y=x^2-4x+3 can be factored as y=(x-1) (x-3)
so the vertex of the parabola is at x=2, midway between 1 and 3. To get the
left side of the parabola you need to force the x-coordinate to be 2 or less. That
is you need x to be 2 less something that is certain, or if nothing else 2 less
something that is never negative. In this manner let x=2-t^2 and whatever value
t has, x will be not exactly or equivalent to 2. Presently substitute x=2-t^2
into y=x^2-4x+3 to find y as far as t.
