# How do you find the vertex of the parabola y=2x^2-5?

Jun 17, 2016

The vertex is at $\left(0 , - 5\right)$

#### Explanation:

The equation of parabola is $y = 2 {\left(x - 0\right)}^{2} - 5$ and the equation in general form is $y = 2 {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex.So comparing with general form the vertex is at $\left(0 , - 5\right)$ graph{2x^2-5 [-20, 20, -10, 10]}[Ans]

Jun 17, 2016

The vertex is at the point $\left(0 , - 5\right)$

#### Explanation:

The parabola equation is of the form:

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

The vertex is at the point $\left(h , k\right)$

The given equation is

$y = 2 {x}^{2} - 5$

Manipulate your equation so that you obtain a form similar to the general one.

$y = 2 {\left(x - 0\right)}^{2} + \left(- 5\right)$

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

By identification, $h = 0$ and $k = - 5$

So the vertex is at the point $\left(0 , - 5\right)$

graph{2*x^2-5 [-10, 10, -5, 5]}