Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 2 - Section 2.6 - Function Notation and Linear Functions - 2.6 Exercises: 31

Answer

$-\frac{p^{2}}{9}+\frac{4p}{3}+1$

Work Step by Step

We are given that $g(x)=-x^{2}+4x+1$. In order to evaluate the function at $x=\frac{p}{3}$, we can plug in $\frac{p}{3}$ to the function and evaluate directly. $g(\frac{p}{3})=-x^{2}+4x+1=-(\frac{p}{3})^{2}+4(\frac{p}{3})+1=-(\frac{p^{2}}{9})+\frac{4p}{3}+1=-\frac{p^{2}}{9}+\frac{4p}{3}+1$
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