Answer
The vertex form of the function is $f(x)=(x-5)^{2}+(-8)$. The vertex is $(5,-8)$.
Work Step by Step
$ f(x)=x^{2}-10x+17\qquad$ ...write in form of $x^{2}+bx=c$ (add $-17$ to each side).
$ f(x)-17=x^{2}-10x\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-10}{2})^{2}=(-5)^{2}=25\qquad$ ...complete the square by adding $25$ to each side of the expression
$ f(x)-17+25=x^{2}-10x+25\qquad$ ... write $x^{2}-10x+25$ as a binomial squared.
$ f(x)+8=(x-5)^{2}\qquad$ ...add $-8$ to each side of the expression
$ f(x)=(x-5)^{2}-8\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$f(x)=(x-5)^{2}+(-8)$
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, $h=5,\ k=-8$, so the vertex is $(5,-8)$
