Precalculus (6th Edition) Blitzer

The required value is $\left( 2,\ 12 \right)$
The quadratic function in its standard form can be written as $f\left( x \right)=a{{\left( x-h \right)}^{2}}+k,\,\,\,\,\,\,\,\,a\ne 0$. Here, $a,\ h,\ k$ are constants and $x$ is a variable. The graph of $f\left( x \right)$ is a parabola which is symmetric about the line $x=h$. Now, the coordinates of the vertex of the parabola are $\left( h,\ k \right)$. Consider the given function $f\left( x \right)=-3{{\left( x-2 \right)}^{2}}+12$. Compare the function with the standard form. Now, the point $\left( h,\ k \right)$ is $\left( 2,\ 12 \right)$