How do you find the vertex of $y = 4 {(x - 3)}^{2} - 1$ $y=4(x-3)^2-1$ ?

Question

2633 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-10T00:24:21

This parabola is already in vertex form ...

Vertex form is:

$y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ with vertex $(h, k)$ $(h,k)$

For this problem, vertex $= (3, - 1)$ $=(3,-1)$

hope that helped

enter image source here

How do you find the vertex of y=4(x−3)2−1y=4(x-3)^2-1?