How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(x^3-x)/(x^3-4x)#?

Question

Lily Anderson · Accepted Answer

The vertical asymptotes are #x=2# and #x=-2#
The horizontal asymptote is #y=1# Let's do some simplification #f(x)=(x^3-x)/(x^3-4x)=(x(x^2-1))/(x(x^2-4))=((x+1)(x-1))/((x+2)(x-2))# As we cannot divide by #0#, The degree of the numerator is identical to the degree of the denominator, so there are no oblique asymptotes. Limit #f(x)=x^2/x^2=1# #x->+-oo# So a horizontal asymptote is #y=1# graph{(x^2-1)/(x^2-4) [-10, 10, -5, 5]}

Serenity Jones · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. For horizontal asymptotes, compare the degrees of the numerator and denominator. If they are equal, divide the leading coefficients. 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 greater, there's no horizontal asymptote. For oblique asymptotes, perform polynomial long division. If the quotient is a polynomial, it represents the oblique asymptote. If there's a remainder, there's no oblique asymptote.