# What is the axis of symmetry and vertex for the graph y=1/20x^2?

May 23, 2017

vertex: $\left(0 , 0\right)$; axis of symmetry: $x = 0$

#### Explanation:

Given: $y = \frac{1}{20} {x}^{2}$

Find the vertex : When $y = A {x}^{2} + B x + C = 0$ the vertex is $\left(h , k\right)$, where $h = \frac{- B}{2 A}$:

$h = - \frac{0}{2 \cdot \frac{1}{20}} = 0$

$k = f \left(h\right) = \frac{1}{20} {\left(0\right)}^{2} = 0$

$\text{vertex} : \left(0 , 0\right)$

Find the axis of symmetry , $x = h$:

axis of symmetry, $x = 0$