What is the vertex of y = (x-5)^2 +2?

1 Answer
Oct 31, 2015

Vertex =(5,2)

Explanation:

In a parabolic function y=a(x-p)^2+q, the vertex is (-p,q).
So in your function of y=(x-5)^2+2,

p=-(-5)
q=(+2)

So the point of the vertex would be (5,2)

Here is the graph:
graph{y=(x-5)^2+2 [-10, 10, -5, 5]}