Rational Functions and Zeros — AP® Precalculus

Part 1 · Hook

Why this matters

A dose of medicine works only while its concentration stays above a threshold. Asking "when is the concentration at least

8

?" is solving an inequality with a rational function. The answer comes from where the function is zero, where it is undefined, and its sign in between.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\lim_{x \to \infty} \frac{5x + 2}{x^2 + 1}$ ?

$0$
$5$
$\infty$
$2$

Show the answer

The denominator has the higher degree.

Correct: $0$ :
$5$ :
$\infty$ :
$2$ :

2. Solve $3 - 2x > 7$ .

$(-\infty, -2)$
$(-2, \infty)$
$(-\infty, 2)$
$(-\infty, -2]$

Show the answer

$-2x > 4$ , flip: $x < -2$ .

Correct: $(-\infty, -2)$ :
$(-2, \infty)$ :
$(-\infty, 2)$ :
$(-\infty, -2]$ :

Part 4 · See it

See it first

The graph of r(x) = (x + 1)(x − 3)/(x − 1). Its zeros are x = −1 and x = 3; the dashed line x = 1 marks the input that is not in the domain. The output is negative for x < −1, positive for −1 < x < 1, negative for 1 < x < 3 and positive for x > 3. — $r(x)$ $= \frac{(x + 1)(x - 3)}{x - 1}$ : zeros at $-1$ and $3$ ; the dashed line marks $x = 1$ , not in the domain. The sign changes at all three inputs. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The numerator of $r$ is $0$ at an input in the domainthat input is a real zero of $r$
The denominator is $0$ at an inputthat input is not in the domain, so it is never a zero
You mark zeros and excluded inputs on a number linethey split it into intervals where $r$ keeps one sign: test one input in each
The inequality is $\le$ or $\ge$ include the zeros with brackets; excluded inputs always get parentheses

Part 6 · Key ideas

Key ideas

Real zeros of $\frac{p(x)}{q(x)}$ : the zeros of $p$ that are in the domain (not zeros of $q$ ).
A sign chart splits the line at the zeros of both numerator and denominator; one test value per interval gives its sign.
Solve a rational inequality from the chart; include zeros only for $\le$ or $\ge$ , never an excluded input.
With a calculator: graph, find every zero, then read the sign between them.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{x^2 - 4}{x - 2}$ has zeros $2$ and $-2$ , because $x^2 - 4 = 0$ there.

What actually happens: At $x = 2$ the denominator is $0$ , so $2$ is not in the domain and not a zero. The only zero is $-2$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Test values for a sign chart

Values of r(x) = (x - 2)(x + 3)/(x + 1) at one test input in each interval between -3, -1 and 2
Interval	$(-\infty, -3)$	$(-3, -1)$	$(-1, 2)$	$(2, \infty)$
Test input $x$	-4	-2	0	3
$r(x)$	-2	4	-6	1.5
Sign of $r(x)$	$-$	$+$	$-$	$+$

1. The table uses $r(x)$ $= \frac{(x - 2)(x + 3)}{x + 1}$ . What are the real zeros of $r$ ?

$x = -3$ and $x = 2$
$x = -3$ , $x = -1$ and $x = 2$
$x = 3$ and $x = -2$
$x = -1$

Show the answer

The real zeros of a rational function are the zeros of the numerator that are in its domain: $x = 2$ and $x = -3$ . The denominator's zero, $-1$ , is excluded.

Correct: $x = -3$ and $x = 2$ : Right: the numerator is $0$ there, and both are in the domain.
$x = -3$ , $x = -1$ and $x = 2$ : At $x = -1$ the denominator is $0$ : $r(-1)$ is undefined, not $0$ .
$x = 3$ and $x = -2$ : The factor $x - 2$ is $0$ at $x = 2$ , and $x + 3$ at $x = -3$ .
$x = -1$ : $-1$ is the input not in the domain.

2. The table uses $r(x)$ $= \frac{(x - 2)(x + 3)}{x + 1}$ . Solve $r(x) > 0$ .

$(-3, -1) \cup (2, \infty)$
$(-\infty, -3) \cup (-1, 2)$
$[-3, -1] \cup [2, \infty)$
$(2, \infty)$

Show the answer

Read the signs: $+$ on $(-3, -1)$ and $(2, \infty)$ . Strict inequality, so the zeros are left out; $-1$ is never included because it is not in the domain.

