Mathematics

Rational functions

When $x$ sits in the denominator, there are places where the function does not exist. And lines the graph only ever approaches.

What you need first

Hyperbola & reciprocal function Fraction equations Limits, intuitively

As long as $x$ only appears on top, every number is allowed. As soon as it turns up in the denominator, things get interesting: $f(x) = \frac{1}{x - 2}$ has no result for $x = 2$ , because then a 0 would sit under the fraction bar. That single forbidden place shapes the entire graph, and that is what this topic is about.

The domain

The first question for any such function is: for which $x$ does the denominator become zero? Those numbers are excluded, everything else is allowed. For $f(x) = \frac{x + 1}{x^2 - 9}$ you set $x^2 - 9 = 0$ and get $x = 3$ and $x = -3$ . So the domain is all real numbers except $-3$ and 3.

The vertical asymptote

What happens near the forbidden place? Take $f(x) = \frac{1}{x - 2}$ and substitute numbers creeping ever closer to 2. At $x = 2.1$ the denominator is 0.1 and the value is 10. At $x = 2.01$ the value is already 100, at $x = 2.001$ a full 1000. The smaller the denominator, the bigger the fraction. So the graph shoots vertically upwards without ever reaching the place $x = 2$ .

From the left the same thing happens with the opposite sign: at $x = 1.9$ the denominator is $-0.1$ and the value $-10$ . The vertical line at $x = 2$ is called an , and the graph hugs it from both sides without ever touching it. It is drawn dashed, because it is not part of the graph but only a guide line.

Coordinate grid

P (3 | 2)

Walk the point along a gap in the domain. For

y = \frac{6}{x}

set (6|1), (3|2), (2|3) and (1|6) one after another: the closer

x

gets to 0, the more steeply the value climbs. At

x = 0

there is no point at all, that is where the vertical asymptote sits. Then set (−1|−6) and (−2|−3), the branch on the other side, and it runs away downwards.

The horizontal asymptote

The second question goes the other way: what happens for very large $x$ ? For $f(x) = \frac{1}{x - 2}$ the denominator is then huge and the fraction tiny. At $x = 1000$ the value is about 0.001. So the graph approaches the height 0 without reaching it, and the $x$ axis is a horizontal asymptote.

Whether a horizontal asymptote exists at all is decided by comparing the degrees. If the degree on top is smaller than below, the fraction tends to 0. If both degrees are equal, the graph approaches the ratio of the leading coefficients: for $\frac{3x^2 + 1}{x^2 - 4}$ that is 3. If the degree on top is larger, there is no horizontal asymptote, because the fraction keeps growing without bound.

\frac{3x^2 + 1}{x^2 - 4} \;\longrightarrow\; \frac{3}{1} = 3 \quad \text{für sehr große } |x|

With equal degrees only the two leading coefficients matter.

Exercises

0 of 6 solved

Time to try it yourself. You can't break anything, every attempt counts.

How do you spot a gap in the domain?

At which place is $f(x) = \frac{4}{x - 7}$ undefined?

Which vertical asymptotes does $f(x) = \frac{x}{x^2 - 16}$ have?

A line the graph gets ever closer to without reaching it is called an ….

Work out $f(4)$ for $f(x) = \frac{x + 2}{x - 2}$ .

Match each function to its horizontal asymptote.

\frac{1}{x + 3}

\frac{5x}{x - 1}

\frac{2x^2}{4x^2 + 1}

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