Rational functions and zeros — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of rational function zeros, factoring to identify real zeros, distinguishing between zeros and discontinuities, multiplicity effects on graph behavior, and solving for zeros via algebraic and graphical methods.

You should already know: Factoring of polynomial numerators and denominators, definition of function zeros, basic limit evaluation at discontinuities.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Rational functions and zeros?

A rational function is defined as the ratio of two polynomials $R(x)=\frac{N(x)}{D(x)}$, where $N(x)$ is the numerator polynomial and $D(x)$ is a non-zero denominator polynomial, with domain all real $x$ except roots of $D(x)$. Zeros of $R(x)$ are the input values $x$ where the output $R(x)=0$, which by the zero product property only occurs when the numerator $N(x)=0$ and the denominator $D(x)\ne 0$ at that same input (if both are zero, it is a discontinuity, not a zero). According to the AP Precalculus CED, this topic falls within Unit 1: Polynomial and Rational Functions, which accounts for 27-32% of total exam weight, and rational function zeros specifically appear in both multiple-choice (MCQ) and free-response (FRQ) sections. Questions often test the ability to distinguish between extraneous roots that are actually discontinuities versus valid zeros, and how the multiplicity of a zero changes the shape of the graph crossing or touching the x-axis at that point. Mastery of this topic is foundational for analyzing discontinuities, asymptotes, and solving rational equations and inequalities later in the course.

2. Finding Zeros of Rational Functions Algebraically

The core rule for finding rational function zeros comes directly from the properties of real numbers: a non-zero fraction equals zero if and only if its numerator is zero and its denominator is non-zero. To find all real zeros algebraically, follow this structured process: 1. Fully factor both the numerator $N(x)$ and denominator $D(x)$ of the rational function. 2. Find all real roots of $N(x)=0$; these are your candidate zeros. 3. Eliminate any candidate that is also a root of $D(x)=0$, because that input is not in the domain of $R(x)$, so it cannot be a zero. If a candidate is a root of both, it is a point discontinuity (hole), not a zero. 4. Any remaining candidate roots are valid zeros of $R(x)$. This works because the only way $\frac{N(x)}{D(x)}=0$ for real numbers with $D(x)\ne 0$ is when $N(x)=0$. The most common early mistake is skipping the elimination step and counting any root of the numerator as a zero, ignoring domain restrictions.

Worked Example

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

1. Factor numerator and denominator completely: Numerator $x^3-2x^2-3x=x(x^2-2x-3)=x(x-3)(x+1)$. Denominator $x^2-9=(x-3)(x+3)$.
2. Identify candidate zeros by solving $N(x)=0$: Candidates are $x=0,x=3,x=-1$.
3. Eliminate candidates not in the domain: Roots of $D(x)=0$ are $x=3$ and $x=-3$, so $x=3$ is eliminated because it is not in the domain.
4. Confirm remaining candidates: $R(0)=\frac{0}{-9}=0$ and $R(-1)=\frac{0}{-8}=0$, so both are valid. 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

Just like polynomial zeros, 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 defined as the exponent of the $(x-a)$ factor for zero $x=a$ in the fully simplified numerator. The graph behavior near a zero $x=a$ follows the same rules as for polynomials, because the denominator does not change sign around $a$ (since $a$ is not a root of the denominator): 1. Odd multiplicity: The sign of $R(x)$ changes when moving from left to right across $a$, so the graph crosses the x-axis directly at $x=a$. 2. Even multiplicity: The sign of $R(x)$ stays the same on both sides of $a$, so the graph touches the x-axis at $x=a$ and turns around, rather than crossing. If the numerator has a higher multiplicity of a root than the denominator, the remaining multiplicity after canceling is the multiplicity of the zero.

