Question

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

Answer 1

Vertex is `(0,0)`

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 `(h,k)`

The equation `y= 1/5 x^2 => y= 1/5 (x-color(red)0)^2 + color(red)0`

Thus the vertex is `(0,0)` , and graph will look like this

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