How do you find the vertical, horizontal or slant asymptotes for #y = (2e^x) / (e^x - 5)#?

1 Answer
Oct 21, 2016

Answer:

We have a vertical asymptote at #x=ln5=1.6094#

and horizontal asymptotes at #y=0# and #y=2#

Explanation:

We have #y=(2e^x)/(e^x-5)#

As #x->-oo#, #y->0/(0-5)=0#

dividing numerator and denominator by #e^x#, we get

#y=2/(1-5/e^x)#

As such, when #x->oo#, #y->2/(1-0)=2#

Hence, horizontal asymptotes are #y=0# and #y=2#

also #ye^x-5y=2e^x# or #ye^x-2e^x=5y#

i.e. #e^x=(5y)/(y-2)# or #x=ln((5y)/(y-2))=ln(5/(1-2/y))#

and as #y->+-oo#, #x->ln5=1.6094#,

we have a vertical asymptote at #x=ln5=1.6094#

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