Answer
The vertex form of the function is $f(x)=(x-2)^{2}+(-8).$ The vertex is $(2,-8)$.
Work Step by Step
$ f(x)=x^{2}-4x-4\qquad$ ...prepare to complete the square.
$ f(x)+?=(x^{2}-4x+?)-4\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-4}{2})^{2}=(-2)^{2}=4\qquad$ ...add $4$ to each side of the expression
$ f(x)+4=(x^{2}-4x+4)-4\qquad$ ... write $x^{2}-4x+4$ as a binomial squared.
$ f(x)+4=(x-2)^{2}-4\qquad$ ...add $-4$ to each side of the expression
$ f(x)+4-4=(x-2)^{2}-4-4\qquad$ ...simplify.
$ f(x)=(x-2)^{2}-8\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$f(x)=(x-2)^{2}+(-8)$
The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph.
Here, $h=2,\ k=-8$, so the vertex is $(2,-8)$
