AP PRECALCULUS • POLYNOMIAL AND RATIONAL FUNCTIONS

Rational Functions and Holes

Understanding removable discontinuities where common factors cancel in rational expressions.

Historical Context & Motivation

The study of rational functions — ratios of polynomial expressions — traces its roots to ancient Greek mathematics, where scholars like Euclid and Diophantus grappled with ratios and proportions that occasionally became undefined. As algebra matured through the Islamic Golden Age and the European Renaissance, mathematicians encountered expressions where division by zero created conceptual difficulties. The notion that a function could be "almost defined" at a point — behaving perfectly on either side yet possessing a pinpoint gap — became a central concern as analysis developed. These removable discontinuities, colloquially called holes, illuminate a subtle distinction between a function's algebraic form and its geometric behavior, a distinction that became foundational for calculus and modern analysis.

~300 BCE
Euclid's Ratios
Euclid's Elements formalized ratios of magnitudes, laying groundwork for rational relationships while carefully avoiding division by zero.
1637
Descartes' Analytic Geometry
René Descartes unified algebra and geometry, enabling polynomial and rational expressions to be visualized as curves with gaps and asymptotes.
1748
Euler's Introductio
Leonhard Euler systematically studied rational functions, classifying their behavior at poles and removable singularities in his landmark treatise.
1821
Cauchy's Rigorous Limits
Augustin-Louis Cauchy formalized the epsilon-delta definition of limits, giving precise meaning to removable discontinuities and distinguishing them from essential ones.

The central question that emerges from this history is deceptively simple: when a factor appears in both the numerator and denominator of a rational expression, does canceling it truly "fix" the function, or does a trace of the original restriction remain? Understanding this question — and recognizing that the answer involves a point missing from the domain even after simplification — is essential for mastering rational functions on the AP Precalculus exam and for building the conceptual foundation that calculus demands.

Core Principles & Definitions

A rational function is any function that can be written as the quotient of two polynomials, f(x) = p(x)/q(x), where q(x) is not the zero polynomial. The domain excludes every x-value that makes q(x) = 0. Among those excluded values, some correspond to vertical asymptotes while others correspond to holes (removable discontinuities). The distinction depends entirely on whether the factor causing q(x) = 0 also appears in p(x).

1

Rational Function

f(x) = p(x)/q(x), where p and q are polynomials and q(x) ≠ 0. The domain is all real numbers except where q(x) = 0.
2

Hole (Removable Discontinuity)

Occurs at x = a when (x − a) is a common factor of both p(x) and q(x). After cancellation, the function is defined everywhere nearby but not at x = a itself.
3

Vertical Asymptote

Occurs at x = a when (x − a) is a factor of q(x) but NOT of p(x) after all common factors are cancelled. The function grows without bound near x = a.
4

Common Factor Test

Factor both p(x) and q(x) completely. Shared linear factors identify holes; remaining factors of q(x) identify vertical asymptotes.
5

Coordinates of a Hole

If x = a produces a hole, the y-coordinate is found by evaluating the simplified (cancelled) function at x = a. The hole is at (a, f_simplified(a)).
KEY TAKEAWAY
KEY TAKEAWAY

Visual Explanation

The following diagram compares the graph of f(x) = (x² − 1)/(x − 1) with its simplified form g(x) = x + 1. Algebraically, x² − 1 factors as (x − 1)(x + 1), so f(x) = (x − 1)(x + 1)/(x − 1). After cancellation we obtain g(x) = x + 1, but f is undefined at x = 1 because the original denominator is zero there. The graph of f is identical to the line y = x + 1 except for an open circle at (1, 2).

The cyan line represents y = x + 1. The open circle (pink) at (1, 2) marks the hole — the single point where f(x) is undefined despite the limit existing.

Notice that the graph is a perfectly straight line everywhere except at x = 1. The open circle is the standard graphical convention for indicating that a point is excluded from the function's range at that input. If someone were to trace the curve from left to right, they would pass seamlessly through the region near x = 1 — the gap is infinitesimally small and invisible to the naked eye without the explicit marker. This is precisely what makes holes subtle: the limit as x approaches 1 equals 2, yet f(1) itself does not exist.

