How do you find the asymptotes for #f(x) =(x^3-1)/(x^2-6)#?

1 Answer
Jan 10, 2017

Answer:

The vertical asymptotes are #x=-sqrt6# and #x=sqrt6#
The slant asymptote is #y=x#
No horizontal asymptote

Explanation:

Let's factorise the denominator

#x^2-6=(x-sqrt6)(x+sqrt6)#

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

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

The vertical asymptotes are #x=-sqrt6# and #x=sqrt6#

As the degree of the numerator is #># than the degree of the denominator, there is a slant asymptote.

We perform a long division

#color(white)(aaaa)##x^3##color(white)(aaaaa)##-1##color(white)(aaaaa)##∣##x^2-6#

#color(white)(aaaa)##x^3-6x##color(white)(aaaa)####color(white)(aaaa)##∣##x#

#color(white)(aaaaa)##0+6x-1##color(white)(aaaa)#

Therefore,

#f(x)=(x)+(6x-1)/(x^2-6)#

#lim_(x->-oo)(f(x)-x)=lim_(x->-oo)(6x-1)/(x^2-6)=lim_(x->-oo)(6x)/x^2=lim_(x->-oo)6/x=0^-#

#lim_(x->+oo)(f(x)-x)=lim_(x->+oo)(6x-1)/(x^2-6)=lim_(x->+oo)(6x)/x^2=lim_(x->+oo)6/x=0^+#

So, the slant asymptote is #y=x#

graph{(y-(x^3-1)/(x^2-6))(y-x)=0 [-28.86, 28.86, -14.43, 14.45]}