jmc

The denominator factors as $x^2-4x+4=(x-2{)}^2,$ so the vertical asymptote is $x=2.$

Since $$y=\frac{x^2-4x+3}{x^2-4x+4}=\frac{(x^2-4x+4)-1}{x^2-4x+4}=1-\frac{1}{x^2-4x+4}.$$Thus, the horizontal asymptote is $y=1,$ and the intersection of the two asymptote is $\boxed{(2,1)}.$