How do you find any asymptotes of #g(x)=(2x-3)/(x^2-6x+9)#?

1 Answer
Nov 1, 2016

Answer:

The vertical asymptote is #x=3#
There is no oblique asymptote

Explanation:

Let's factorise the denominator
#x^2-6x+9=(x-3)^2#
Therefore #g(x)=(2x-3)/(x-3)^2#
As you cannot divide by #0#, so #x!=3#
So #x=3# is a vertical asymptote
Limit #g(x)=2/x=0^-#
#x->-oo#
Limit #g(x)=2/x=0^+#
#x->+oo#

graph{(2x-3)/(x-3)^2 [-10, 10, -5, 5]}