## Precalculus (10th Edition)

maximum: $16$
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 $f(2)=-3(2)^2+12(2)+4=16.$