Concepts Examples 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. End Behavior The end behavior of the graph of a polynomial function ƒ1x2 is determined by the dominating term, or term of greatest degree. If axn is the dominating term of ƒ1x2, then the end behavior is as follows. a > 0 n odd a < 0 n odd a > 0 n even a < 0 n even Graphing a Polynomial Function To graph a polynomial function ƒ, follow these steps. Step 1 Find the real zeros of ƒ. Plot the corresponding x-intercepts. Step 2 Find ƒ102 = a0. Plot the corresponding y-intercept. Step 3 Use end behavior, whether the graph crosses, bounces on, or wiggles through the x-axis at the x-intercepts, and selected points as necessary to complete the graph. Intermediate Value Theorem If ƒ1x2 is a polynomial function with only real coefficients, and if for real numbers a and b the values of ƒ1a2 and ƒ1b2 are opposite in sign, then there exists at least one real zero between a and b. Boundedness Theorem Let ƒ1x2 be a polynomial function of degree n Ú 1 with real coefficients and with a positive leading coefficient. Suppose ƒ1x2 is divided synthetically by x - c. (a) If c 70 and all numbers in the bottom row of the synthetic division are nonnegative, then ƒ1x2 has no zero greater than c. (b) If c 60 and the numbers in the bottom row of the synthetic division alternate in sign (with 0 considered positive or negative, as needed), then ƒ1x2 has no zero less than c. The graph of ƒ1x2 = 4x5 - 2x3 + 3x2 + x - 10 has at most four turning points (because 5 - 1 = 4). The end behavior of ƒ1x2 = 3x5 + 2x2 + 7 is . The end behavior of ƒ1x2 = -x4 - 3x3 + 2x - 9 is . Graph ƒ1x2 = 1x + 221x - 121x + 32. The x-intercepts correspond to the zeros of ƒ, which are -2, 1, and -3. Because ƒ102 =21-12132 = -6, the y-intercept is 10, -62. The dominating term is x1x21x2, or x3, so the end behavior is . Begin at either end of the graph with the correct end behavior, and draw a smooth curve that crosses the x-axis at each zero, has a turning point between successive zeros, and passes through the y-intercept. For the polynomial function ƒ1x2 = -x4 + 2x3 + 3x2 + 6, ƒ13.12 = 2.0599 and ƒ13.22 = -2.6016. Because ƒ13.12 70 and ƒ13.22 60, there exists at least one real zero between 3.1 and 3.2. Show that ƒ1x2 = x3 - x2 - 8x + 12 has no zero greater than 4 and no zero less than -4. 4)1 -1 -8 12 4 12 16 1 3 4 28 All signs positive -4)1 -1 -8 12 -4 20 -48 1 -5 12 -36 Alternating signs x y 0 –6 –2 1 f(x) = (x + 2)(x – 1)(x + 3) 423 CHAPTER 3 Test Prep
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