How do you find the horizontal asymptote for #(x-3) / (x+5)#?

1 Answer
Mar 22, 2018

Answer:

#y=1#

Explanation:

There are two ways of solving this.
1. Limits:

#y=lim_(xto+-oo)(ax+b)/(cx+d)=a/c#, therefore horizontal asymptote occurs when #y=1/1=1#

2. Inverse:
Let's take the inverse of #f(x)#, this is because the #x# and #y# asymptotes of #f(x)# will be the #y# and #x# asymptotes for #f^-1(x)#

#x=(y-3)/(y+5)#
#xy+5x=y-3#
#xy-y=-5x-3#
#y(x-1)=-5x-3#
#y=f^-1(x)=-(5x+3)/(x-1)#

The vertical asymptote is the same as the horizontal asymptote of #f(x)#

The vertical asymptote of #f^-1(x)# is #x=1#, therefore the horizontal asymptote of #f(x)# is #y=1#