# What is the vertex of  y= 1/5x^2 ?

Nov 26, 2015

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

#### Explanation:

The standard equation for a parabola (non-conic) is
y= a(x-h)^2 +k ; => a != 0 , h, k are real number
the vertex is $\left(h , k\right)$

The equation $y = \frac{1}{5} {x}^{2} \implies y = \frac{1}{5} {\left(x - \textcolor{red}{0}\right)}^{2} + \textcolor{red}{0}$

Thus the vertex is $\left(0 , 0\right)$ , and graph will look like this

graph{1/5x^2 [-10, 10, -5, 5]}