How do you find the asymptotes for #r(x) = (-1) / (x+ 1)^2#?

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Answer 1 · 2016-02-01T10:58:25

The asymptotes are: #x=-1# and #y=0#

From the given: #r(x)=-1/(x+1)^2#

First question: What will make #r(x)=oo#?
Answer: when #(x+1)^2=0#

Therefore , just solve #(x+1)^2=0# for #x#:

#(x+1)^2=0#

#x+1=0#
#x=-1# Asymptote

Second question: What happens to the value of #r(x)# if #x# is made to approach #oo#.

if #x=+-oo#

#r(x)=-1/(+-oo+1)^2#

#r(x)=0# is an asymptote, whether #x=+oo# or #x=-oo#

Have a nice day!!! from the Philippines.