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

Question

1328 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-26T13:56:50

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#