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

Question

Benjamin Achtenberg · Accepted Answer

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

Ariana Alvarez · Answer

undefined To find the asymptotes for $ f(x) = -\frac{1}{(x+1)^2} $:

1. Vertical asymptotes occur where the function approaches positive or negative infinity as $ x $ approaches a certain value. Set the denominator equal to zero and solve for $ x $ to find potential vertical asymptotes.

$ x + 1 = 0 $
$ x = -1 $

So, there is a vertical asymptote at $ x = -1 $.

2. Horizontal asymptotes occur when $ x $ approaches positive or negative infinity, and the function approaches a constant value. For rational functions like this one, the horizontal asymptote can be found by comparing the degrees of the numerator and denominator.

In this case, the degree of the numerator is 0 and the degree of the denominator is 2. Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is at $ y = 0 $.

Therefore, the asymptotes for $ f(x) = -\frac{1}{(x+1)^2} $ are:
- Vertical asymptote: $ x = -1 $
- Horizontal asymptote: $ y = 0 $