# How do you find the axis of symmetry and vertex point of the function: y=8(x-10)^2-16?

##### 1 Answer
Sep 29, 2015

Axis of Symmetry: $x = 10$
Vertex: $\left(10 , - 16\right)$

#### Explanation:

For a quadratic equation in vertex form $y = a {\left(x - h\right)}^{2} + k$, the axis of symmetry is $x = h$ and the vertex is $\left(h , k\right)$.

Now take a look at your equation:
$y = a {\left(x - h\right)}^{2} + k$
y=8(x−10)^2−16

Here:
$a = 8$
$h = 10$
$k = - 16$

Axis of Symmetry
$x = h$
$\textcolor{b l u e}{x = 10}$

Vertex
$\left(h , k\right)$
color(red)((10,-16)