# What is the vertex of  y= 2|x|^2 – 1 ?

Jan 30, 2016

Vertex : $\left(0 , - 1\right)$

#### Explanation:

$y = 2 {\left\mid x \right\mid}^{2} - 1$

This should give us a parabola and this equation is same as

$y = 2 {x}^{2} - 1$ as ${\left\mid x \right\mid}^{2}$ and ${x}^{2}$ would give the same value as on squaring we would get only the positive value.

The vertex of $y = 2 {x}^{2} - 1$ can be found by comparing it with the vertex form
$y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex

$y = 2 {\left(x - 0\right)}^{2} - 1$
$y = a {\left(x - h\right)}^{2} + k$

We can see $h = 0$ and $k = - 1$

Vertex is $\left(0 , - 1\right)$