Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 37: 30



Work Step by Step

We are given $$h(z)=-2 z^{2}+12 z+3$$ We factor the equation to get it into parabola/quadratic form: \begin{align*} h(z)&=-2 z^{2}+12 z+3\\ &= -2[z^2-6z-3/2]\\ &= -2[(z-3)^2-9-3/2] \\ &=-2(z-3)^{2}+21 \end{align*} Compare with the quadratic function $$f(x)=a(x-h)^{2}+k$$ We see that the vertex is $(h,k)= (3,21)$. Since $a<0$, then the maximum value is $f(3)=21$
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