Answer
$(-1,4)$ is the vertex.
Work Step by Step
$f(x) = -3x^2+6x+4$
$f(x) = (-3x^2+6x)+4$
$f(x) = -3(x^2+2x)+4$
$f(x) = -3(x^2+2x+(2/2)^2)+4$
$f(x) = -3(x^2+2x+1)+4$
$f(x) = -3(x+1)^2+4$
The vertex is of the form $(h,k)$, and this quadratic equation is of the form $f(x)=a(x−h)^2+k$.
$(-1,4)$ is the vertex.