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

2 Answers
Jun 17, 2016

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]

Jun 17, 2016

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]}