Question

Find the vertex of y=(x-3)^2

Answer 1

The vertex form of parabola is,

$y=a(x-p)^2+q$

Where $(p,q)$ is vertex point.

In this case our parabola is given by,

$y=1(x-3)^2+0$

Hence the vertex is at $(3,0)$.

Hope this helps.