# What is the vertex of y= -1/16(2x-4)^2+8?

Dec 30, 2015

$\left(2 , 8\right)$

#### Explanation:

This is almost in vertex form, except for that there is a $2$ being multiplied by the $x$.

$y = a {\left(x - h\right)}^{2} + k$

$y = - \frac{1}{16} \left(2 x - 4\right) \left(2 x - 4\right) + 8$

$y = - \frac{1}{4} {\left(x - 2\right)}^{2} + 8$

(Since the $2 x - 4$ term is squared, a $2$ is factored from each term.)

This is now in vertex form.

The center is at $\left(h , k\right) \rightarrow \left(2 , 8\right)$.

graph{-1/16(2x-4)^2+8 [-13.78, 14.7, -2.26, 11.98]}