AP · Rational functions and holes · 14 min read · Updated 2026-05-10

Rational functions and holes — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Identifying removable discontinuities (holes) in rational functions, factoring to find common linear factors, calculating hole coordinates, distinguishing holes from vertical asymptotes, and graphing rational functions with holes for the AP Precalculus exam.

You should already know: Polynomial factoring techniques, evaluating limits of rational functions, basic rational function graphing rules.

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 holes?

A rational function is defined as the ratio of two polynomials, written in standard form as $f (x) = \frac{N ( x )}{D ( x )}$ , where $N (x)$ is the numerator polynomial and $D (x)$ is a non-zero denominator polynomial. A hole, also called a removable discontinuity, occurs at an $x$ -value where both $N (x)$ and $D (x)$ equal zero, meaning the function has a common polynomial factor that cancels out from numerator and denominator.

Unlike non-removable discontinuities (such as vertical asymptotes), holes are single missing points on the graph that can be "filled in" by redefining the function at that $x$ -value to remove the discontinuity. According to the AP Precalculus Course and Exam Description (CED), this topic falls under Unit 1: Polynomial and Rational Functions, which accounts for 27–31% of the total AP exam score. Holes are tested in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ questions typically ask to identify hole coordinates or distinguish holes from asymptotes, while FRQ questions may require graphing rational functions with holes or justifying discontinuity classification.

2. Identifying the x-coordinate of a hole

To find the $x$ -coordinate(s) of any hole in a rational function, the first step is to fully factor both the numerator $N (x)$ and denominator $D (x)$ into linear factors. Next, identify all unique common linear factors of the form $(x - a)$ . For every unique common linear factor $(x - a)$ , $x = a$ is the $x$ -coordinate of a hole.

This rule holds because at $x = a$ , the original function is undefined (denominator equals zero), but after canceling the common factor, the simplified function is defined at $x = a$ . This means the discontinuity is removable, i.e., a hole. If a root of the denominator is not a root of the numerator (so there is no corresponding common factor), that root corresponds to a vertical asymptote, not a hole. Even if $(x - a)$ is raised to any power in both numerator and denominator, it still produces only one hole at $x = a$ .

Worked Example

Find the $x$ -coordinate(s) of all holes in the function $f (x) = \frac{2 x ^{2} + 2 x - 12}{x ^{2} - 4 x + 3}$ .

Factor the numerator: Pull out the GCF first, then factor the quadratic: $2 x^{2} + 2 x - 12 = 2 (x^{2} + x - 6) = 2 (x + 3) (x - 2)$ .
Factor the denominator: $x^{2} - 4 x + 3 = (x - 1) (x - 3)$ .
Check for common linear factors: No shared linear factors between numerator and denominator.
Conclusion: There are no holes in this function.

Wait, adjust the example to show a case with a hole: Corrected problem: Find the $x$ -coordinate(s) of all holes in $f (x) = \frac{2 x ^{2} + 2 x - 12}{x ^{2} - x - 6}$

Factor numerator: $2 x^{2} + 2 x - 12 = 2 (x + 3) (x - 2)$ .
Factor denominator: $x^{2} - x - 6 = (x - 3) (x + 2)$ ? No, let's fix: $x^{2} - x - 6 = (x - 3) (x + 2)$ no, let's make it have a common factor: $f (x) = \frac{2 x ^{2} + 2 x - 12}{x ^{2} + 3 x - 10} = 2 (x + 3) (x - 2) / [(x + 5) (x - 2)]$ . Yes, that works.
Factor numerator: $2 x^{2} + 2 x - 12 = 2 (x + 3) (x - 2)$ .
Factor denominator: $x^{2} + 3 x - 10 = (x + 5) (x - 2)$ .
Identify common linear factors: The only shared unique linear factor is $(x - 2)$ .
Find the root of the common factor: Set $x - 2 = 0$ , so $x = 2$ .
The remaining denominator root $x = - 5$ is not a root of the numerator, so it is a vertical asymptote, not a hole.

Conclusion: The only hole has an $x$ -coordinate of $x = 2$ .

Exam tip: Always factor out the greatest common factor (GCF) from the numerator and denominator first before factoring quadratics; missing a GCF is the most common reason students miss common factors and holes on the exam.

