# How do you find the vertex and intercepts for y=2(x+2)^2-1?

Jun 5, 2017

Vertex (-2, -1)

#### Explanation:

Coordinate of vertex:
x + 2 = 0 --> x = - 2
y-coordinate of vertex: -1
Vertex (-2, -1)
To find the 2 x-intercepts, make y= 0, and solve the quadratic equation:
$2 {\left(x + 2\right)}^{2} - 1 = 0$
${\left(x + 2\right)}^{2} = \frac{1}{2}$
$\left(x + 2\right) = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$
$x = - 2 \pm \frac{\sqrt{2}}{2} = \frac{- 4 \pm \sqrt{2}}{2}$