Rational Functions and End Behavior — AP® Precalculus Notes

Divide one polynomial by another and you get a rational function. Its ends can do three things: level off at a constant, shrink to $0$ , or keep growing, sometimes along a slanted line. This topic predicts which from the leading terms and writes the answer in limit notation.

Rational functions

A rational function has the form $r(x) = \frac{p(x)}{q(x)}$ , where $p$ and $q$ are polynomials. Examples: $\frac{3x^2 - 3}{x^2 + 2}$ , $\frac{5x + 2}{x^2 + 1}$ , average cost $C(n) = \frac{500 + 4n}{n}$ .

The quotient of the leading terms

Topic 1.6 showed that for large $|x|$ a polynomial behaves like its leading term. Apply that to the top and bottom separately, and a rational function behaves like the quotient of the two leading terms.

Degrees	Quotient of leading terms	End behavior	Example
top less than bottom	goes to $0$	horizontal asymptote $y = 0$	$\frac{5x + 2}{x^2 + 1} \sim \frac{5}{x}$
equal	a constant $\frac{a}{b}$	horizontal asymptote $y = \frac{a}{b}$	$\frac{3x^2 - 3}{x^2 + 2} \sim 3$
top one more	linear, $mx$	slant asymptote	$\frac{x^2 + 3x + 1}{x} \sim x$
top two or more more	$x^2$ or higher	like a polynomial	$\frac{x^3 + 1}{x - 2} \sim x^2$

Horizontal asymptotes

If $\lim_{x \to \infty} r(x) = c$ or $\lim_{x \to -\infty} r(x) = c$ , the line $y = c$ is a horizontal asymptote: the graph gets as close to it as you like far out.

Worked example 1. Describe the end behavior of $s(x)$ $= \frac{3x^2 - 3}{x^2 + 2}$ .

Leading terms: $\frac{3x^2}{x^2} = 3$ . So $\lim_{x \to \infty} s(x) = 3$ and $\lim_{x \to -\infty} s(x) = 3$ ; the horizontal asymptote is $y = 3$ .

Which side? $3x^2 - 3 = 3(x^2 + 2) - 9$ , so $s(x) = 3 - \frac{9}{x^2 + 2}$ , always a little below $3$ .

Figure 1.

\frac{2x^2 - 8}{x^2 + 1}

approaches

y = 2

at both ends. LevlPrep original diagram.

An asymptote describes the ends, not the middle. A graph may cross its horizontal asymptote: $\frac{5x + 2}{x^2 + 1}$ is $0$ at $x = -0.4$ and still has the asymptote $y = 0$ .

Slant asymptotes

When the top's degree is one more than the bottom's, the quotient of the leading terms is linear and the graph approaches a slanted line. To find it, rewrite the function as a line plus a fraction that goes to $0$ .

Worked example 2. Find the slant asymptote of $k(x)$ $= \frac{x^2 + 2x + 3}{x + 1}$ .

Write the top around the bottom: $x^2 + 2x + 3$ $= (x + 1)^2 + 2$ .

So $k(x) = x + 1 + \frac{2}{x + 1}$ . As $x \to \pm\infty$ , $\frac{2}{x + 1} \to 0$ , and the graph approaches $y = x + 1$ .

With a single-term denominator it is quicker: $\frac{x^2 + 3x + 1}{x}$ $= x + 3 + \frac{1}{x}$ , slant asymptote $y = x + 3$ .

In context

The poster shop's average cost $C(n)$ $= \frac{500 + 4n}{n}$ $= 4 + \frac{500}{n}$ has $\lim_{n \to \infty} C(n) = 4$ . In words: for very large orders, the average cost per poster gets close to $4, because the $500 setup is spread over more and more posters.