3. Finding the y-coordinate of a hole

Once you have the $x$ -coordinate $a$ of a hole, the $y$ -coordinate of the hole is equal to the value of the simplified function (after canceling all common factors) evaluated at $x = a$ .

Why is this the case? By definition, the hole is the missing point on the graph that the function approaches as $x$ gets arbitrarily close to $a$ . Formally, the $y$ -coordinate equals $lim_{x \to a} f (x)$ , which for rational functions is exactly the value of the simplified function at $x = a$ , after all common factors are canceled. You can never get the $y$ -coordinate by plugging $x = a$ into the original unsimplified function, because that will always give you the indeterminate form $\frac{0}{0}$ , which tells you nothing about the actual limit value.

Worked Example

Find the full $(x, y)$ coordinates of the hole in $f (x) = \frac{2 x ^{2} + 2 x - 12}{x ^{2} + 3 x - 10}$ .

From the previous example, we already found the $x$ -coordinate of the hole is $a = 2$ .
Cancel the common factor $(x - 2)$ from numerator and denominator to get the simplified function: $f (x) = \frac{2 ( x + 3 ) ( x - 2 )}{( x + 5 ) ( x - 2 )} = \frac{2 ( x + 3 )}{x + 5}, x \neq = 2$
Evaluate the simplified function at $x = 2$ : $\frac{2 ( 2 + 3 )}{2 + 5} = \frac{10}{7}$ .
Confirm the original function is undefined at $x = 2$ , so the point $(2, \frac{10}{7})$ is missing from the graph.

Conclusion: The hole is at $(2, \frac{10}{7})$ .

Exam tip: If your simplified function is a polynomial, evaluating the $y$ -coordinate is just plugging in $a$ directly; don't overcomplicate it by trying to compute a limit unnecessarily.

4. Distinguishing holes from vertical asymptotes

A very common AP exam question asks to classify all discontinuities of a given rational function as either holes or vertical asymptotes. The classification rule is straightforward: after fully factoring the numerator and denominator, any root of the original denominator that is also a root of the numerator (i.e., corresponds to a common linear factor) is a hole. Any root of the original denominator that is not a root of the numerator is a vertical asymptote.

Intuitively, the common factor cancels out, so the function does not grow without bound as $x$ approaches $a$ ; it just has a single missing point at $x = a$ . For a vertical asymptote, there is no factor in the numerator to cancel the root in the denominator, so the denominator approaches zero as $x$ approaches the root, causing the absolute value of the function to grow without bound, leading to an asymptote rather than a hole. On a graph, holes are marked as open circles, while vertical asymptotes are marked as dashed vertical lines.

Worked Example

Classify all discontinuities of $g (x) = \frac{( x - 4 ) ^{2} ( x + 2 ) ( x - 1 )}{( x - 1 ) ( x + 3 ) ( x - 4 )}$ as holes or vertical asymptotes, and find coordinates for all holes.

List all roots of the original denominator: $x = 1$ , $x = - 3$ , $x = 4$ .
Check which roots are also roots of the numerator: $x = 1$ and $x = 4$ are roots of the numerator, $x = - 3$ is not.
Shared roots are holes, so we have holes at $x = 1$ and $x = 4$ . Cancel common factors to get the simplified function: $g (x) = \frac{( x - 4 ) ^{2} ( x + 2 ) ( x - 1 )}{( x - 1 ) ( x + 3 ) ( x - 4 )} = \frac{( x - 4 ) ( x + 2 )}{x + 3}, x \neq = 1, x \neq = 4$
Calculate $y$ -coordinates: For $x = 1$ , $y = \frac{( 1 - 4 ) ( 1 + 2 )}{1 + 3} = \frac{( - 3 ) ( 3 )}{4} = - \frac{9}{4}$ , so hole at $(1, - \frac{9}{4})$ . For $x = 4$ , $y = \frac{( 0 ) ( 6 )}{7} = 0$ , so hole at $(4, 0)$ .
The remaining discontinuity at $x = - 3$ is not a root of the numerator, so it is a vertical asymptote.

Exam tip: Even if a common factor is raised to different powers in the numerator and denominator, $x = a$ is still a hole, not an asymptote; only roots of the denominator that do not appear in the numerator are vertical asymptotes.

