How do you find vertical, horizontal and oblique asymptotes for #(2x^3+4x-8)/(x^3-6x)#?

1 Answer
Dec 1, 2016

Answer:

The vertical asymptotes are #x=0#, #x=sqrt6# and #x=-sqrt6#
No oblique asymptote
the horizontal asymptote is #y=2#

Explanation:

Let #f(x)=(2x^3+4x-8)/(x^3-6x)#

The denominator is

#=x^3-6x=x(x-sqrt6)(x+sqrt6)#

The domain of #f(x)# is #D_f(x)=RR-{0, +-sqrt6} #

As you cannot divide by #0#, #x!=0#, #x!=sqrt6# and #x!=-sqrt6#

So the vertical asymptotes are #x=0#, #x=sqrt6# and #x=-sqrt6#

As the degree of the numerator #=# the degree of the denominator, there is no oblique asymptote.

To calculate the limits as #x->+-oo#, we take the terms of highest degree in the numerator and the denominator.

#lim_(x->+-oo)f(x)=lim_(x->+-oo)(2x^3)/x^3=2#

So, the horizontal asymptote is #y=2#

graph{(2x^3+48-8)/(x^3-6x) [-28.86, 28.87, -14.43, 14.45]}