I'm trying to understand why the following method seems to work in finding the asymptotes of a general cubic. Take this cubic equation: $$x^3+2xy^2-\frac{1}{4}y^3+6x^2-4xy+2y^2+7x-2y+1=0$$ This looks like this: cubic
Dividing through by $x^3$, gives $$1+\frac{2y^2}{x^2}-\frac{y^3}{4x^3}+\frac{6}{x}-\frac{4y}{x^2}+\frac{2y^2}{x^3}+\frac{7}{x^2}-\frac{2y}{x^3}+\frac{1}{x^3}=0$$
With the understanding that $x$ goes to infinity, we drop terms where the power of $x$ in the denominator is greater than the power of $y$ in the numerator or where the numerator is a constant, and get
$$1+\frac{2y^2}{x^2}-\frac{y^3}{4x^3}=0$$
Here's where I just kind off wave my hands and get the right answer (which is confusing). I identify $\frac{y^3}{x^3}$ with $y$ and $\frac{y^2}{x^2}$ with $x$ which gives $$1+2x-\frac{y}{4}=0,$$ which appears to be the asymptote of the cubic. cubic with asymptote Insight is appreciated.
