1. MAC 1140 Module 5 Nonlinear Functions and Equations

2. Rev.S082 Learning Objectives Upon completing this module, you should be able to 1.identify intervals where a function is increasing or decreasing. 2.find extrema of a function. 3.identify symmetry in a graph of a function. 4.determine if a function is odd, even, or neither. 5.understand the graphs of polynomial functions. 6.evaluate and graph piecewise-defined functions. 7.divide polynomials. 8.understand the division algorithm, remainder theorem, and factor theorem. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

3. Rev.S083 Learning Objectives (Cont.) 9.factor higher degree polynomials. 10.analyze polynomials with multiple zeros. 11.solve polynomial equations. 12.perform arithmetic operations on complex numbers. 13.solve quadratic equations having complex solutions. 14.apply the fundamental theorem of algebra. 15.factor polynomials having complex zeros. 16.solve polynomial equations having complex solutions. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

4. Rev.S084 Nonlinear Functions and Equations http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. 4.1Nonlinear Functions and Their Graphs 4.2Polynomial Functions and Models 4.3Real Zeros of Polynomial Functions 4.4The Fundamental Theorem of Algebra There are four sections in this module:

5. Rev.S085 Let’s get started by looking at some polynomial functions. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

6. Rev.S086 Polynomial Functions Polynomial functions are frequently used to approximate data. A polynomial function of degree 2 or higher is a nonlinear function. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

7. Rev.S087 Is the Function Increasing or Decreasing? The concept of increasing and decreasing relate to whether the graph of a function rises or falls. Moving from left to right along a graph of an increasing function would be uphill. Moving from left to right along a graph of a decreasing function would be downhill. We speak of a function f increasing or decreasing over an interval of its domain. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

8. Rev.S088 Is the Function Increasing or Decreasing? (Cont.) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

9. Rev.S089 Example Use the graph of shown below and interval notation to identify where f is increasing or decreasing. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Solution Moving from left to right on the graph of f, the y-values decreases until x = 0, increases until x = 2, and decreases thereafter. Thus, in interval notation f is decreasing on

10. Rev.S0810 Extrema of Nonlinear Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Graphs of polynomial functions often have “hills” or “valleys”. The “highest hill” on the graph is located at (–2, 12.7). This is the absolute maximum of g. There is a smaller peak located at the point (3, 2.25). This is called the local maximum.

11. Rev.S0811 Extrema of Nonlinear Functions (Cont.) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Maximum and minimum values that are either absolute or local are called extrema. A function may have several local extrema, but at most one absolute maximum and one absolute minimum. It is possible for a function to assume an absolute extremum at two values of x. The absolute maximum is 11. It is a local maximum as well, because near x = –2 and x = 2 it is the largest y-value.

12. Rev.S0812 Absolute and Local Extrema http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. The absolute maximum is the maximum y-value on the graph y = f(x).

13. Rev.S0813 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. The monthly average ocean temperature in degrees Fahrenheit at Bermuda can be modeled by where x = 1 corresponds to January and x = 12 to December. The domain of f is D = {x|1 }. (Source: J. Williams, The Weather Almanac 1995.) a)Graph f in [1, 12, 1] by [50, 90, 10]. b)Estimate the absolute extrema. Interpret the results. Solution

14. Rev.S0814 Example (Cont.) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. b)Many graphing calculators have the capability to find maximum and minimum y-values. An absolute minimum of about 61.5 corresponds to the point (2.01, 61.5). This means the monthly average ocean temperature is coldest during February, when it reaches An absolute maximum of about 82 corresponds to the point (7.61, 82.0), meaning that the warmest average ocean temperature occurs in August when it reaches a maximum of

15. Rev.S0815 What is an Even Function? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. If a graph was folded along the y-axis, and the right and left sides would match, the graph would be symmetric with respect to the y-axis. A function whose graph satisfies this characteristic is called an even function.

16. Rev.S0816 What is an Odd Function? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. When the symmetry occurs in respect to the origin. If the graph could rotate, the original graph would reappear after half a turn (clockwise or counter-clockwise.) This represents an odd function.

