Free AP Precalculus Polynomial and Rational Functions

AP Precalculus · Polynomial and Rational Functions · free lesson

Rational Functions: End Behavior and Zeros

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Rational functions

A rational function is a quotient of two polynomials, $r(x) = \frac{p(x)}{q(x)}$ , where $q$ is not the zero polynomial. Its domain is every real number except the zeros of the denominator $q$ . This topic is about two things a formula of this kind tells at once: what $r$ does far out (its end behavior), and where $r(x) = 0$ (its zeros).

End behavior: the quotient of the leading terms

For large $\lvert x \rvert$ each polynomial behaves like its leading term (topic 05), so the quotient behaves like the quotient of the leading terms. Take

\begin{aligned} r(x) &= \frac{ {2x^{2} - 3x - 2} }{ {x^{2} + 1} } \end{aligned}

Far out, $r(x)$ behaves like $\frac{2x^2}{x^2} = {2}$ . So

\begin{aligned} \lim_{x \to \infty} r(x) &= 2 \\ \lim_{x \to -\infty} r(x) &= 2 \end{aligned}

and the line $y = 2$ is a horizontal asymptote: the graph comes as close to it as we like at both ends. Watch the graph draw itself. In the middle it crosses its horizontal asymptote; an asymptote describes the ends only, and the graph may cross it any number of times nearer the origin.

Where it crosses is found by solving $r(x) = 2$ : the engine gives $x = -\frac{4}{3}$ .

Three cases, by the degrees

Let $p$ have degree $m$ and leading coefficient $a$ , and $q$ degree $n$ and leading coefficient $b$ . The quotient of the leading terms is $\frac{a x^m}{b x^n}$ , and three cases follow.

degrees	quotient of leading terms	end behavior
$m < n$	goes to 0	horizontal asymptote $y = 0$
$m = n$	the number $\frac{a}{b}$	horizontal asymptote $y = \frac{a}{b}$
$m > n$	$\frac{a}{b}x^{m - n}$ , unbounded	no horizontal asymptote: the ends behave like the polynomial $\frac{a}{b}x^{m - n}$

Drag the slider to change the leading coefficient $k$ of the numerator in $w(x) = \frac{k x^2 - 4}{x^2 + 1}$ . The degrees are equal, so the horizontal asymptote is $y = \frac{k}{1} = k$ and it moves with the slider.

When the numerator's degree is exactly one more than the denominator's, the ends follow a slanted line, a slant asymptote, found by dividing (topic 08). For $s(x) = \frac{x^3 - x}{x^2 + 1}$ the quotient of the leading terms is $\frac{x^3}{x^2} = x$ , so the ends behave like $x$ :

The engine gives the slant asymptote $y = x$ , and $\lim_{x \to \infty} s(x) = {\infty}$ , $\lim_{x \to -\infty} s(x) = {-\infty}$ .

Zeros

A quotient is 0 only when its numerator is 0 and its denominator is not. So the zeros of $r$ are the zeros of the numerator that are in the domain. Take

\begin{aligned} t(x) &= \frac{x^2 - 4}{x^2 - 5x + 6} = \frac{(x - 2)(x + 2)}{(x - 2)(x - 3)} \end{aligned}

The numerator is 0 at 2 and at $-2$ . But 2 also makes the denominator 0, so 2 is not in the domain (the domain leaves out $2$ and $3$ ), and $t(2)$ does not exist. The only zero is $x = -2$ . What happens at $x = 2$ is the subject of topic 07.

Rational inequalities

The sign of $r(x)$ can change only at a zero of the numerator or of the denominator. Mark those numbers on the line, test one value in each interval, and read off where $r(x) \ge 0$ or $r(x) < 0$ . To solve $r(x) \ge 0$ for $r(x) = \frac{x + 1}{x - 3}$ , the break points are $-1$ (a zero, included) and 3 (not in the domain, never included):

interval	$x < -1$	$-1 < x < 3$	$x > 3$
test value	$r(-2) = \frac{1}{5}$	$r(0) = -\frac{1}{3}$	$r(4) = 5$
sign	positive	negative	positive

So $\frac{x + 1}{x - 3} \ge 0$ for $x \le -1$ or $x > 3$ .

Warning.

Four slips to avoid: giving $\frac{a}{b}$ as the asymptote when the degrees differ; giving $y = 0$ when the numerator's degree is the larger; listing a zero of the denominator as a zero of $r$ ; and closing the interval at a number that makes the denominator 0.

Worked examples

Each example is one kind of question the AP exam asks about the end behavior and the zeros of a rational function. Every number in the working is found by the engine.

Example.

