Asymptotes of a rational function We have a function $$f(x) =\dfrac{2x^4+4x^3+3x^2+4x-4}{x^3-x^2-6x}$$
How would I systematically go about finding the asymptotes of this function? I know how to find the asymptotes of for example log functions or functions with a square root in it, but I don't really know how to find them for this function. 
 A: Euclidean division of the numerator by the denominator: if 
$$N(x)=(ax+b)D(x)+R(x)\quad \deg R(x)<\deg D(x),$$
the $y=ax+b$ is the equation of the asymptote. 
Naturally there is an asymptote only if  $\deg D(x)\le \deg N(x)\le\deg D(x)+1$.
A: The problem can be simplified by dividing the numerator and denominator by $(x+2)$:
$$
\dfrac{2x^3+3x-2}{x^2-3x} \\
$$


*

*Vertical asymptotes occur where $y$ tends to infinity, which happens at the roots of the denominator:
$$
x^2 - 3x = x (x - 3) = 0 \\
$$
There are two roots so the vertical asymptotes are $x = 0$ and $x = 3$.

*Horizontal asymptotes occur where $x$ tends to infinity, which becomes clear by dividing the numerator and the denominator by the highest power of $x$:
$$
\lim_{x\rightarrow \infty} \dfrac{2x^3+3x-2}{x^2-3x} = \lim_{x\rightarrow \infty} \dfrac{2+3/x^2-2/x^3}{1/x-3/x^2} = \infty \\
$$
There is no limit so there are no horizontal asymptotes.

*Oblique asymptotes occur where the graph approaches the line $(mx+b)$, which becomes clear by dividing the numerator and denominator:
$$
\dfrac{2x^3+3x-2}{x^2-3x} = 2x + 6 \text{ remains } 21x - 2 \\
$$
There is a whole part so the oblique asymptote is $y = 2x + 6$.
