Rational functions and zeros

AP Precalculus· AP Precalculus CED — Polynomial and Rational Functions· 14 min read

1. Definition of Rational Functions and Zeros★★☆☆☆⏱ 3 min

A rational function is defined as the ratio of two polynomials, where the denominator is a non-zero polynomial.

R(x) = \frac{N(x)}{D(x)}

📘 Definition

Zero of a rational function

An input value $x = a$ where $R (a) = 0$ , which requires $N (a) = 0$ and $D (a) \neq = 0$ . If $N (a) = 0$ and $D (a) = 0$ , $a$ is a discontinuity, not a zero.

Example:

For $R (x) = \frac{x}{x - 1}$ , $x = 0$ is a zero because $N (0) = 0$ and $D (0) = - 1 \neq = 0$

This topic falls within Unit 1 of the AP Precalculus CED, which accounts for 27-32% of total exam weight, and appears in both multiple-choice and free-response sections. Mastery is foundational for analyzing discontinuities, asymptotes, and solving rational equations later in the course.

2. Finding Zeros Algebraically★★☆☆☆⏱ 5 min

A non-zero fraction equals zero if and only if its numerator is zero and its denominator is non-zero. Follow this structured process to find all real zeros:

Fully factor both the numerator $N (x)$ and denominator $D (x)$ of the rational function.
Find all real roots of $N (x) = 0$ ; these are your candidate zeros.
Eliminate any candidate that is also a root of $D (x) = 0$ , since it is not in the domain of $R (x)$ .
Any remaining candidates are valid zeros of $R (x)$ .

📐 Worked Example

Find all real zeros of $R (x) = \frac{x ^{3} - 2 x ^{2} - 3 x}{x ^{2} - 9}$

1
Factor numerator and denominator completely:
2
$x^3 - 2x^2 - 3x = x(x^2 - 2x - 3) = x(x-3)(x+1) \\ x^2 - 9 = (x-3)(x+3)$
3
Identify candidate zeros from roots of the numerator: $x = 0, x = 3, x = - 1$
4
Eliminate candidates that make the denominator zero: $x = 3$ is a root of $D (x)$ , so it is eliminated because it is not in the domain.
5
Confirm remaining candidates: $R (0) = 0$ and $R (- 1) = 0$ , so both are valid.
6
Final answer: Real zeros are $x = 0$ and $x = - 1$

Exam tip:

On AP MCQ, answer options almost always include the extraneous root as a distractor, so always cross off any candidate zero that makes the denominator zero before selecting your answer.

3. Multiplicity of Zeros and Graph Behavior★★★☆☆⏱ 4 min

Zeros of rational functions inherit their multiplicity from the multiplicity of the corresponding root in the numerator, after all common factors with the denominator have been canceled. Multiplicity is the exponent of the $(x - a)$ factor for zero $x = a$ in the fully simplified numerator, and determines graph behavior just like for polynomial zeros.

Odd multiplicity: The sign of $R (x)$ changes when moving across the zero, so the graph crosses the x-axis directly at $x = a$ .
Even multiplicity: The sign of $R (x)$ stays the same on both sides of the zero, so the graph touches the x-axis at $x = a$ and turns around.

📐 Worked Example

Given $R (x) = \frac{( x + 2 ) ^{4} ( x - 3 ) ^{2} ( x + 1 )}{( x + 2 ) ( x - 5 )}$ , find all real zeros, state the multiplicity of each, and describe graph behavior at each zero.

1
Simplify by canceling common factors: the common $(x + 2)$ term cancels one power from numerator and denominator. The original domain excludes $x = - 2$ and $x = 5$ , so $x = - 2$ is not a zero.
2
Identify valid zeros from the simplified numerator that are in the domain: $x = - 1$ and $x = 3$
3
State multiplicity: $x = - 1$ has multiplicity 1 (odd), $x = 3$ has multiplicity 2 (even)
4
Describe behavior: At $x = - 1$ , the graph crosses the x-axis; at $x = 3$ , the graph touches the x-axis and turns around.

Exam tip:

When asked to describe graph behavior on FRQ, you must connect the behavior to odd/even multiplicity explicitly to earn full credit — just stating "crosses" or "touches" is not enough.

4. Graphical and Numerical Identification★★☆☆☆⏱ 3 min

AP Precalculus often asks to identify zeros from a graph or table, even when the numerator cannot be easily factored. Graphically, a zero is a closed x-intercept: a point where the graph intersects the x-axis ( $y = 0$ ) and the point is included in the domain. An open circle on the x-axis indicates a hole (discontinuity), which is not a zero even if it lies on the x-axis.

Numerically, the Intermediate Value Theorem tells us an odd multiplicity zero exists between two consecutive $x$ -values where the sign of $R (x)$ changes. A sign change can also occur across a vertical asymptote, so you must confirm no asymptote falls between the test points.

