Question

How do you find the vertex and the intercepts for `f(x)=5x^2`?

Answer 1

Vertex: `(0,0)`
`x`-intercept: `y=0`
`y`-intercept: `x=0`

Explanation:

The vertex form of a quadratic equation is expressed as

`y=a(x-h)^2+k`

Where `(h,k)` is the vertex. That vertex is a maximum if the coefficient of `x^2` term is negative. But the vertex is a minimum if the coefficient of the `x^2` term is positive.

For the equation we are given, we can easily add zeros to put it into vertex form

`y=5x^2` becomes

`y=5(x-0)^2+0`

This quadratic equation has a vertex at `(0,0)` and, because the coefficient `5` of the `x^2` term is positive, we know this vertex is a minimum.

The `y`-intercept occurs when `x=0`, which is again the origin at `(0,0)`. The `x`-intercept occurs when `y=0`, which can only happen when `x=0`. So the only `x` or `y` intercept happens at the vertex, all of which converge at the origin: `(0,0)`.

The graph is as follows:

graph{5x^2[-2,2,-1.5,10]}