Graph rational function f(x) = -x^2+4x+3 over x-3 , I found out that the Vertical asymtote is 3. plugged 3 back into equation as f(3) = 24/0 is what I got but I do not know what to do with it or how to continue graphing this function.
Vertical asymptote is x=3 and Oblique asymptote is y=1-x, which is found like this: y=(-x 2+4x+3)/(x-3)=-x+1+6/(x-3) since 6/(x-3) <> 0, y=-x+1+6/(x-3) <>-x+1
