# How do you find the vertex and intercepts for y = 6(x - 2)^2 ­- 8?

Nov 25, 2015

vertex $\left(2 , - 8\right)$

#### Explanation:

Any quadratic equation in the form
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$
is in "vertex form" with its vertex at $\left(a , b\right)$

$y = 6 {\left(x - 2\right)}^{2} - 8 \iff y = 6 {\left(x - 2\right)}^{2} + \left(- 8\right)$
is in this form
and therefore has its vertex at $\left(2 , - 8\right)$