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

1 Answer
Lucy
Jul 26, 2018

Below

Explanation:

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

For vertical asymptotes, #x-2 !=0# since the denominator cannot equal to 0 as the function will be undefined at that point. To find at what point the function is undefined, we can go #x-2=0# so #x=2# is our vertical asymptote.

For horizontal asymptotes, we let #x ->oo#. As #x ->oo#, #2/(x-2)->0#. Hence, #f(oo)=1+0=1#. Therefore, there is an horizontal asymptote at #y=1#

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

Plotting your intercepts and drawing in your asymptotes (remember that the asymptotes influence the endpoints of your graph ONLY and nothing else)

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

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