Let $f(x) \,=\, \frac{P(x)}{Q(x)}$ be a fractional expression where $P(x)$ and $Q(x)$ are polynomials with real coefficients such that their quotient can not be reduced to a polynomial. We suppose that $P(x)$ and $Q(x)$ have no common zeros.

If the division of the polynomials is performed, then a result of the form $$f(x) \;=\; H(x)+\frac{R(x)}{Q(x)}$$ is gotten, where $H(x)$ and $R(x)$ are polynomials such that $$\deg{R(x)} < \deg{Q(x)}$$

The graph of the rational function $f$ may have asymptotes:

1. Every zero $a$ of the denominator $Q(x)$ gives a vertical asymptote $x = a$ .
2. If $\deg{H(x)} < 1$ (i.e. $0$ or $-\infty$ ) then the graph has the horizontal asymptote $y = H(x)$ .
3. If $\deg{H(x)} = 1$ then the graph has the skew asymptote $y = H(x)$ .

Proof of 2 and 3. We have $\displaystyle f(x)\!-\!H(x) = \frac{R(x)}{Q(x)}\,\to 0$ as $|x|\to\infty$ .

Remark. Here we use the convention that the degree of the zero polynomial is $-\infty$ .