Question

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

Answer 1

They are all at `(0,0)`.

Explanation:

Finding the vertex:
Use the formula `(-b/(2a), f(-b/(2a)))`, given the form of the parabola `f(x)=ax^2+bx+c`. In your case, `a=2,b=0` and `c=0`.
Here, `-b/(2a)=-0/(2*2)=0` and `f(-b/(2a))=f(0)=2(0)^2=0`.
Thus, the vertex is located at `(0,0)`.

Finding the X-intercept:
Plug in `0` for `y`.
`0 = 2x^2=>0=x^2=>0=x`
Then plug in `0` for `y` to determine the point of the intercept. We already know that if we plug in `0` for either point, we will receive `0` as our answer, so the X-intercept is `(0,0)`.

Finding the Y-intercept:
This is very similar to finding the X-intercept, in that we now plug in `0` for `x`, which we already know will result in the point `(0,0)`.

Alternative method: Graph the parabola and look at the vertex and intercepts.