# How do you find the vertex and intercepts for y=2(x+2)^2-13?

Dec 2, 2015

Find vertex and intercepts for y = 2(x + 2)^2 - 13

#### Explanation:

y = 2(x + 2)^2 - 13.
This is the intercept form, therefor the coordinates of vertex are: Vertex (-2, -13).
To find y-intercept, make x = 0 --> $y = 2 {\left(2\right)}^{2} - 13 = 8 - 13 = - 5$
To find x-intercepts, make y = 0 --> $0 = 2 {\left(x + 2\right)}^{2} - 13$
${\left(x + 2\right)}^{2} = \frac{13}{2} - \to x + 2 = \pm \frac{\sqrt{13}}{\sqrt{2}}$
$x = - 2 \pm \frac{\sqrt{26}}{2}$