How do you find the vertex of a parabola y=2(x+3)^2+6?

Jun 28, 2015

The vertex of $y = 2 {\left(x + 3\right)}^{2} + 6$ is at $\left(- 3 , 6\right)$

Explanation:

Any parabolic equation of the form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$
is said to be in vertex form with it's vertex at $\left(a , b\right)$

$y = 2 {\left(x + 3\right)}^{2} + 6$ is almost in vertex form.

It can easily be converted to vertex form as:
$\textcolor{w h i t e}{\text{XXXX}}$$y = 2 {\left(x - \left(- 3\right)\right)}^{2} + 6$