Answer
The vertex form of the function is $y=2(x-7)^{2}+1.$ The vertex is $(7,1)$.
Work Step by Step
$ y=2x^{2}-28x+99\qquad$ ...factor out $2$ from the first two terms.
$ y=2(x^{2}-14x)+99\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-14}{2})^{2}=7^{2}=49\qquad$ ...complete the square by adding$ 2\cdot 49$ to each side of the expression
$ y+2\cdot 49=2(x^{2}-14x)+99+2\cdot 49\qquad$ ... ...factor out $2$ from the first and third term on the right side of the expression.
$ y+98=2(x-14x+49)^{2}+99\qquad$ ... write $x-14x+49$ as a binomial squared.
$ y+98=2(x-7)^{2}+99\qquad$ ...solve for $y$ by adding $-98$ to each side
$ y+98-98=2(x-7)^{2}+99-98\qquad$ ...simplify.
$y=2(x-7)^{2}+1$
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, $h=7,\ k=1$, so the vertex is $(7,1)$
