Rational Functions and Zeros - AP Precalculus

SECTION 1

Historical Context & Motivation

The study of rational functions — functions formed as ratios of two polynomials — has roots stretching back to antiquity, where Greek mathematicians first grappled with proportions and ratios that could become undefined. The concept of a function having zeros, points where the output equals zero, emerged naturally from the broader quest to solve equations and understand curves. As algebra matured through Arabic and European traditions, mathematicians realized that the zeros of a rational function depend entirely on the zeros of its numerator polynomial, provided no cancellation with the denominator occurs.

~300 BCE • Euclid's Ratios

Euclid formalized the theory of proportions in Elements Book V, establishing foundational reasoning about ratios that would later inform the study of rational expressions.

1637 • Descartes' La Géométrie

René Descartes unified algebra and geometry, enabling polynomial equations to be visualized as curves. Zeros of polynomials gained geometric meaning as x-intercepts on coordinate axes.

1748 • Euler's Introductio

Leonhard Euler systematically classified functions including rational functions, studying their behavior near points of discontinuity and establishing the formal notion of vertical asymptotes.

1821 • Cauchy's Rigorous Analysis

Augustin-Louis Cauchy provided rigorous definitions of limits, enabling precise characterization of how rational functions behave near zeros, poles, and removable discontinuities.

The central question driving this topic is deceptively simple: given a rational function r(x) = p(x)/q(x), where exactly does the function's output equal zero, and how do we distinguish true zeros from removable discontinuities (holes)? Answering this question requires a careful analysis of both the numerator and denominator polynomials, their factored forms, and the interplay between them — skills that form the backbone of AP Precalculus and extend directly into calculus.

SECTION 2

Core Principles & Definitions

A rational function is any function that can be expressed as r(x) = p(x)/q(x), where p(x) and q(x) are polynomial functions and q(x) ≠ 0. The domain of a rational function excludes all x-values for which q(x) = 0. A zero (or real zero) of a rational function is an x-value in the domain where r(x) = 0. Since a fraction equals zero only when its numerator equals zero, finding the zeros of a rational function reduces to finding the zeros of p(x), then excluding any values simultaneously zeroing out q(x).

Zeros of the Numerator

The real zeros of r(x) come from solving p(x) = 0. Each solution is a zero of r(x) only if it does not also make q(x) = 0.

Zeros of the Denominator

Values where q(x) = 0 are excluded from the domain. They produce vertical asymptotes or holes, depending on whether the factor cancels with p(x).

Holes (Removable Discontinuities)

If a linear factor (x − a) appears in both p(x) and q(x), the common factor cancels algebraically, creating a hole at x = a rather than a zero or vertical asymptote.

Multiplicity

A zero of even multiplicity causes the graph to touch the x-axis and bounce back, while a zero of odd multiplicity causes the graph to cross through the axis.

SECTION 3

Visualizing Zeros, Holes, and Asymptotes

The following diagram illustrates the graph of a rational function with distinct zeros, a vertical asymptote, and a hole. Consider the function r(x) = (x − 1)(x + 2) / [(x + 2)(x − 3)]. After canceling the common factor (x + 2), the simplified form is (x − 1)/(x − 3), valid for x ≠ −2. The graph reveals a zero at x = 1 where the curve crosses the x-axis, a vertical asymptote at x = 3, and a hole at x = −2.

The cyan curve shows r(x). The solid cyan dot at (1, 0) marks the zero. The open amber circle at x = −2 marks the hole. The dashed red line at x = 3 is the vertical asymptote, and the dashed green line at y = 1 is the horizontal asymptote.

Notice the critical distinction: both x = 1 and x = −2 make the numerator zero, but only x = 1 is a true zero of r(x). At x = −2, the factor (x + 2) cancels between numerator and denominator, removing that point from the graph entirely rather than producing an x-intercept. Meanwhile, x = 3 zeros out only the denominator after cancellation, producing a vertical asymptote where the function grows without bound.

SECTION 4

Mathematical Framework

Defining Rational Functions and Their Zeros

GENERAL RATIONAL FUNCTION

r(x) = p(x) / q(x) = (aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ⋯ + a₁x + a₀) / (bₘxᵐ + bₘ₋₁xᵐ⁻¹ + ⋯ + b₁x + b₀)

where p(x) is a polynomial of degree n with leading coefficient aₙ, q(x) is a polynomial of degree m with leading coefficient bₘ, and q(x) ≠ 0.

CONDITION FOR A ZERO

r(c) = 0 ⟺ p(c) = 0 AND q(c) ≠ 0

A value x = c is a zero of r(x) if and only if c is a root of the numerator and c is not a root of the denominator. If both p(c) = 0 and q(c) = 0, then x = c is a candidate for a hole, not a zero.

