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1. Chapter 3 Polynomial and Rational Functions

2. 3.1 Quadratic Functions and Models

3. 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 an, an 1, …a1, and a0 are real numbers, with an 0. • Examples

4. Quadratic Functions • A function f is a quadratic function if f(x) = ax2 + bx + c, where a, b, and c are real numbers with a 0. • The simplest quadratic function is f(x) = x2. 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.

5. Examples • Graph each function. Give the domain and range. • a) • b) • c)

6. The domain is (, ). The range is [7, ) The vertex is (3, 7) a) x y 0 2 1 3 2 6 3 7 5 3 6 2 Solutions

7. b) The graph is broader than y = x2 it is reflected across the x-axis. The vertex is (0, 0) The domain is (, ) The range is (, 0]. Solutions continued

8. 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]. Solutions continued

9. Completing the Square • Graph f(x) = x2 4x + 7 by completing the square and locating the vertex. • Solution: • The vertex is (2, 3) and the axis is the line x = 2.

10. Find additional ordered pairs that satisfy the equation. Plot the points and connect in a smooth curve. x y 0 7 1 4 2 3 3 4 4 7 Completing the Square continued

11. Example • Graph f(x) = 3x2 + 6x  1 by completing the square and locating the vertex.

12. The vertex is (1, 2). Find additional points by substituting x-values into the original equation. x y 0 1 1 2 2 1 Example continued

13. Graph of a Quadratic Function • The quadratic function defined by f(x) = ax2 + 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.

14. 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 = x2 if |a| < 1 and narrower if |a| > 1. • 4. The y-intercept is f(0) = c. • 5. If b2 4ac  0, the x-intercepts are If b2  4ac < 0, there are no x-intercepts.

15. Example • Find the axis and vertex of the parabola having equation f(x) = 4x2 + 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).

16. 3.2 Synthetic Division

17. 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).

18. 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.

19. 2x3 4x2 x2 x2  2x 2x  32 2x  4  28 2 4 1 1  2 2  32 2  4  28 Example

20. 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. Example continued

21. 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 2x2 + x + 2  Remainder

22. Caution • Note: To avoid errors, use 0 as coefficient for any missing terms, including a missing constant, when setting up the division.

23. Example Use synthetic division to divide 2x3 3x2  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

24. 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

25. Example continued Add 7 and 6 to obtain 1. 6 9  6 2 3  21 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

26. 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) = 2x3  3x2  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

27. Applying the Remainder Theorem Example: Let f(x) = 3x4 7x3  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

28. 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.

29. Example Decide whether the given number k is a zero of f(x). • f(x) = 2x3 2x2  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) = 2x3  2x2  34x  30 • f(x) = 2x3 + 4x2  x + 5 k = 2 Solution: Since the remainder is 7, not 0, 2 is not a zero of f(x) = 2x3 + 4x2  x + 5 . In fact, f(2 ) = 7.

30. 3.3 Zeros of Polynomial Functions

31. 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.

32. Example Determine whether x 3 is a factor of f(x) for f(x) = x3  7x2 + 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

33. 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 x2  4x  1, and f(x) = (x  3) (x2  4x  1).

34. 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.

35. Using the Rational Zeros Theorem Example: Do each of the following for the polynomial function defined by f(x) 6x3 5x2  7x + 4. • List all possible rational zeros. • Find all rational zeros and factors f(x) into linear factors.

36. Using the Rational Zeros Theorem continued Solution: • For a rational number to be a zero, p must be a factor of a0 = 4 and q must be a factor of a4 = 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

37. Using the Rational Zeros Theorem continued The new quotient for the polynomial is 6x2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) = 6x3  5x2 7x + 4.

38. 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) = x3 + 3x2 + 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.

39. 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.

40. 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) =

41. 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).

42. 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)(x2 + 25) = x3  3x2 + 25x  75. • Any nonzero multiple also satisfies the given conditions.

43. 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.

44. Example • Determine the possible number of positive real number zeros and negative real number zeros of f(x) = x6 + 7x4 x3  2x2 + 6x  5. • Solution: First consider the possible number of positive zeros by observing that f(x) has three variations in signs. + x6 + 7x4 x3  2x2 + 6x  5 123

45. 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 = x6 + 7x4 + x3  2x2  6x  5. • Since there is only one variation in sign, f(x) has only 1 negative real zero.

46. 3.4 Polynomial Functions: Graphs, Applications, and Models

47. x f(x) x g(x) 1 2 1 1/3 0 0 0 0 1 2 1 1/3 2 16 1.5 2.53 Graphing Functions of the Formf(x) = axn • Graph each function.

48. a) f(x) = x5 + 1 The graph of f(x) = x5 + 1 will be the same as that of f(x) = x5, but translated up 1 unit. b) g(x) = (x 1)6 The graph is translated 1 unit to the right. Horizontal and Vertical TranslationsGraph the function.

49. 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.