the graph of the function. Slide

66 Graphing Rational Functions 1. Find the real zeros of the denominator. Determine the domain of the function and sketch any vertical asymptotes. 2. Find the horizontal or oblique asymptote, if there is one, and sketch it. 3. Find the zeros of the function. The zeros are found by determining the zeros of the numerator. These are the first coordinates of the x-intercepts of the graph. 4. Find f(0). This gives the y-intercept (0, f(0)), of the function. 5. Find other function values to determine the general shape. Then draw the graph. Slide

67 Example Graph x 3 f( x). 2 2x 5x 3 1. Find the zeros by solving: 2 2x 5x x 5x 3 (2x 1)( x 3) The zeros are 1/2 and 3, thus the domain excludes these values. The graph has vertical asymptotes at x = 3 and x = 1/2. We sketch these with dashed lines. 2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, y = 0, is the horizontal asymptote. Slide

68 Example continued 3. To find the zeros of the numerator, we solve x + 3 = 0 and get x = 3. Thus, 3 is the zero of the function, and the pair ( 3, 0) is the x-intercept. 4. We find f(0): f ( 0) ( 0) 5( 0) Thus (0, 1) is the y-intercept. Slide

69 Example continued 5. We find other function values to determine the general shape of the graph and then draw the graph. x f(x) 1/2 2/3 1 7/9 Slide

70 More Examples Graph the following functions. a) x 3 f( x) x 2 b) f( x) 8 x 2 x 2 9 c) f( x) x 2 x 1 Slide

71 Graph a f( x) x 3 x 2 1. Vertical Asymptote x = 2 2. Horizontal Asymptote y = 1 3. x-intercept (3, 0) 4. y-intercept (0, 3/2) Slide

72 Graph b f( x) 8 x 2 x Vertical Asymptote x = 3, x = 3 2. Horizontal Asymptote y = 1 3. x-intercepts ( 2.828, 0) 4. y-intercept (0, 8/9) Slide

73 Graph c f( x) x 2 x 1 1. Vertical Asymptote x = 1 2. Oblique Asymptote y = x 1 3. x-intercept (0, 0) 4. y-intercept (0, 0) Slide

74 4.6 Polynomial and Rational Inequalities Solve polynomial and rational inequalities.

75 Polynomial Inequalities A quadratic inequality can be written in the form ax 2 + bx + c > 0, where the symbol > could be replaced with either <,, or. A quadratic inequality is one type of polynomial inequality. Examples of polynomial inequalities: x x 2 4 x 6 0 8x 2x 6x 5 Slide

76 Example Solve: x 2 3x 4 > 0. Let s look at the graph of f(x) = x 2 3x 4. x 2 3x 4 > 0 x 2 3x 4 > 0 (0, 1) (0, 4) x 2 3x 4 < 0 The zeros are 4 and 1. Thus the x-intercepts of the graph are (0, 1) and (4, 0). Slide

77 Example The zeros divide the x-axis into three intervals: 1 4 The sign of the function is the same for all values of x in a given interval. We can choose a test value for x from each interval and find the sign of f(x). Test values: f( 2) = 6 f(0) = 4 f(5) = The solution set consists of the intervals where the sign of f(x) is positive. So the solution set is {x x < 1 or x > 4}. Slide

78 To Solve a Polynomial Inequality 1. Find an equivalent inequality with P(x) on one side and 0 on one side. 2. Solve the related polynomial equation; that is solve for f(x) = Use the solutions to divide the x-axis into intervals. Then select a test value from each interval and determine the polynomial s sign on the interval. 4. Determine the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is or. Slide

79 Example Solve: 4x 3 7x 2 15x. We need to find all the zeros of the function so we solve the related equation x 7x 15x x 7x 15x 0 x x x 2 (4 7 15) 0 x(4x 5)( x 3) 0 The zeros are 0, 3 and 5/4. Thus the x-intercepts of the graph are (0, 0), (3, 0) and ( 5/4, 0). Slide

80 Example continued The zeros divide the x-axis into four intervals. For all x- values within a given interval, the sign of 4x 3 7x 2 15x 0 must be either positive or negative. To determine which, we choose a test value for x from each interval and find f(x). Interval (, 5/4) ( 5/4, 0) (0, 3) Test Value f( 2) = 30 f( 1) = 4 f(1) = 18 Sign of f(x) Negative Positive Negative (3, ) f(4) = 84 Positive Since we are solving 4x 3 7x 2 15x 0, the solution set consists of only two of the four intervals, those in which the sign of f(x) is negative {x < x < 5/4 or 0 < x < 3}. Slide

81 To Solve a Rational Inequality 1. Find an equivalent inequality with 0 on one side. 2. Change the inequality symbol to an equals sign and solve the related equation, that is, solve f(x) = Find the values of the variable for which the related rational function is not defined. 4. The numbers found in steps (2) and (3) are called critical values. Use the critical values to divide the x-axis into intervals. Then test an x-value from each interval to determine the function s sign in that interval. 5. Select the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. If the inequality symbol is or, then the solutions to step (2) should be included in the solution set. The x-values found in step (3) are never included in the solution set. Slide

82 Example x 3 Solve. 2 0 x 1 The denominator tells us that f(x) is not defined for x = 1 and x = 1. Next, solve the related equation. x x 1 2 x 3 2 ( x 1) 0( x 1) 2 x 1 x 3 0 x 3 Slide

83 Example continued The critical values are 3, 1, and 1. These values divide the x-axis into four intervals. We use a test value to determine the sign of f(x) in each interval. Interval (, 3) ( 3, 1) ( 1, 1) (1, ) Test Value f( 4) = 1/15 f( 2) = 1/3 f(0) = 3 f(2) = 5/3 Sign of f(x) + + Slide

84 Example continued Function values are positive in the intervals ( 3, 1) and (1, ). Since 1 is not in the domain of f, it cannot be part of the solution set. Note that 3 does satisfy the inequality. The solution set is [ 3, 1) (1, ). Slide