Correct: $(-3, -1) \cup (2, \infty)$ : Right: the intervals marked $+$ , with the endpoints left out because $r$ must be strictly positive.
$(-\infty, -3) \cup (-1, 2)$ : These are the intervals where $r$ is negative.
$[-3, -1] \cup [2, \infty)$ : At $-3$ and $2$ , $r = 0$ , which is not $> 0$ ; and $-1$ is not in the domain.
$(2, \infty)$ : The interval $(-3, -1)$ is positive too: $r(-2) = 4$ .

3. The table uses $r(x)$ $= \frac{(x - 2)(x + 3)}{x + 1}$ . Solve $r(x) \le 0$ .

$(-\infty, -3) \cup (-1, 2)$
$(-\infty, -3] \cup [-1, 2]$
$(-\infty, -3] \cup (-1, 2]$
$[-3, -1) \cup [2, \infty)$

Show the answer

Negative on $(-\infty, -3)$ and $(-1, 2)$ ; zero at $-3$ and $2$ (included for $\le$ ); undefined at $-1$ (excluded).

$(-\infty, -3) \cup (-1, 2)$ : $\le$ includes the zeros: $r(-3) = r(2) = 0$ .
$(-\infty, -3] \cup [-1, 2]$ : $x = -1$ is not in the domain, so it takes a parenthesis.
Correct: $(-\infty, -3] \cup (-1, 2]$ : Right: the negative intervals plus the zeros, without $-1$ .
$[-3, -1) \cup [2, \infty)$ : These are where $r \ge 0$ .

Graph

The graph of $q(x)$ $= \frac{x^3 - 4x + 1}{x^2 + 2}$

4. Let $q(x)$ $= \frac{x^3 - 4x + 1}{x^2 + 2}$ . Find the largest zero of $q$ , to three decimal places.

Calculator allowed. Give the answer correct to three decimal places.

Show the answer

Graph $y = q(x)$ and use the zero feature near $x = 2$ : $x \approx 1.861$ (the true value is $1.86080...$ , so $1.860$ truncated is also accepted).

Answer: 1.861

5. Let $q(x)$ $= \frac{x^3 - 4x + 1}{x^2 + 2}$ . Which describes where $q(x) > 0$ ?

$(-2.115, 0.254) \cup (1.861, \infty)$
$(-\infty, -2.115) \cup (0.254, 1.861)$
$(1.861, \infty)$
$(-2.115, \infty)$

Show the answer

The denominator is always positive, so $q$ has the sign of $x^3 - 4x + 1$ . Its zeros (about $-2.115$ , $0.254$ , $1.861$ ) split the line; test $q(0) = 0.5$ and $q(1) < 0$ .

Correct: $(-2.115, 0.254) \cup (1.861, \infty)$ : Right: the graph is above the axis between the first two zeros and after the third.
$(-\infty, -2.115) \cup (0.254, 1.861)$ : These are where the graph is below the axis.
$(1.861, \infty)$ : $q(0) = 0.5 > 0$ , so part of the solution is between $-2.115$ and $0.254$ .
$(-2.115, \infty)$ : Between $0.254$ and $1.861$ the graph dips below the axis: $q(1) = -\frac{2}{3}$ .

6. What is the real zero of $g(x) = \frac{2x - 5}{x + 4}$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Set the numerator to $0$ : $2x - 5 = 0$ , $x = \frac{5}{2}$ . It is in the domain ( $\frac{5}{2} \ne -4$ ).

Answer: $\frac{5}{2}$ ≈ 2.500

Part 9 · Summary

Summary

The real zeros of a rational function are the numerator's zeros that are in its domain. Zeros and excluded inputs split the number line; a test value in each interval gives the sign there. Solve a rational inequality from that sign chart, including zeros only for

\le

or

\ge

and never an excluded input.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Test values for a sign chart

The graph of q(x)q(x)=x3−4x+1x2+2= \frac{x^3 - 4x + 1}{x^2 + 2}

Summary

What comes next

Connections

The graph of $q(x)$ $= \frac{x^3 - 4x + 1}{x^2 + 2}$