# How do you find the axis of symmetry and vertex point of the function: y=1/20x^2?

If there are only ${x}^{2}$'s involved, the answer is allways
$x = 0 \mathmr{and} \left(0 , 0\right)$
A non-square $x$ or a number in the function would displace both the axis and the vertex. The 1/20 only makes it flatter as compared to the standard ${x}^{2}$-graph.