Start by remembering that you'll get a vertical asymptote where you get an undefined ...or infinite result. In rational functions, this generally happens when the denominator goes to 0 for some value .... you can't divide by 0, even with polynomials!
I'd do this by factoring the denominator into x2-3x+2 = (x-1)(x-2) Check that this works.
Now, if either quantity in parentheses there goes to 0, the whole denominator is 0, and we need to know the values for which this happens to identify the asymptotes. Since If (x-1) = 0 makes the bottom all 0 just as (x-2) = 0 does, we can solve these two simple equations for x to get x = 1 and x = 2. Check that x=1 or 2 makes the denominator go to 0. There are the x values where your asymptote will be.
BUT WAIT. The numerator can be factored, too. Into (x-2)(x+2). So the function actually looks like:
f(x) = [(x-2)(x+2)]/[(x-1)(x-2)] So the (x-2) terms cancel and so does that asymptote. You have a vertical asymptote at x = 1 only.
f(x) is now ... [(x+2)]/[(x-1)]
Now, the horizontal asymptote is identified by the ratio of the coefficients on the leading x-terms. HEAVY, I KNOW.
So, you have a horizontal asymptote if and only if the highest power in the numerator is the same as the highest power in the denominator. So, here you have x up top and x down below, so you'll have a horizontal asymptote. Both the top and bottom x have coefficients of 1.
x = 1*x, right? The ratio of 1/1 is 1. So you'll have a horizontal asymptote at y = 1.
To graph, draw in the asymptotes, first, then pick some points in the regions away from the asymptotes on the x-axis. Plug and chug!
PS: My answer is correct. idk what these other guys were doing.