How do you find a horizontal asymptote #(x^2 - 5x + 6)/(x - 3#?

1 Answer
May 26, 2015

We can first rewrite the function as

#((x-2)(x-3))/((x-3))# and cancel the #(x-3)#'s

And this will turn the function into #x-2# which has no asymptotes at all.

One thing though:
#x=3# is not allowed , as this would make the numerator #=0#
(Actually it would turn the fraction into #0/0#, which is not defined at all).