Mathematical Framework

The algebraic procedure for identifying and locating holes in rational functions involves three systematic steps: factoring, identifying common factors, and evaluating the simplified function. Below are the key equations and relationships that govern this process.

GENERAL RATIONAL FUNCTION
f(x) = p(x) / q(x)
where p(x) and q(x) are polynomials, q(x) ≠ 0. The domain is {x ∈ ℝ : q(x) ≠ 0}.
FACTORED FORM WITH COMMON FACTOR
f(x) = (x − a)ᵏ · P(x) / [(x − a)ᵐ · Q(x)]
If k ≥ m, then x = a produces a hole (removable discontinuity). If k < m, a vertical asymptote remains at x = a after cancellation. P(a) ≠ 0 and Q(a) ≠ 0.
HOLE COORDINATES
Hole at (a, L) where L = lim_{x→a} f(x) = P(a) / Q(a)
After canceling all common (x − a) factors, substitute x = a into the simplified expression. This limit L gives the y-coordinate of the hole.
MULTIPLICITY COMPARISON
Numerator multiplicity k vs. Denominator multiplicity m at x = a
If k ≥ m: hole (removable). If k < m: vertical asymptote persists with multiplicity (m − k). This distinction is critical for functions with repeated factors.
AP Exam Alert

Classifying Discontinuities in Rational Functions

Rational functions can exhibit multiple types of discontinuities simultaneously. A single function may have one or more holes alongside one or more vertical asymptotes, and the end behavior may include horizontal or oblique asymptotes. The diagram below provides a decision flowchart for classifying each zero of the denominator.

This flowchart guides the classification of each zero of q(x). The green path leads to holes (removable), while the red path leads to vertical asymptotes (non-removable). The orange case handles the less common scenario where a common factor partially cancels.
Comparison of holes and vertical asymptotes in rational functions
FeatureHole (Removable)Vertical Asymptote (Non-removable)
Common factor?Yes — (x − a) divides both p(x) and q(x)No — (x − a) divides only q(x)
Limit at x = aFinite: L = P(a)/Q(a)Does not exist (±∞)
Graph behaviorOpen circle at (a, L)Curve approaches ±∞ near x = a
Domain effectx = a excluded from domainx = a excluded from domain
After cancellationSimplified function is defined at x = aFactor remains in denominator

Worked Example

Consider the rational function f(x) = (2x² − 2)/(x² − 3x + 2). We will find all discontinuities, classify each as a hole or vertical asymptote, determine the coordinates of any holes, and describe the end behavior.

1
Step 1 — Factor the numeratorFactor out the greatest common factor from 2x² − 2: this gives 2(x² − 1). Recognizing x² − 1 as a difference of squares, we obtain 2(x − 1)(x + 1).
p(x) = 2(x − 1)(x + 1)
2
Step 2 — Factor the denominatorFactor x² − 3x + 2. We seek two numbers that multiply to 2 and add to −3: those are −1 and −2. Therefore x² − 3x + 2 = (x − 1)(x − 2).
q(x) = (x − 1)(x − 2)
3
Step 3 — Identify common factorsComparing the factored forms, the common factor is (x − 1). This appears once in both the numerator and denominator (multiplicity 1 in each). Therefore x = 1 produces a hole. The remaining denominator factor (x − 2) has no match in the numerator, so x = 2 produces a vertical asymptote.
4
Step 4 — Simplify and find hole coordinatesCancel (x − 1): f(x) simplifies to 2(x + 1)/(x − 2) for x ≠ 1. To find the y-coordinate of the hole, evaluate the simplified expression at x = 1: 2(1 + 1)/(1 − 2) = 2(2)/(−1) = −4.
Hole at (1, −4)
5
Step 5 — Determine end behaviorSince the degree of p(x) equals the degree of q(x) (both degree 2), the horizontal asymptote is the ratio of leading coefficients: y = 2/1 = 2. As x → ±∞, f(x) → 2.
Horizontal asymptote: y = 2; Vertical asymptote: x = 2
Domain Statement

Common Errors & Misconceptions

Students frequently make predictable errors when working with holes in rational functions. Understanding these pitfalls — and why each is wrong — strengthens conceptual mastery and prevents avoidable exam mistakes.

