
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)