## Calculus (3rd Edition)

$21$
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 $f(3)=21$