How do you identify all horizontal and slant asymptote for #f(x)=4/(x-2)^3#?

Question

Lily Anderson · Accepted Answer

The vertical asymptote is #x=2#
The horizontal asymptote is #y=0# As you cannot divide by #0#, you have a vertical asymptote #x=2# There is no slant asymptote as the degree of the numerator is smaller than the degree of the denominator. #lim_(n rarr -oo )f(x)=lim_(narr-oo)4/x^3=0^-# #lim_(n rarr +oo )f(x)=lim_(narr+oo)4/x^3=0^+# So there is a horizontal asymptote #y=0# graph{4/(x-2)^3 [-16.02, 16.01, -8.01, 8.01]}

Benjamin Achtenberg · Answer

undefined To identify the horizontal and slant asymptotes for the function $ f(x) = \frac{4}{(x-2)^3} $:

1. Horizontal asymptote:
   - For rational functions, the horizontal asymptote can be determined by comparing the degrees of the numerator and denominator.
   - If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.
   - If the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.
   - In this case, the degree of the numerator is 0, and the degree of the denominator is 3, indicating that the denominator grows faster.
   - Therefore, the horizontal asymptote is $ y = 0 $.

2. Slant asymptote (if applicable):
   - Slant asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator.
   - To find the slant asymptote, perform polynomial long division or synthetic division.
   - Divide the numerator by the denominator, and the quotient obtained will be the equation of the slant asymptote.
   - In this case, since the degree of the numerator is less than the degree of the denominator, there is no slant asymptote.

Therefore, for the function $ f(x) = \frac{4}{(x-2)^3} $, the only asymptote is the horizontal asymptote $ y = 0 $.