## Precalculus (10th Edition)

maximum: $12$
Let's compare $f(x)=-x^2+8x-4$ to $f(x)=ax^2+bx+c$. We can see that a=-1, b=8, c=-4. $a\lt0$, hence the graph opens down, thus it's vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{8}{2\cdot(-1)}=4.$ Hence the maximum value is $f(2)=-4^2+8(4)-4=12.$