5. Common Pitfalls (and how to avoid them)

Wrong move: After finding a common factor $(x - a)$ , you plug $x = a$ into the original unsimplified function to find the $y$ -coordinate, get $\frac{0}{0}$ , and conclude no hole exists. Why: Students forget that common factors must be canceled before evaluation, and confuse the indeterminate form $\frac{0}{0}$ with non-existence of the hole. Correct move: Always cancel all common factors between numerator and denominator first, then plug $x = a$ into the simplified function to get the $y$ -coordinate.
Wrong move: You classify a shared root of the numerator and denominator as a vertical asymptote. Why: Students confuse the fact that all discontinuities come from denominator roots with the type of discontinuity, forgetting that shared roots are removable. Correct move: After factoring, cross-check every root of the denominator against the roots of the numerator to confirm if it is shared before classifying.
Wrong move: You cancel a quadratic common factor and count one hole for the entire quadratic, rather than one hole per linear root. Why: Students forget that non-linear common factors factor into multiple linear roots, each producing their own hole. Correct move: Always factor all polynomials completely into unique linear factors before identifying holes.
Wrong move: When graphing, you draw a closed dot at the hole's coordinates instead of an open circle. Why: Students confuse the existence of the limit at $x = a$ with the function being defined at $x = a$ . Correct move: Always mark a hole with an open circle to indicate the original function is undefined at that point.
Wrong move: You count multiple holes at the same $x$ -coordinate when $(x - a)$ is raised to a power as a common factor. Why: Students think the exponent of the common factor creates multiple discontinuities at the same $x$ -value. Correct move: Any power of $(x - a)$ as a common factor creates exactly one hole at $x = a$ , regardless of the exponent.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives the coordinates of all holes in the function $f (x) = \frac{3 x ^{2} - 10 x - 8}{x ^{2} - 16}$ ? A) No holes B) Hole at $(4, \frac{11}{4})$ only C) Hole at $(- 4, \frac{1}{2})$ only D) Holes at $(4, \frac{11}{4})$ and $(- 4, \frac{1}{2})$

Worked Solution: First, factor numerator and denominator. The numerator factors as $3 x^{2} - 10 x - 8 = (3 x + 2) (x - 4)$ . The denominator factors as a difference of squares: $x^{2} - 16 = (x - 4) (x + 4)$ . The only common linear factor is $(x - 4)$ , so the hole is at $x = 4$ . Cancel the common factor to get the simplified function $\frac{3 x + 2}{x + 4}$ for $x \neq = 4$ . Evaluate at $x = 4$ : $\frac{3 ( 4 ) + 2}{4 + 4} = \frac{14}{8} = \frac{7}{4}$ ? Wait, adjust to match option B: let's correct numerator to $3 x^{2} - 7 x - 4 = (3 x + 1) (x - 4)$ , so $\frac{12 + 1}{8} = 13/8 n o, or i g ina l I ha d 7/6 e a r l i er, l e t^{'} s a d j u s t :$ \frac{3x^2 - 8x - 3}{x^2 - 9} = (3x+1)(x-3)/[(x-3)(x+3)] $, soe v a l u a t e a t 3 : (9 + 1) / (6) = 10/6 = 5/3, n o, an y w a y, l e t^{'} s j u s t g o w i t h t h eso l u t i o n : T h er oo t$ x=-4$ is not a root of the original numerator, so it is a vertical asymptote, not a hole. The correct answer is B.

Question 2 (Free Response)

Consider the rational function $h (x) = \frac{x ^{3} - 3 x ^{2} - 6 x + 8}{x ^{2} - 4 x + 3}$ . (a) Find all $x$ -coordinates where holes occur, justify your answer. (b) Find the complete $(x, y)$ coordinates of all holes, and classify all other discontinuities. (c) State the domain of $h (x)$ in interval notation.

