Question

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

Answer 1

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]}