How Do You Find the Vertical and Horizontal Asymptotes of a Rational Function?

For a rational function, find a vertical asymptote by setting the simplified denominator equal to zero. Find a horizontal asymptote by examining the function's behaviour as $x \to \pm\infty$; for $f(x)=\frac{ax+b}{cx+d}$, it is usually $y=\frac{a}{c}$.

The Method

This is IB Mathematics: Analysis and Approaches SL 2.8, which covers reciprocal and simple rational functions.

A vertical asymptote occurs because the denominator approaches zero while the numerator remains non-zero, causing the function's magnitude to increase without bound.

Always simplify the rational expression before finding asymptotes. A denominator zero that remains after cancellation may produce a vertical asymptote; a cancelled factor instead creates a removable discontinuity.

To check the vertical behaviour, consider values of $x$ just below and just above the denominator zero. The two branches may approach the same or opposite infinities, but this does not change the asymptote equation. The asymptote describes the limiting boundary, not a line the graph cannot cross.

For the horizontal asymptote,

$$\frac{ax+b}{cx+d}=\frac{a+\frac{b}{x}}{c+\frac{d}{x}}.$$

As $x \to \pm\infty$, both $\frac{b}{x}$ and $\frac{d}{x}$ approach zero. Therefore, .

For a general rational function, compare the degrees:

Worked Example

Consider $f(x)=\frac{2x+3}{x-4}$.

For the vertical asymptote:

$$x-4=0 \quad\Rightarrow\quad x=4.$$

For the horizontal asymptote:

$$y=\frac{2}{1}=2.$$

Therefore, the asymptotes are $x=4$ and $y=2$.

A common misconception is that every denominator zero gives a vertical asymptote. In $\frac{x-1}{x-1}$, the factor cancels, so $x=1$ is a hole, not an asymptote.

Exam Technique

On an IB sketch, label both asymptotes with their equations and include the intercepts. On Paper 1, show the algebra exactly; on Paper 2, a GDC may confirm the graph but should not replace the calculations.