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

1 Answer
Jul 31, 2017

Answer:

#"vertical asymptote at "x=-1#
#"horizontal asymptote at "y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "(x+1)^2=0rArrx=-1" is the asymptote"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by "x^2#

#f(x)=-(1/x^2)/(x^2/x^2+(2x)/x^2+1/x^2)=-(1/x^2)/(1+2/x+1/x^2)#

as #xto+-oo,f(x)to0/(1+0+0)#

#rArry=0" is the asymptote"#
graph{-1/(x+1)^2 [-10, 10, -5, 5]}