## Precalculus (10th Edition)

The graph has a maximum value of $18$.
RECALL: (1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens: (i) upward when $a \gt0$ and has its vertex as its minimum; or (ii) downward when $a\lt0$ and has its vertex as its maximum. (2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$ Let's compare $f(x)=-2x^2+12x$ to $f(x)=ax^2+bx+c$. We can see that $a=-2$, $b=12$, $c=0$. $a\gt0$, hence the graph opens down, hence it's vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot (-2)}=3.$ Thus, the maximum value is $f(3)=-2(3)^2+12(3)=18.$