How do you graph #y=(x+4)/(x-3)# using asymptotes, intercepts, end behavior?

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Answer 1 · 2018-01-11T15:25:44

See below.

#y =(x+4)/(x-3)#

This is undefined at #x=3#, so vertical asymptote occurs here.

#(x+4)/(x-3)#

Divide by #x#

#(x/x+4/x)/(x/x-3/x)=(1+4/x)/(1-3/x)#

as #x->oocolor(white)(88888)# , #(1+4/x)/(1-3/x)=(1+0)/(1-0)=1#

as #x->-oocolor(white)(888)# , #(1+4/x)/(1-3/x)=(1+0)/(1-0)=1#

#:.#

#y=1# is a horizontal asymptote.

GRAPH:

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