College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 5 - Section 5.1 - Polynomial Functions and Models - 5.1 Assess Your Understanding - Page 340: 80

Answer

Using the smallest degree possible, we get $f(x)=x^2(x+3)(x+1)(x-2)$

Work Step by Step

Thanks to the real zeros we see on the graph, we can find the function through its factored form: $f(x)=a(x+3)(x+1)(x-0)^2(x-2)$ $f(x)=ax^2(x+3)(x+1)(x-2)$ That's because when any (x-r) equals zero, f(x) equals zero. The multiplicity of (x+3), (x+1), and (x-2) could be any positive odd integer since the graph crosses the x-axis at (-3,0), (-1,0), and (2,0) respectively. The multiplicity of (x) could be any positive even integer since the graph touches the x-axis at (0,0). Because the exercise asks for the smallest degree possible, we'll set the odd multiplicities to 1 and the even multiplicities to 2. The only thing left is to find 'a'. So, using the given point, we can replace the values of 'x' and f(x) and solve for 'a': $16=a(-2)^2(-2+3)(-2+1)(-2-2)$ $16=a(4)(1)(-1)(-4)$ $16=16a$ $16/16=16a/16$ $a=1$ Thus, we get: $f(x)=x^2(x+3)(x+1)(x-2)$
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