Originally Posted by

**ticbol** Umm, so the "completing the square" was done to determine the vertex of the parabola.

Then your teacher's method is okay.

But you used 0 = x^2 -4x +3.

That is not a parabola. Where is the y or f(x) there?

The parabola is y or f(x) = x^2 -4x +3.

So your teacher "completed the square" of the RHS to arrive at

y = (x-2)^2 -1

A standard form of the equation of a vertical parabola ....where the x is squared....is

(y -k) = a(x -h)^2

where (h,k) is the vertex.

So, in the parabola y = (x-2)^2 -1,

rearranging it,

(y +1) = (x -2)^2

the h is 2, and the k is -1.

Hence, the vertex is (2,-1)