1. Horizontal asymptotes by the degrees. Find the horizontal asymptote, if there is one, of each function.

$f(x) = \frac{3x - 5}{x^2 + 2}$ , $\quad g(x) = \frac{6x^2 + x}{2x^2 - 7}$ , $\quad h(x) = \frac{x^3}{4x + 1}$

$f$ : the numerator has degree 1 and the denominator degree 2. The quotient of the leading terms, $\frac{3x}{x^2} = \frac{3}{x}$ , goes to 0, so the horizontal asymptote is $y = 0$ .
$g$ : equal degrees. The quotient of the leading terms is $\frac{6x^2}{2x^2}$ , so the horizontal asymptote is $y = 3$ .
$h$ : the numerator has the larger degree. The quotient of the leading terms is $\frac{x^3}{4x} = \frac{x^2}{4}$ , which is unbounded: there is no horizontal asymptote, and $\lim_{x \to \infty} h(x) = {\infty}$ , $\lim_{x \to -\infty} h(x) = {\infty}$ , as for $\frac{1}{4}x^2$ .

For $f$ the usual slip is to answer $y = 3$ , the quotient of the leading coefficients; that rule holds only when the degrees are equal.

Example.

2. End behavior in limit notation. Describe the end behavior of $r(x) = \frac{5 - 2x^3}{x^2 + 3x}$ with limits.

The leading terms are $-2x^3$ and $x^{2}$ , so for large $\lvert x \rvert$ , $r(x)$ behaves like $\frac{-2x^3}{x^2} = -2x$ . As $x \to \infty$ , $-2x$ decreases without bound; as $x \to -\infty$ , it increases without bound:

\begin{aligned} \lim_{x \to \infty} r(x) &= -\infty \\ \lim_{x \to -\infty} r(x) &= \infty \end{aligned}

The numerator's degree is one more than the denominator's, so the ends follow a slant asymptote; dividing (topic 08) gives $y = -2x + 6$ . Its slope $-2$ is the quotient of the leading coefficients, and it falls to the right, as the limits say.

Example.

3. Zeros, with a factor in common. Find the zeros of $r(x) = \frac{x^2 + x - 6}{x^2 - 4}$ .

Factor the numerator and the denominator:

\begin{aligned} r(x) &= \frac{ {(x + 3)(x - 2)} }{ {(x + 2)(x - 2)} } \end{aligned}

The numerator is 0 at $x = -3$ and at $x = 2$ . The denominator is 0 at $-2$ and $2$ , so those are not in the domain. Since 2 makes the denominator 0, $r(2)$ does not exist and 2 is not a zero. The only zero is $x = -3$ .

A student who lists 2 as a zero has forgotten to check the domain. (The graph has a hole at $x = 2$ ; topic 07 finds its height.)

Example.

4. A rational inequality. Solve $\frac{x - 2}{(x + 1)(x - 4)} \le 0$ .

The break points are the zero of the numerator, $2$ , and the zeros of the denominator, $-1$ and $4$ . They cut the line into four intervals; test one value in each.

interval	$x < -1$	$-1 < x < 2$	$2 < x < 4$	$x > 4$
test value	$r(-2) = -\frac{2}{3}$	$r(0) = \frac{1}{2}$	$r(3) = -\frac{1}{4}$	$r(5) = \frac{1}{2}$
sign	negative	positive	negative	positive

The quotient is negative on $x < -1$ and on $2 < x < 4$ , and it is 0 at $x = 2$ , which the inequality includes. The numbers $-1$ and $4$ are never included: $r$ has no value there. So the solution is $x < -1$ or $2 \le x < 4$ .

Example.

5. Where the graph crosses its horizontal asymptote. Does the graph of $r(x) = \frac{2x^2 + 5x}{x^2 + 1}$ cross its horizontal asymptote? If so, where?

The degrees are equal, so the horizontal asymptote is $y = \frac{2}{1} = {2}$ . Set $r(x) = 2$ and solve; multiplying by $x^{2} + 1$ , which is never 0, gives $2x^2 + 5x = 2x^2 + 2$ :

\begin{aligned} x &= \frac{2}{5} \end{aligned}

So the graph crosses the line $y = 2$ at $\left({\frac{2}{5}}, {2}\right)$ , then approaches it from above as $x \to \infty$ . A graph never crosses a vertical asymptote; a horizontal one may be crossed.

Example.

6. A context: salt in a tank. A tank holds 100 gallons of water with 30 pounds of salt dissolved in it. Brine with 0.4 pound of salt per gallon is pumped in at 5 gallons per minute and nothing flows out. After $t$ minutes the concentration of salt, in pounds per gallon, is $C(t) = \frac{30 + 2t}{100 + 5t}$ . What happens to the concentration in the long run, and when is it 0.35 pound per gallon?

Numerator and denominator have degree 1, so $C(t)$ approaches the quotient of the leading coefficients:

\begin{aligned} \lim_{t \to \infty} C(t) &= \frac{2}{5} = {0.4} \end{aligned}

In context: as time goes on the concentration approaches $0.4$ pound per gallon, the concentration of the brine being pumped in. It starts at $C(0) = {0.3}$ and rises toward that value but never reaches it. Solving $C(t) = 0.35$ gives $t = 20$ minutes.

Practice

Answer AP-style questions on rational functions: end behavior and zeros. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise rational functions: end behavior and zeros with new numbers