Answer
See below.
Work Step by Step
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, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot(-3)}=2.$ Hence the maximum value is $f(2)=-3(2)^2+12(2)+4=16.$