Factored Form Analysis

FACTORED RATIONAL FUNCTION

r(x) = a · (x − z₁)^(k₁)(x − z₂)^(k₂)⋯(x − zⱼ)^(kⱼ) / [(x − d₁)^(m₁)(x − d₂)^(m₂)⋯(x − dᵢ)^(mᵢ)]

Here z₁, z₂, …, zⱼ are the roots of p(x) with multiplicities k₁, k₂, …, kⱼ, and d₁, d₂, …, dᵢ are the roots of q(x) with multiplicities m₁, m₂, …, mᵢ. If zₛ = dₜ for some s and t, the common factor must be analyzed for cancellation.

The factored form is the essential tool for identifying zeros. After factoring both p(x) and q(x) completely, cancel all common factors. Each remaining factor of the form (x − c) in the numerator yields a zero at x = c. The multiplicity of a zero determines the graph's behavior at that x-intercept: if the multiplicity kₛ is odd, the graph crosses the x-axis at x = zₛ; if kₛ is even, the graph touches the axis and reverses direction. This behavior is identical to polynomial zero behavior because the denominator is nonzero at a true zero, so it acts as a nonzero scaling factor near that point.

HOLE CONDITION

If (x − a) divides both p(x) and q(x), then x = a is a removable discontinuity (hole) at the point (a, r̃(a))

where r̃(x) is the simplified form of r(x) after canceling (x − a). The y-coordinate of the hole is found by evaluating the simplified function at x = a.

SECTION 5

Classifying Features from Factored Form

Every x-value that makes either the numerator or denominator zero falls into exactly one of three categories. The following decision diagram systematizes the classification process, which is the single most important skill for analyzing rational functions on the AP exam.

Decision flowchart for classifying x = a. Begin by factoring both polynomials fully. Each critical x-value is classified as a zero, a vertical asymptote, or a hole based on where the factor appears and multiplicities.

Feature	Condition	Graph Behavior
Zero	p(c) = 0 and q(c) ≠ 0	Graph crosses or touches x-axis at (c, 0)
Vertical Asymptote	q(c) = 0 and p(c) ≠ 0 (after cancellation)	Graph grows without bound: r(x) → ±∞
Hole	Common factor (x − c) cancels completely	Open circle at (c, r̃(c)); function undefined
Zero + Hole	Multiplicity of (x − c) in p(x) > in q(x), and remaining factor in numerator yields zero	After cancellation, simplified function has zero at x = c, but original is undefined there

Summary of rational function feature classification

SECTION 6

Worked Example

Worked example

1
Step 1 — State the Function

Analyze the rational function r(x) = (2x³ − 2x² − 12x) / (x³ − 9x). Find all zeros, vertical asymptotes, and holes.
2
Step 2 — Factor the Numerator

Factor out the GCF from the numerator: 2x³ − 2x² − 12x = 2x(x² − x − 6). Factor the quadratic: x² − x − 6 = (x − 3)(x + 2). So p(x) = 2x(x − 3)(x + 2).

p(x) = 2x(x − 3)(x + 2)
3
Step 3 — Factor the Denominator

Factor out x: x³ − 9x = x(x² − 9). Recognize the difference of squares: x² − 9 = (x − 3)(x + 3). So q(x) = x(x − 3)(x + 3).

q(x) = x(x − 3)(x + 3)
4
Step 4 — Identify and Cancel Common Factors

The common factors between p(x) and q(x) are x and (x − 3). Cancel these: r(x) = 2x(x − 3)(x + 2) / [x(x − 3)(x + 3)] = 2(x + 2) / (x + 3), valid for x ≠ 0 and x ≠ 3. The canceled factors produce holes at x = 0 and x = 3.

r̃(x) = 2(x + 2) / (x + 3), holes at x = 0 and x = 3
5
Step 5 — Find the Zeros

Set the simplified numerator equal to zero: 2(x + 2) = 0 → x = −2. Check: x = −2 is in the domain (q(−2) = (−2)(−2 − 3)(−2 + 3) = (−2)(−5)(1) = 10 ≠ 0). Therefore x = −2 is the only zero, and the graph crosses the x-axis at (−2, 0).

Zero: x = −2, giving the point (−2, 0)
6
Step 6 — Find Vertical Asymptotes

After cancellation, the remaining denominator factor is (x + 3). Setting x + 3 = 0 gives x = −3. Since (x + 3) does not cancel, x = −3 is a vertical asymptote.

Vertical asymptote: x = −3
7
Step 7 — Find Hole Coordinates

For the hole at x = 0: evaluate r̃(0) = 2(0 + 2)/(0 + 3) = 4/3. Hole at (0, 4/3). For the hole at x = 3: evaluate r̃(3) = 2(3 + 2)/(3 + 3) = 10/6 = 5/3. Hole at (3, 5/3).

