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]}