Question

How do you find the vertex and intercepts for `y = 1/20x^2`?

Answer 1

Vertex is at `(0,0)`
y-intercept at `0`
x-intercept (only one) at `0`

Explanation:

One way to look at this is to considered the standard vertex form:
`color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)`
with vertex at `(color(red)(a),color(blue)(b))`

the given equation: `y=1/20x^2` can be written in explicit standard vertex form as:
`color(white)("XXX")y=color(green)(1/20)(x-color(red)(0))^2+color(blue)(0)`
with vertex at `(color(red)(0),color(blue)(0))`

The y-intercept is the value of `y` when `x=0`
`color(white)("XXX")y=1/20(0)^2=0`

The x-intercept(s) is(are) the value(s) of `x` when `y=0`
`color(white)("XXX")0=1/20x^2 rarr x=0`