Holes: (0, 4/3) and (3, 5/3)

SECTION 7

Rational vs. Polynomial Zeros: Strengths & Limitations

Understanding the zeros of rational functions requires distinguishing their behavior from the more straightforward zeros of polynomial functions. While both types of functions can be analyzed through factoring, rational functions introduce additional complexity through domain restrictions, potential cancellation, and the interplay between numerator and denominator behaviors.

Property	Polynomial f(x)	Rational r(x) = p(x)/q(x)
Domain	All real numbers (−∞, ∞)	All reals except where q(x) = 0
Finding zeros	Solve f(x) = 0; every root is a zero	Solve p(x) = 0, then verify q(x) ≠ 0 at each root
Max number of real zeros	Equal to deg(f)	Equal to deg(p), minus any canceled factors
Multiplicity behavior	Directly determines crossing/bouncing	Same, but only for zeros after cancellation
Possible complications	None — zeros are straightforward	Holes, asymptotes, sign changes near discontinuities

Comparison of zero-finding for polynomial and rational functions

SECTION 8

Connections to Calculus and Advanced Topics

The analysis of rational function zeros connects directly to several calculus concepts. In differential calculus, the zeros of a rational function determine its x-intercepts, which are critical reference points for curve sketching. In integral calculus, the technique of partial fraction decomposition requires fully factoring both the numerator and denominator — exactly the skills developed here. Removable discontinuities (holes) foreshadow the limit concept: the limit of r(x) as x approaches a hole exists even though the function value does not, a distinction that becomes central in the definition of continuity.

AP Precalculus Concept	Calculus Extension
Zeros of r(x) via factored numerator	Finding x-intercepts for curve sketching; sign analysis for first/second derivative tests
Holes (removable discontinuities)	Limits: lim as x→a of r(x) = r̃(a); definition of continuity
Vertical asymptotes from q(x) = 0	Infinite limits; improper integrals and convergence
Complete factored form of r(x)	Partial fraction decomposition for integration of rational functions

How precalculus rational function skills extend to calculus

ℹ

AP Exam Tip

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following best explains why x = 4 is NOT a zero of r(x) = (x² − 16) / (x − 4)?

A. Because x² − 16 does not equal zero when x = 4
B. Because x = 4 makes the denominator zero, so it is not in the domain of r(x)
C. Because the function simplifies to x + 4, which equals 8 at x = 4
D. Because r(x) has no real zeros at all
E. Because the numerator has a higher degree than the denominator

Show answer (B)

The correct answer is B. Although x = 4 makes the numerator x² − 16 = 0, it simultaneously makes the denominator x − 4 = 0, excluding it from the domain. A zero must be in the domain. Choice A is wrong because x² − 16 does equal zero at x = 4 (16 − 16 = 0). Choice C correctly describes the simplified form and its value but does not explain why x = 4 is not a zero — the fundamental reason is the domain exclusion, not the y-value of the simplified form. Choice D is wrong because x = −4 is a valid zero (numerator zero, denominator nonzero). Choice E is irrelevant; degree comparison determines end behavior, not zeros.

Question 2: BASIC CALCULATION

Find all real zeros of r(x) = (3x² − 12) / (x² + 5x + 6).

A. x = 2 and x = −2
B. x = 2 only
C. x = −2 only
D. x = −2 and x = −3
E. No real zeros

Show answer (B)

The correct answer is B. Factor: numerator = 3(x² − 4) = 3(x − 2)(x + 2); denominator = (x + 2)(x + 3). The factor (x + 2) is common and cancels, creating a hole at x = −2. After cancellation, r̃(x) = 3(x − 2)/(x + 3). Setting the simplified numerator to zero: x − 2 = 0, so x = 2. At x = 2, q(2) = (4)(5) = 20 ≠ 0, confirming it is in the domain. Choice A is wrong because x = −2 is a hole, not a zero. Choice C is wrong — x = −2 is excluded from the domain. Choice D lists denominator roots. Choice E is wrong because x = 2 is a valid zero.

Question 3: INTERMEDIATE

Consider g(x) = (x³ − 4x) / (x² − x − 2). Determine the number of real zeros, vertical asymptotes, and holes of g(x).