📐 Worked Example

The graph of $R (x)$ has an open circle at $(- 2, 0)$ , crosses the x-axis at $(1, 0)$ with a closed dot, and touches the x-axis at $(4, 0)$ with a closed dot. Identify all real zeros of $R (x)$ .

1
Open circles mark points not in the domain, so $x = - 2$ is a hole, not a zero, and is eliminated.
2
The crossing at $(1, 0)$ is a closed point in the domain with $y = 0$ , so $x = 1$ is a valid zero.
3
The turning point touching the x-axis at $(4, 0)$ is a closed point in the domain with $y = 0$ , so $x = 4$ is a valid zero.
4
Final answer: Real zeros are $x = 1$ and $x = 4$

✓ Quick check

Test your understanding with this AP-style multiple choice question:

Which of the following gives all real zeros of the function $R (x) = \frac{x ^{3} - 4 x ^{2} + 3 x}{x ^{2} - 2 x - 3}$ ?
- A) $x = 0$ only
- B) $x = 0, x = 1, x = 3$
- C) $x = 0, x = 1$ only
- D) No real zeros
Reveal answer
C) $x=0, x=1$ only —
Factoring gives numerator $x (x - 1) (x - 3)$ and denominator $(x - 3) (x + 1)$ . $x = 3$ makes the denominator zero, so it is eliminated as a candidate zero, leaving only $x = 0$ and $x = 1$ .

Exam tip:

If you use a graphing calculator to find zeros on exam day, always plug the x-value back into the denominator to confirm it is non-zero and not a hole.

5. Common Pitfalls

Wrong move:

Calling $x = a$ a zero because it makes the numerator zero, even if it also makes the denominator zero.

Why:

Students forget to check domain restrictions after finding roots of the numerator, and extraneous roots are standard exam distractors.

Correct move:

After finding all candidate zeros from the numerator, test each candidate by plugging into the denominator, eliminate any candidate that gives a denominator of zero.

Wrong move:

Using the original multiplicity of a root in the numerator before canceling common factors with the denominator.

Why:

Students confuse the original factored form with the simplified form, leading to wrong multiplicity predictions for graph behavior.

Correct move:

Always cancel all common factors between numerator and denominator first, then count the exponent of the $(x - a)$ factor in the simplified numerator to get multiplicity.

Wrong move:

Counting a hole on the x-axis as a valid zero.

Why:

A hole at $y = 0$ looks like an x-intercept on a rough sketch, so students misidentify it.

Correct move:

On a graph, any x-intercept with an open circle is not a zero; always confirm the point is in the domain.

Wrong move:

Assuming all sign changes in a table of rational function values correspond to a zero.

Why:

A sign change can also occur across a vertical asymptote, not just a zero.

Correct move:

When identifying a zero from a sign change, confirm that the x-interval does not contain a vertical asymptote between the two test points.

Wrong move:

Stating that a rational function must have at least one real zero.

Why:

Students generalize the odd-degree polynomial rule to rational functions, which do not follow this requirement.

Correct move:

If no roots of the numerator are in the domain, explicitly state that the rational function has no real zeros.

6. Quick Reference Cheatsheet

Category	Rule / Formula	Notes
General rational function	$R (x) = \frac{N ( x )}{D ( x )}$	$N (x), D (x)$ are polynomials, $D (x) \neq = 0$
Condition for a real zero	$R (a) = 0 ⟺ N (a) = 0 and D (a) \neq = 0$	Applies to all real $a$ in the domain
Algebraic zero finding	Factor $N (x), D (x)$ ; 2. Find roots of $N (x) = 0$ ; 3. Eliminate roots of $D (x) = 0$	All remaining roots are valid zeros
Multiplicity of a zero	Multiplicity = exponent of $(x - a)$ in simplified $N (x)$ after canceling	Multiplicity determines graph behavior
Odd multiplicity zero	Graph crosses the x-axis	$R (x)$ changes sign across the zero
Even multiplicity zero	Graph touches the x-axis and turns	$R (x)$ has the same sign on both sides of the zero
Graphical zero identification	Zero = closed x-intercept, not an open circle	Open circles on the x-axis are holes, not zeros

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2024 · AP Precalculus
MCQ identifying valid zeros
2023 · AP Precalculus
FRQ on multiplicity and graph behavior

What's Next

Mastery of rational function zeros is a prerequisite for the next key topics in AP Precalculus Unit 1. Without correctly identifying zeros and distinguishing them from discontinuities, you will not be able to correctly sketch rational function graphs or solve rational inequality problems, which frequently appear on both MCQ and FRQ sections of the exam. This topic extends polynomial zero concepts to rational functions, laying the groundwork for limits of rational functions and end behavior analysis later in the course, and prepares you for key calculus concepts like the first derivative test for extrema.