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 factor $(x+2)$ cancels one power from numerator and denominator, giving simplified numerator $(x+2{)}^3(x-3{)}^2(x+1)$. The original domain excludes $x=-2$ and $x=-5$, so $x=-2$ is not in the domain and cannot be a zero.
2. Identify valid zeros: Roots of the simplified numerator in the domain are $x=-1$ 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 of Zeros

AP Precalculus often asks to identify zeros from a graph or table, so you need to be able to recognize zeros even when the numerator cannot be factored easily. Graphically, a zero of a rational function 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 though it lies on the x-axis, because it is not in the domain. Numerically, from a table of values, the Intermediate Value Theorem tells us that an odd multiplicity zero exists between two consecutive $x$-values where the sign of $R(x)$ changes. Note that a sign change can also occur across a vertical asymptote, so you must confirm no asymptote falls between the test points. For even multiplicity zeros, $R(x)$ has the same sign on both sides, so you look for $R(x)$ values very close to zero at a given input.

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 on the x-axis, 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. Final answer: Real zeros are $x=1$ and $x=4$.

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 (and how to avoid them)

• 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 the extraneous root is a standard exam distractor. 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. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives all real zeros of the function $R(x)=\frac{x^3-4x^2+3x}{x^2-2x-3}$? A) $x=0$ only B) $x=0,x=1,x=3$ C) $x=0,x=1$ only D) No real zeros

Worked Solution: First, factor the numerator and denominator completely: the numerator factors to $x(x-1)(x-3)$, and the denominator factors to $(x-3)(x+1)$. Candidate zeros from the numerator are $x=0,x=1,x=3$. Next, eliminate any candidate that makes the denominator zero: the denominator equals zero at $x=3$ and $x=-1$, so $x=3$ is eliminated as it is not in the domain. The remaining valid zeros are $x=0$ and $x=1$. The correct answer is C.

Question 2 (Free Response)

Let $R(x)=\frac{(x-2{)}^2(x+3)(x-1{)}^3}{(x-2)(x+4)}$. (a) Find all real zeros of $R(x)$. (b) For each zero, state its multiplicity and describe the behavior of the graph of $R(x)$ at that zero. (c) How many times does the graph of $R(x)$ cross the x-axis at valid zeros?

Worked Solution: (a) First, cancel the common factor of $(x-2)$ between numerator and denominator. The original domain excludes $x=2$ and $x=-4$, so $x=2$ is not in the domain and cannot be a zero. After canceling, the roots of the simplified numerator that are in the domain are $x=-3$ and $x=1$. Final answer for (a): Real zeros are $x=-3$ and $x=1$. (b) $x=-3$ has multiplicity 1, which is odd. Therefore, the graph crosses the x-axis at $x=-3$. $x=1$ has multiplicity 3, which is odd. Therefore, the graph crosses the x-axis at $x=1$. (c) Both valid zeros have odd multiplicity, so the graph crosses the x-axis twice at valid zeros. Final answer for (c): 2 crossings.

Question 3 (Application / Real-World Style)

The concentration $C(t)$ (in milligrams per liter, mg/L) of a drug in a patient's bloodstream $t$ hours after injection is given by the rational function: $C(t)=\frac{0.5t-0.2t^2}{t^2+1}$ for $t\ge 0$. At what times $t\ge 0$ is the concentration of the drug equal to zero? Interpret your result in context.

Worked Solution: To find zeros, set the numerator equal to zero and factor: $0.5t-0.2t^2=t(0.5-0.2t)=0$, giving candidates $t=0$ and $t=\frac{0.5}{0.2}=2.5$. The denominator $t^2+1$ is always positive for all real $t$, so no candidates are eliminated. Both $t=0$ and $t=2.5$ are non-negative, so both are valid zeros. Interpretation: At the moment of injection ($t=0$ hours), there is no drug in the patient's bloodstream yet, and after 2.5 hours, all of the drug has left the bloodstream.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of rational function zeros is a prerequisite for the next key topics in Unit 1: vertical and horizontal asymptotes, discontinuity classification, and solving rational equations and inequalities. 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 AP Precalculus 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. Understanding how zeros relate to sign changes also prepares you for calculus topics like the first derivative test for extrema.

• Vertical asymptotes and holes of rational functions
• Solving rational equations and inequalities
• Graphing rational functions