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

Question

1419 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-01T09:45:22

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

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]}

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