232 CHAPTER 4 Polynomial and Rational Functions Using the Conjugate Pairs Theorem A polynomial function f of degree 5 whose coefficients are real numbers has the zeros 1, i5 , and +i 1 . Find the remaining two zeros. Solution EXAMPLE 1 Since f has coefficients that are real numbers, complex zeros appear as conjugate pairs. It follows that − i5, the conjugate of i5 , and −i 1 , the conjugate of +i 1 , are the two remaining zeros. Now Work PROBLEM 9 2 Find a Polynomial Function with Specified Zeros Finding a Polynomial Function Whose Zeros Are Given Find a polynomial function f of degree 4 whose coefficients are real numbers that has the zeros 1, 1, and − +i 4 . Solution EXAMPLE 2 Since − +i 4 is a zero, by the Conjugate Pairs Theorem, − −i 4 is also a zero of f . From the Factor Theorem, if ( ) = f c 0, then −x c is a factor of ( ) f x . So f can now be written as ( ) ( )( ) ( ) [ ] ( ) [ ] = − − − − + − − − f x a x x x i x i 1 1 4 4 where a is any nonzero real number. Then [ ] ( ) ( ) ( )[ ] ( )( ) ( ) ( ) ( ) ( )( ) ( ) [ ] ( ) [ ] ( ) [ ] ( ) [ ] ( ) ( ) = − − − − + − − − = − + + − + + = − + + − = − + + + − − = − + + + = + + − − − + + + = + + − + f x a x x x i x i a x x x i x i a x x x i a x x x x a x x x x a x x x x x x x x a x x x x 1 1 4 4 2 1 4 4 2 1 4 2 1 8 16 1 2 1 8 17 8 17 2 16 34 8 17 6 2 26 17 2 2 2 2 2 2 2 2 4 3 2 3 2 2 4 3 2 =−1 2i Exploration Graph the function f found in Example 2 for = a 1. Does the value of a affect the zeros of f ? How does the value of a affect the graph of f ? What information about f is sufficient to uniquely determine a? Result An analysis of the polynomial function f tells us what to expect: • At most three turning points. • For large x , the graph resembles the graph of = y x .4 • A repeated real zero at 1, so the graph will touch the x-axis at 1. • The only x-intercept is 1; the y-intercept is 17. Figure 33 shows the complete graph on a TI-84 Plus CE. (Do you see why? The graph has exactly three turning points.) The value of a causes a stretch or compression; a reflection also occurs if < a 0. The zeros are not affected. If any point other than an x-intercept on the graph of f is known, then a can be determined. For example, if (2, 3) is on the graph, then ( ) ( ) = = f a 2 3 37 , so = a 3 37. Why won’t an x-intercept work? Figure 33 ( ) = + + − + f x x x x x 6 2 26 17 4 3 2 25 25 50 3
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