Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Chapter Test - Page 188: 10

Answer

$2x^2-4x-30$

Work Step by Step

Step 1. Based on the given graph, identify zeros $x=-3,5$ and we can write a general form as $y=a(x+3)(x-5)$ Step 2. As $(0,-30)$ is on the curve, we have $-30=a(3)(-5)$, thus $a=2$ Step 3. Thus function is $y=f(x)=2(x+3)(x-5)=2(x^2-2x-15)=2x^2-4x-30$
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