How do you graph #f(x)=x/(x-2)# using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Jul 12, 2018

Below

Explanation:

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

As #x -> oo#, #y ->1#
ie As x approaches a very big number, #2/(x-2)# will become 0
For the vertical asymptote, #x-2 !=0# since the denominator cannot equal to 0.

Therefore, there is a horizontal asymptote at #y=1# and a vertical asymptote at #x=2#

For intercepts,
When #y=0#, #x=0#
When #x=0#, #y=0#

Then, after plotting your intercepts and your asymptotes, you can hopefully see the outline of your graph. Your end points should be approaching the asymptotes but never touching your asymptotes.
graph{x/(x-2) [-10, 10, -5, 5]}