How do you find the asymptotes for #(x^2 - 5x + 6)/(x - 3)#?

1 Answer
Jul 29, 2016

Answer:

This function has no asymptotes. It has a hole at #(3, 1)#.

Explanation:

#(x^2-5x+6)/(x-3)=(color(red)(cancel(color(black)((x-3))))(x-2))/color(red)(cancel(color(black)((x-3))))=x-2#

with exclusion #x != 3#

This is a straight line with a hole at #(3, 1)#

It has no asymptotes.