Frequent mistakes and their corrections
Common ErrorWhy It's WrongCorrect Approach
Claiming f(a) exists after canceling the common factorCancellation produces a new function g(x) that agrees with f(x) everywhere except x = a. The original f(a) remains undefined.State that the limit equals L but f(a) is undefined; the hole is at (a, L).
Forgetting to factor before cancelingWithout complete factoring, hidden common factors go undetected, causing misidentification of a hole as a vertical asymptote.Always factor numerator and denominator fully — including GCF, difference of squares, and grouping.
Labeling every zero of q(x) as a vertical asymptoteZeros of q(x) that are also zeros of p(x) may be holes, not asymptotes. The classification depends on common factors.Test each zero of q(x) individually against the factored numerator before classifying.
Including the hole in the range of fSince f(a) is undefined, the value L = P(a)/Q(a) is not in the range of the original function (unless achieved at another x-value).Check whether L is attained at any other input before listing it in the range.
KEY TAKEAWAY
KEY TAKEAWAY

Connection to Limits & Calculus

Holes in rational functions are the precalculus gateway to the formal concept of a limit. In AP Calculus, the expression lim_{x→a} f(x) = L precisely captures the idea that f(x) approaches L as x nears a, regardless of whether f(a) exists. When you evaluate the simplified function at the hole to find L, you are computing this limit algebraically. Recognizing removable discontinuities also prepares you for L'Hôpital's Rule, where 0/0 indeterminate forms arise from common factors in numerator and denominator, and for the definition of the derivative, which is itself a limit of a rational-like difference quotient.

How precalculus concepts extend into calculus
ConceptAP Precalculus TreatmentAP Calculus Extension
HoleCancel common factor, evaluate simplified function at x = a to find hole coordinates.Formalize as lim_{x→a} f(x) = L; classify as removable discontinuity using epsilon-delta.
Vertical AsymptoteIdentify non-cancelable zeros of q(x); note f(x) → ±∞.Compute one-sided limits: lim_{x→a⁺} and lim_{x→a⁻}; classify infinite discontinuity.
End BehaviorCompare degrees for horizontal/slant asymptote.Compute lim_{x→±∞} f(x) using dominant terms; connect to polynomial long division.
0/0 FormRecognize as indicator of a common factor; factor and simplify.Apply L'Hôpital's Rule or algebraic manipulation for general 0/0 indeterminate forms.

Mastering holes now pays dividends later. The ability to factor, cancel, and compute the resulting limit is exactly the skill set you will rely on when computing derivatives from the definition, evaluating integrands at points of indeterminacy, and analyzing the continuity of piecewise-defined functions.

Practice Problems

1
Which of the following best describes why the graph of f(x) = (x² − 4)/(x − 2) has a hole rather than a vertical asymptote at x = 2?
2
Find the coordinates of the hole in f(x) = (x² + x − 6)/(x² − 4).
3
The function g(x) = (x³ − x)/(x² − 1) has how many holes and how many vertical asymptotes?
PROBLEM 4APPLIED
A chemical reaction's rate is modeled by R(c) = (c² − 9c + 18)/(c² − 6c), where c is the concentration in mol/L and c > 0. (a) Identify all values of c excluded from the domain. (b) Classify each excluded value as a hole or vertical asymptote. (c) Find the coordinates of any holes and interpret the result in context. (d) Determine the horizontal asymptote and explain what it means for large concentrations.
PROBLEM 5CRITICAL THINKING
Consider f(x) = (x² − 2x + 1)/(x³ − 3x² + 3x − 1). (a) Factor the numerator and denominator completely. (b) Determine whether x = 1 is a hole or a vertical asymptote. Justify using multiplicities. (c) If f is redefined so that f(1) = L to make the function continuous, explain why no such finite L exists.
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