How do you find the vertical, horizontal or slant asymptotes for #(e^x)/(7+e^x)#?

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Answer 1 · 2016-05-06T16:51:18

We do not have any other asymptote, just two horizontal asymptotes, #x=0# and #x=1#

Dividing numerator and denominator by #e^x# in #e^x/(7+e^x)#

we get #1/(7e^(-x)+1)#

As #x->oo#, #e^(-x)->1/e^oo=0#

Hence #e^x/(7+e^x)=1/(7e^(-x)+1)->1/(0+1)=1#

Hence the asymptote is #y=1#

When #x->-oo#, #e^x/(7+e^x)=e^(-oo)/(7+e^(-oo))=0/(0+7)=0#

Hence other asymptote is #y=0#

We do not have any other asymptote, just two horizontal asymptotes

graph{e^x/(7+e^x) [-10, 10, -2, 2]}