17. Rev.S0817 Is the Function Even or Odd? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Identify whether the function is even or odd. Solution Since f is a polynomial containing only odd powers of x, it is an odd function. This also can be shown symbolically as follows.

18. Rev.S0818 How to Obtain the Turning Points of a Polynomial Function? A polynomial function f of degree n can be expressed as f(x) = a n x n + … + a 2 x 2 + a 1 x + a 0, where each coefficient a k is a real number, a n  0, and n is a nonnegative integer. A turning point occurs whenever the graph of a polynomial function changes from increasing to decreasing or from decreasing to increasing. Turning points are associated with “hills” or “valleys” on a graph. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

19. Rev.S0819 Next, let’s look at the characteristics of different types of polynomial functions. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

20. Rev.S0820 Constant Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. * No x-intercepts. * No turning points.

21. Rev.S0821 Linear Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree 1. One x-intercepts. No turning points.

22. Rev.S0822 Quadratic Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree 2 - parabola that opens up or down. Zero, one or two x-intercepts. Exactly ONE turning point, which is also the VERTEX.

23. Rev.S0823 Cubic Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree 3. Up to three x-intercepts. Up to two turning points.

24. Rev.S0824 Quartic Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree 4. Up to four x-intercepts. Up to three turning points.

25. Rev.S0825 Quintic Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree 5. Up to five x-intercepts. Up to four turning points.

26. Rev.S0826 End Behavior of Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

27. Rev.S0827 Main Characteristics of Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Degree, x-intercepts, and turning points The graph of a polynomial function of degree n  1 has at most n x-intercepts and at most n  1 turning points.

28. Rev.S0828 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Use the graph of the polynomial function shown. a) How many turning points and x-intercepts are there? b) Is the leading coefficient a positive or negative? Is the degree odd or even? c) Determine the minimum degree of f.Solution a) There are three turning points corresponding to the one “hill” and two “valleys”. There appear to be 4 x-intercepts.

29. Rev.S0829 Example (Cont.) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. b) Is the leading coefficient a positive or negative? Is the degree odd or even? The left side and the right side rise. Therefore, a > 0 and the polynomial function has even degree. c) Determine the minimum degree of f. The graph has three turning points. A polynomial of degree n can have at most n  1 turning points. Therefore, f must be at least degree 4.

30. Rev.S0830 Another Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Graph f(x) = 2x 3  5x 2  5x + 7, and then complete the following. a) Identify the x-intercepts. b) Approximate the coordinates of any turning points to the nearest hundredth. c) Use the turning points to approximate any local extrema. Solution a) The graph appears to intersect the x-axis at the points (  1.3, 0), (0.89, 0) and (2.9, 0)

31. Rev.S0831 Let’s Practice One More Time http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Let f(x) = 3x 4 + 5x 3  2x 2. a) Give the degree and leading coefficient. b) State the end behavior of the graph of f. Solution a) The term with the highest degree is 3x 4 so the degree is 4 and the leading coefficient is 3. b) The degree is even and the leading coefficient is positive. Therefore the graph of f rises to the left and right. More formally,

32. Rev.S0832 Piecewise-Defined Polynomial Functions http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Example Evaluate f(x) at  6, 0, and 4. Solution To evaluate f(  6) we use the formula  5x because  6 is <  5. f(  6) =  5(  6) = 30 Similarly, f(0) = x 3 + 1 = (0) 3 + 1 = 1 f(4) = 3  x 2 = 3  (4) 2 =  13

33. Rev.S0833 Let’s Look at One More Example Complete the following. a) Sketch the graph of f. b) Determine if f is continuous on its domain. c) Evaluate f(1). Solution a) Graph as shown to the right. b) The domain is not continuous since there is a break in the graph. c) f(1) = 4  x 2 = 4 – (1) 2 = 3 http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

34. Rev.S0834 Review of Dividing Polynomials http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Divide x 3 + 2x 2  5x  6 by x  3. Check the result. Solution The quotient is x 2 + 5x + 10 with a remainder of 24. Check:

