How do you find the asymptotes for #f(x) = (x^3 + x^2 - 6x) /( 4x^2 - 8x - 12)#?

Question

Serenity Johnson · Accepted Answer

Asymptotes are #x=3# , #x=-1# and #x/4#

Start by finding the factors for both numerator and denominator. #f(x) = (x^3+x^2-6x)/(4x^2-8x-12)#

#f(x) = (x(x^2+x - 6))/(4(x^2-2x-3))#

#f(x) = (x(x+3)(x-2))/(4(x-3)(x+1))#

This confirms that there are no removable asymptotes as there are no factors that cancel out.

The denominator approaches zero, and therefore the function value #->oo# as #x->3# or # x->-1# so there are vertical asymptotes at #x=3# and #x=-1#

As #x->oo# the function # f(x) -> (xcancel(x^2+x - 6))/(4cancel(x^2-2x-3)) -> x/4#

So there is a slant asymptote of #x/4#

Anna Allen · Answer

undefined To find the asymptotes of the function $ f(x) = \frac{x^3 + x^2 - 6x}{4x^2 - 8x - 12} $:

1. Horizontal asymptotes:
   As $ x $ approaches positive or negative infinity, the terms with the highest degree in the numerator and denominator dominate. Divide the leading terms of the numerator and denominator to find horizontal asymptotes, if any.

2. Vertical asymptotes:
   Vertical asymptotes occur where the denominator equals zero, but the numerator does not. To find them, solve $ 4x^2 - 8x - 12 = 0 $ for $ x $.

3. Oblique (slant) asymptotes:
   Oblique asymptotes occur when the degree of the numerator is one greater than the degree of the denominator. To find them, perform polynomial long division to divide the numerator by the denominator. The quotient represents the oblique asymptote.

Performing these steps will provide the asymptotes for the function $ f(x) $.