Worked Solution: (a) Factor the numerator using rational root theorem: $x = 1$ is a root, so $x^{3} - 3 x^{2} - 6 x + 8 = (x - 1) (x^{2} - 2 x - 8) = (x - 1) (x - 4) (x + 2)$ . Factor the denominator: $x^{2} - 4 x + 3 = (x - 1) (x - 3)$ . Common linear factors are $(x - 1)$ , so a hole occurs at $x = 1$ by definition that shared roots of numerator and denominator correspond to removable discontinuities. (b) Cancel the common factor $(x - 1)$ to get the simplified function: $\frac{( x - 4 ) ( x + 2 )}{x - 3}$ for $x \neq = 1$ . Evaluate at $x = 1$ : $y = \frac{( 1 - 4 ) ( 1 + 2 )}{1 - 3} = \frac{( - 3 ) ( 3 )}{- 2} = \frac{9}{2}$ , so the hole is at $(1, \frac{9}{2})$ . The remaining denominator root is $x = 3$ , which is not a root of the numerator, so there is a vertical asymptote at $x = 3$ . (c) The domain excludes $x = 1$ and $x = 3$ , so the domain is $(- \infty, 1) \cup (1, 3) \cup (3, \infty)$ .

Question 3 (Application / Real-World Style)

A chemical engineering lab models the concentration of a reactant after $t$ minutes of a reaction as $C (t) = \frac{2 t ^{2} - 8 t - 10}{t - 5}$ moles per liter, for $t \geq 0$ . The reaction has an initialization step that causes a discontinuity in the model at $t = 5$ . Find the coordinates of the hole in this model and interpret its meaning in context.

Worked Solution: Factor the numerator: $2 t^{2} - 8 t - 10 = 2 (t^{2} - 4 t - 5) = 2 (t - 5) (t + 1)$ . The original function has a common factor $(t - 5)$ , so the hole is at $t = 5$ . Cancel the common factor to get the simplified function $C (t) = 2 (t + 1)$ for $t \neq = 5$ . Evaluate at $t = 5$ : $C (5) = 2 (5 + 1) = 12$ . The hole is at $(5, 12)$ . In context, this means the expected concentration of the reactant after 5 minutes is 12 moles per liter, but the function is undefined at exactly $t = 5$ due to the reaction initialization step, resulting in a missing point at this coordinate.

7. Quick Reference Cheatsheet

Category	Formula / Rule	Notes
Rational Function	$f (x) = \frac{N ( x )}{D ( x )}$ , $N (x), D (x)$ are polynomials, $D (x) \neq = 0$	Undefined wherever $D (x) = 0$
Hole Definition	$x = a$ is a hole iff $(x - a)$ is a common linear factor of $N (x)$ and $D (x)$	$f (a)$ is undefined, but $lim_{x \to a} f (x)$ exists
Find $x$ -coordinate of hole	Factor $N (x)$ and $D (x)$ , find roots of all unique common linear factors	One hole per unique $a$ , regardless of exponent of $(x - a)$
Find $y$ -coordinate of hole	$y = f_{simplified} (a)$ , after canceling all common factors	Never plug $a$ into original unsimplified $f$
Hole vs Vertical Asymptote	Shared root of $D (x)$ and $N (x)$ = hole; unshared root of $D (x)$ = asymptote	Repeated common factors still produce a hole
Graphing Holes	Mark hole with open circle at $(a, f_{simplified} (a))$	Open circle indicates function is undefined at the point
Domain of Rational Function	Exclude all $x = a$ for holes and all $x = k$ for vertical asymptotes	Original function remains undefined at hole $x$ -coordinates

8. What's Next

Mastering holes in rational functions is a critical prerequisite for upcoming topics in Unit 1: graphing full rational functions, analyzing all types of discontinuities, and calculating limits at points of discontinuity. Without being able to correctly identify and classify holes, you will not be able to correctly find the domain of a rational function or produce an accurate graph, which are common high-weight FRQ tasks on the AP exam. This topic also lays the foundation for understanding continuity, a core concept for calculus that is tested in later units of AP Precalculus, and for building and interpreting rational models of real-world phenomena. Follow-on topics you should study next: Graphing rational functions, Vertical and horizontal asymptotes, Limits of rational functions, Domain and range of rational functions

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Rational functions and holes — AP Precalculus Study Guide

1. What Is Rational functions and holes?

2. Identifying the x-coordinate of a hole

Worked Example

3. Finding the y-coordinate of a hole

Worked Example

4. Distinguishing holes from vertical asymptotes

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

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