What are the asymptote(s) and hole(s), if any, of # f(x) =(x^2-3x+2)/(x-3)#?

1 Answer
Feb 3, 2017

Answer:

Vertical asymptote #x=3# and oblique / slanting asymptote #y=x#

Explanation:

As #f(x)=(x^2-3x+2)/(x-3)=((x-1)(x-2))/(x-3)# and as #(x-3)# in denominator does not cancel out with numeraor, we do not ave a hole.

If #x=3+delta# as #delta->0#, #y=((2+delta)(1+delta))/delta# and as #delta->0#, #y->oo#. But if #x=3-delta# as #delta->0#, #y=((2-delta)(1-delta))/(-delta)# and as #delta->0#, #y->-oo#.

Hence #x=3# is a vertical asymptote.

Further #y=(x^2-3x+2)/(x-3)=(x^2-3x)/(x-3)+2/(x-3)#

= #x+2/(x-3)=x+(2/x)/(1-3/x)#

Hence as #x->oo#, #y->x# and we have an oblique or slant asymptote #y=x#
graph{(y-(x^2-3x+2)/(x-3))=0 [-17.34, 22.66, -8.4, 11.6]}