35. Rev.S0835 Synthetic Division http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. A short cut called synthetic division can be used to divide x – k into a polynomial. Steps 1. Write k to the left and the coefficients of f(x) to the right in the top row. If any power does not appear in f(x), include a 0 for that term. 2. Copy the leading coefficient of f(x) into the third row and multiply it by k. Write the result below the next coefficient of f(x) in the second row. Add the numbers in the second column and place the result in the third row. Repeat the process. 3. The last number in the third row is the remainder. If the remainder is 0, then the binomial x – k is a factor of f(x). The other numbers in the third row are the coefficients of the quotient in descending powers.

36. Rev.S0836 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Use synthetic division to divide 2x 3 + 7x 2 – 5 by x + 3. Solution Let k = –3 and perform the following. The remainder is 4 and the quotient is 2x 2 + x – 3. The result can be expressed as –3270–5 –6–39 21 4

37. Rev.S0837 Division Algorithm for Polynomials http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

38. Rev.S0838 What is the Remainder Theorem? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

39. Rev.S0839 What is the Factor Theorem? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

40. Rev.S0840 Example of Applying the Factor Theorem Use the graph of f(x) = x 3  x 2 – 9x + 9 and the factor theorem to list the factors of f(x). Solution The graph shows that the zeros or x-intercepts of f are  3, 1and 3. Since f(  3) = 0, the factor theorem states that (x + 3) is a factor, and f(1) = 0 implies that (x  1) is a factor, and f(3) = 0 implies (x  3) is a factor. Thus the factors are (x + 3)(x  1), and (x  3). http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

41. Rev.S0841 What is a Complete Factored Form? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

42. Rev.S0842 Example of a Complete Factored Form http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. The polynomial f(x) = 2x 3  3x 2  17x + 30 has a zero of 2. Express f(x) in complete factored form. Solution If 2 is a zero, by the factor theorem x  2 is a factor. Use synthetic division. 22– 3–1730 42–30 21–150

43. Rev.S0843 What is the Rational Zero Test? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Note: p is a factor of the last term - constant term. q is a factor of the leading coefficient of the first term - the highest power term. For Example: f(x) = 6x 4 + 7x 3  12x 2  3x + 2.

44. Rev.S0844 An Example of Finding All Rational Zeros http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Find all rational zeros of f(x) = 6x 4 + 7x 3  12x 2  3x + 2. Write in complete factored form. Solution Any rational zero must occur in the list

45. Rev.S0845 An Example of Finding All Rational Zeros (Cont.) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. xf(x)f(x)xf(x)f(x)xf(x)f(x) 10½  5/4 1/61.20 11 88 ½½ 0  1/6 2.14 21001/302/3  2.07 22 0  1/3 1.48  2/3  2.22 Evaluate f(x) at each value in the list. From the table there are four rational zeros of 1,  2,  1/2, and 1/3. The complete factored form is:

46. Rev.S0846 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Find all real solutions to the equation 4x 4 – 17x 2 – 50 = 0. Solution The expression can be factored similar to a quadratic equation. The only solutions are since the equation x 2 = –2 has no real solutions.

47. Rev.S0847 Another Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Solve the equation x3 – 2.1x2 – 7.1x + 0.9 = 0 graphically. Round any solutions to the nearest hundredth. Solution Since there are three x-intercepts the equation has three real solutions. x .012,  1.89, and 3.87

48. Rev.S0848 Quick Review on Complex Number A complex number can be written in standard form as a + bi, where a and b are real numbers. The real part is a and the imaginary part is b. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

49. Rev.S0849 Examples Write each expression in standard form. Support your results using a calculator. a)(  4 + 2i) + (6  3i)b) (  9i)  (4  7i) Solution a) (  4 + 2i) + (6  3i) =  4 + 6 + 2i  3i = 2  i b) (  9i)  (4  7i) =  4  9i + 7i =  4  2i http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

