How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= -1/(x+1)^2#?

Question

Cora Alexander · Accepted Answer

vertical asymptote: #x=-1#
horizontal asymptote: #f(x)=0#
oblique asymptote: does not exist

Finding the Vertical Asymptote Given, #f(x)=-1/(x+1)^2# Set the denominator equal to #0# and solve for #x#. #(x+1)^2=0# #x+1=0# #color(green)(|bar(ul(color(white)(a/a)color(black)(x=-1)color(white)(a/a)|)))# Finding the Horizontal Asymptote Given, #f(x)=-1/(x+1)^2# If you foil the denominator, you will notice that the degree of the denominator is #2#. #f(x)=-1/(x^2+2x+1)# Since the degree of the denominator is greater than the degree of the numerator, the horizontal asymptote is #f(x)=0#. #color(green)(|bar(ul(color(white)(a/a)color(black)(f(x)=0)color(white)(a/a)|)))# Finding the Oblique Asymptote Given, #f(x)=-1/(x+1)^2# There would be a slant asymptote if the degree of the leading term in the numerator is #1# value larger than the degree of the leading term in the denominator. In your case, we see the opposite — ie. the degree in the denominator is greater than the degree in the numerator. #:.#, the oblique asymptote does not exist.

Lincoln Anderson · Answer

undefined To find the vertical, horizontal, and oblique asymptotes of the function $ f(x) = -\frac{1}{(x+1)^2} $, follow these steps:

1. Vertical Asymptotes:
   Set the denominator equal to zero and solve for $ x $. The vertical asymptote(s) occur where the function is undefined due to division by zero.
   $$ (x + 1)^2 = 0 $$
   $$ x = -1 $$
   Therefore, the vertical asymptote is at $ x = -1 $.

2. Horizontal Asymptote:
   Determine the behavior of the function as $ x $ approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. In this case, the degree of the numerator is 0, and the degree of the denominator is 2, so the horizontal asymptote is at $ y = 0 $.

3. Oblique Asymptote (if applicable):
   To check for an oblique asymptote, divide the numerator by the denominator using polynomial long division. If the result is a polynomial plus a proper fraction, there is an oblique asymptote.
   $$ \frac{-1}{(x+1)^2} = -\frac{1}{x^2+2x+1} $$
   Perform long division or synthetic division to divide $ -1 $ by $ x^2 + 2x + 1 $.
   $$ -1 \div (x^2+2x+1) = -1 $$
   Therefore, there is no oblique asymptote in this case.

To summarize:
- Vertical asymptote: $ x = -1 $
- Horizontal asymptote: $ y = 0 $
- There is no oblique asymptote.