Answer
The vertex form of the function is $y=(x-2)^{2}+(-5).$ The vertex is $(2,-5)$.
Work Step by Step
$ y=x^{2}-4x-1\qquad$ ...prepare to complete the square.
$ y+?=x^{2}-4x+?-1\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-4}{2})^{2}=(-2)^{2}=4\qquad$ ...complete the square by adding $4$ to each side of the expression
$ y+4=x^{2}-4x+4-1\qquad$ ... write $x^{2}-4x+4$ as a binomial squared.
$ y+4=(x-2)^{2}-1\qquad$ ...add $-4$ to each side of the expression
$ y+4-4=(x-2)^{2}-1-4\qquad$ ...simplify.
$ y=(x-2)^{2}-5\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$y=(x-2)^{2}+(-5)$
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, $h=2,\ k=-5$, so the vertex is $(2,-5)$
