Answer
The vertex form of the function is $y=(x-4)^{2}+1.$ The vertex is $(4,1)$.
Work Step by Step
$ y=x^{2}-8x+17\qquad$ ...prepare to complete the square.
$ y+(?)=x^{2}-8x+(?)+17\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-8}{2})^{2}=(-4)^{2}=16\qquad$ ...add $16$ to each side of the expression
$ y+(16)=x^{2}-8x+(16)+17\qquad$ ... write $x^{2}-8x+16$ as a binomial squared.
$ y+16=(x-4)^{2}+17\qquad$ ...add $-16$ to each side of the expression
$ y+16-16=(x-4)^{2}+17-16\qquad$ ...simplify.
$y=(x-4)^{2}+1$
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, $h=4,\ k=1$, so the vertex is $(4,1)$
