Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 3 - Linear and Quadratic Functions - Chapter Review - Review Exercises - Page 163: 16


maximum: $16$

Work Step by Step

Let's compare $f(x)=-3x^2+12x+4$ to $f(x)=ax^2+bx+c$. We can see that $a=-3, b=12, c=4$. $a\lt0$, hence the graph opens down, thus it's vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot(-3)}=2.$ Hence the maximum value is $f(2)=-3(2)^2+12(2)+4=16.$
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