A. 3 zeros, 1 vertical asymptote, 0 holes
B. 1 zero, 2 vertical asymptotes, 1 hole
C. 2 zeros, 1 vertical asymptote, 1 hole
D. 3 zeros, 0 vertical asymptotes, 2 holes
E. 1 zero, 1 vertical asymptote, 2 holes

Show answer (C)

The correct answer is C. Factor the numerator: x³ − 4x = x(x² − 4) = x(x − 2)(x + 2). Factor the denominator: x² − x − 2 = (x − 2)(x + 1). The common factor is (x − 2), creating a hole at x = 2. After cancellation: g̃(x) = x(x + 2)/(x + 1). The zeros come from x(x + 2) = 0, giving x = 0 and x = −2. Both are in the domain (neither makes the original denominator zero after checking). The remaining denominator factor (x + 1) gives a vertical asymptote at x = −1. Total: 2 zeros, 1 VA, 1 hole. Choice A incorrectly counts 3 zeros by treating x = 2 as a zero rather than a hole. Choice B incorrectly counts vertical asymptotes and undercounts zeros. Choice D is wrong — x = 2 is a hole, not a zero, and there is a VA. Choice E undercounts zeros and overcounts holes.

Question 4: APPLIED

A chemical reaction's rate is modeled by R(c) = (6c² − 54) / (c³ − 3c² − 18c), where c represents concentration in mol/L and R is the rate in mol/(L·s). (a) Find the domain of R(c) in the context of concentration (c > 0). (b) Determine all real zeros of R(c). (c) Identify all holes and vertical asymptotes. (d) Interpret the zero(s) in the context of the chemical reaction.

Show answer

(a) Factor the denominator: c³ − 3c² − 18c = c(c² − 3c − 18) = c(c − 6)(c + 3). Setting each factor to zero: c = 0, c = 6, c = −3. Since concentration must be positive, the domain is (0, 6) ∪ (6, ∞). (b) Factor the numerator: 6c² − 54 = 6(c² − 9) = 6(c − 3)(c + 3). The common factor with the denominator is (c + 3). After cancellation: R̃(c) = 6(c − 3)/[c(c − 6)]. Setting the simplified numerator to zero: c − 3 = 0, so c = 3 is the only zero. Verify: the original denominator at c = 3 is 3(3 − 6)(3 + 3) = 3(−3)(6) = −54 ≠ 0. ✓ (c) The canceled factor (c + 3) produces a hole at c = −3, which is outside the physical domain. The remaining denominator factors c and (c − 6) give vertical asymptotes at c = 0 and c = 6. (d) The zero at c = 3 mol/L means the net reaction rate is zero — the system reaches equilibrium at this concentration, with no net change in reactant or product concentrations.

Question 5: CRITICAL THINKING

Construct a rational function r(x) that has exactly one distinct real zero at x = 5, a hole at x = 2, vertical asymptotes at x = −4 and x = 3, and a horizontal asymptote at y = 3. Write r(x) in factored form and justify that it satisfies all five conditions.

Show answer

We need: exactly one real zero at x = 5, a hole at x = 2, vertical asymptotes at x = −4 and x = 3, and a horizontal asymptote at y = 3. Step 1 — Build the numerator. The zero at x = 5 contributes (x − 5). The hole at x = 2 requires a common factor (x − 2). So the numerator contains (x − 5)(x − 2). That gives degree 2. Step 2 — Build the denominator. The hole at x = 2 requires (x − 2). The VAs at x = −4 and x = 3 require (x + 4) and (x − 3). So the denominator is (x − 2)(x + 4)(x − 3). That gives degree 3. Step 3 — Match degrees for HA y = 3. The numerator must also be degree 3. We need one more linear factor in the numerator that does NOT create an additional zero (i.e., it must not cancel from the simplified form and must not introduce a new x-intercept that conflicts). However, any additional linear factor (x − a) in the numerator that doesn't cancel with a denominator factor will create a second zero. Instead, we can use a repeated factor: include (x − 5)² in the numerator so that the zero at x = 5 still counts as exactly one distinct real zero. Numerator: 3(x − 5)²(x − 2) — degree 3, leading coefficient 3. Denominator: (x − 2)(x + 4)(x − 3) — degree 3, leading coefficient 1. So: r(x) = 3(x − 5)²(x − 2) / [(x − 2)(x + 4)(x − 3)]. Step 4 — Verify all conditions. • Zero: After cancellation, r̃(x) = 3(x − 5)² / [(x + 4)(x − 3)]. Setting numerator = 0 gives x = 5 (multiplicity 2), so exactly one distinct real zero at x = 5. ✓ • Hole: (x − 2) cancels. At x = 2, r̃(2) = 3(2 − 5)² / [(2 + 4)(2 − 3)] = 3(9) / [(6)(−1)] = 27/(−6) = −9/2. Hole at (2, −9/2). ✓ • Vertical asymptotes: Denominator of r̃ is zero at x = −4 and x = 3; numerator is nonzero there. VAs at x = −4 and x = 3. ✓ • Horizontal asymptote: deg(num) = deg(den) = 3. Ratio of leading coefficients = 3/1 = 3. HA at y = 3. ✓