Find the parametrization for the left half of the parábola y=x^2-4x+3

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.