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Stitz-Zeager_College_Algebra_e-book

# It follows that pc c c q c r 0 q c r r

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Unformatted text preview: ctor out the 2, however, we get (2x − 1) = 2 x − 2 , and we see the multiplicity of x = 1 is 2 1. Since 1 is an odd number, we know from Theorem 3.3 that the graph of f will cross through 1 the x-axis at 2 , 0 . Since the zero x = −1 corresponds to the factor (x + 1)2 = (x − (−1))2 , we see its multiplicity is 2 which is an even number. As such, the graph of f will touch and rebound from the x-axis at (−1, 0). Though we’re not asked to, we can ﬁnd the y -intercept by ﬁnding f (0) = −3(2(0) − 1)(0 + 1)2 = 3. Thus (0, 3) is an additional point on the graph. Putting this together gives us the graph below. y x 16 Obtaining the factored form of a polynomial is the main focus of the next few sections. 190 3.1.1 Polynomial Functions Exercises 1. For each polynomial given below, ﬁnd the degree, the leading term, the leading coeﬃcient, the constant term and the end behavior. √ (a) f (x) = 3x17 + 22.5x10 − πx7 + 1 (d) s(t) = −4.9t2 + v0 t + s0 3 (b) p(t) = −t2 (3 − 5t)(t2 + t + 4) (c) Z (b...
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