50. Rev.S0850 More Examples Write each expression in standard form. Support your results using a calculator. a) (  2 + 5i) 2 b) Solution a) (  2 + 5i) 2 = (  2 + 5i)(  2 + 5i) = 4 – 10i – 10i + 25i 2 = 4  20i + 25(  1) =  21  20i b) http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

51. Rev.S0851 Quadratic Equations with Complex Solutions We can use the quadratic formula to solve quadratic equations, even if the discriminant is negative. However, there are no real solutions, and the graph does not intersect the x-axis. The solutions can be expressed as imaginary numbers. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

52. Rev.S0852 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Solve the quadratic equation 4x 2 – 12x = –11. Solution Rewrite the equation: 4x 2 – 12x + 11 = 0 a = 4, b = –12, c = 11

53. Rev.S0853 What is the Fundamental Theorem of Algebra? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Fundamental Theorem of Algebra The polynomial f(x) of degree n  1 has at least one complex zero.

54. Rev.S0854 What is the Number of Zeros Theorem? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Number of Zeros Theorem A polynomial of degree n has at most n distinct zeros.

55. Rev.S0855 Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Represent a polynomial of degree 4 with leading coefficient 3 and zeros of  2, 4, i and  i in complete factored form and expanded form. Solution Let a n = 3, c 1 =  2, c 2 = 4, c 3 = i, and c 4 =  i. f(x) = 3(x + 2)(x  4)(x  i)(x + i) Expanded: 3(x + 2)(x  4)(x  i)(x + i) = 3(x + 2)(x  4)(x 2 + 1) = 3(x + 2)(x 3  4x 2 + x  4) = 3(x 4  2x 3  7x 2  2x  8) = 3x 4  6x 3 – 21x 2 – 6x – 24

56. Rev.S0856 What is the Conjugate Zeros Theorem? http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Conjugate Zeros Theorem If a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a  bi is also a zero of f(x).

57. Rev.S0857 Example of Applying the Conjugate Zeros Theorem http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Find the zeros of f(x) = x 4 + 5x 2 + 4 given one zero is  i. Solution By the conjugate zeros theorem it follows that i must be a zero of f(x). (x + i) and (x  i) are factors (x + i)(x  i) = x 2 + 1, using long division we can find another quadratic factor of f(x). The solution is x 4 + 5x 2 + 4 = (x 2 + 4)(x 2 + 1)

58. Rev.S0858 Another Example http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. Solve x 3 = 2x 2  5x + 10. Solution Rewrite the equation: f(x) = 0, where f(x) = x 3  2x 2 + 5x  10 We can use factoring by grouping or graphing to find one real zero. The graph shows a zero at 2. So, x  2 is a factor.

59. Rev.S0859 Another Example (Cont.) Let’s try to solve this by synthetic division: x 3  2x 2 + 5x  10 = (x  2)(x 2 + 5) The solutions are 2 and http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules. 21–25–10 2010 1050

60. Rev.S0860 What have we learned? We have learned to 1.identify intervals where a function is increasing or decreasing. 2.find extrema of a function. 3.identify symmetry in a graph of a function. 4.determine if a function is odd, even, or neither. 5.understand the graphs of polynomial functions. 6.evaluate and graph piecewise-defined functions. 7.divide polynomials. 8.understand the division algorithm, remainder theorem, and factor theorem. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

61. Rev.S0861 What have we learned? (Cont.) 9.factor higher degree polynomials. 10.analyze polynomials with multiple zeros. 11.solve polynomial equations. 12.perform arithmetic operations on complex numbers. 13.solve quadratic equations having complex solutions. 14.apply the fundamental theorem of algebra. 15.factor polynomials having complex zeros. 16.solve polynomial equations having complex solutions. http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.

62. Rev.S0862 Credit Some of these slides have been adapted/modified in part/whole from the slides of the following textbook: Rockswold, Gary, Precalculus with Modeling and Visualization, 3th Edition http://faculty.valenciacc.edu/ashaw/ http://faculty.valenciacc.edu/ashaw/ Click link to download other modules.