Question

How do you find the vertex of the parabola `y=2x^2-5`?

Answer 1

The vertex is at `(0,-5)`

Explanation:

The equation of parabola is `y=2(x-0)^2-5` and the equation in general form is `y=2(x-h)^2+k` where `(h,k)` is the vertex.So comparing with general form the vertex is at `(0,-5)` graph{2x^2-5 [-20, 20, -10, 10]}[Ans]

Answer 2

The vertex is at the point `(0,-5)`

Explanation:

The parabola equation is of the form:

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

The vertex is at the point `(h,k)`

The given equation is

`y=2x^2-5`

Manipulate your equation so that you obtain a form similar to the general one.

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

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

By identification, `h=0` and `k=-5`

So the vertex is at the point `(0,-5)`

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