Rational functions and holes

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

1. Core Concepts: What Is a Hole?★★☆☆☆⏱ 3 min

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📘 Definition

Hole (Removable Discontinuity)

A hole occurs at an $x$ -value $a$ where both the numerator and denominator of a rational function equal zero, meaning there is a common linear factor $(x - a)$ that cancels out. Unlike vertical asymptotes, holes are single missing points that can be 'filled in' to remove the discontinuity.

Example:

The function $f (x) = \frac{2 ( x + 3 ) ( x - 2 )}{( x + 5 ) ( x - 2 )}$ has a hole at $x = 2$ .

A rational function is defined as the ratio of two polynomials $f (x) = \frac{N ( x )}{D ( x )}$ , where $N (x)$ is the numerator polynomial and $D (x)$ is a non-zero denominator polynomial. This topic is tested in both multiple-choice and free-response sections of the AP Precalculus exam, and falls within Unit 1, which makes up 27–31% of your total exam score.

2. Finding the x-coordinate of a Hole★★☆☆☆⏱ 4 min

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To find the $x$ -coordinate(s) of any hole, first fully factor both the numerator and denominator into linear factors. Any unique common linear factor of the form $(x - a)$ gives a hole at $x = a$ . Roots of the denominator that are not roots of the numerator correspond to vertical asymptotes, not holes. Even if $(x - a)$ is raised to a power in both numerator and denominator, it only produces one hole at $x = a$ .

📐 Worked Example

Find the $x$ -coordinate(s) of all holes in the function $f (x) = \frac{2 x ^{2} + 2 x - 12}{x ^{2} + 3 x - 10}$

1
Factor the numerator completely starting with the GCF:
2
$2x^2 + 2x - 12 = 2(x^2 + x - 6) = 2(x+3)(x-2)$
3
Factor the denominator completely:
4
$x^2 + 3x - 10 = (x + 5)(x - 2)$
5
Identify unique common linear factors: the only shared factor is $(x - 2)$ .
6
The remaining denominator root $x = - 5$ is not a root of the numerator, so it is a vertical asymptote.
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Conclusion: The only hole has an $x$ -coordinate of:
8
$x = 2$

3. Finding the y-coordinate of a Hole★★☆☆☆⏱ 3 min

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Once you have the $x$ -coordinate $a$ of a hole, the $y$ -coordinate equals the value of the simplified function (after canceling all common factors) evaluated at $x = a$ . This is equal to $lim_{x \to a} f (x)$ , the value the function approaches as $x$ nears $a$ . Never plug $a$ into the original unsimplified function, as this will always give the indeterminate form $\frac{0}{0}$ .

📐 Worked Example

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

1
We already know the $x$ -coordinate of the hole is $a = 2$ from the previous example.
2
Cancel the common factor $(x - 2)$ from numerator and denominator to get the simplified function:
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$f(x) = \frac{2(x+3)\cancel{(x-2)}}{(x+5)\cancel{(x-2)}} = \frac{2(x+3)}{x+5}, \quad x \neq 2$
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Evaluate the simplified function at $x = 2$ :
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$\frac{2(2+3)}{2+5} = \frac{10}{7}$
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Conclusion: The hole is at:
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$\left(2, \frac{10}{7}\right)$

4. Distinguishing Holes from Vertical Asymptotes★★★☆☆⏱ 5 min

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A very common AP exam question asks to classify all discontinuities of a given rational function. 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. 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.

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

✓ Quick check

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

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 $(- 4, \frac{1}{2})$
Reveal answer
B —
The only common linear factor between numerator and denominator is $(x - 4)$ , so only $x = 4$ is a hole. $x = - 4$ is an unshared root of the denominator, so it is a vertical asymptote. Only option B is correct.

5. Common Pitfalls

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. Quick Reference Cheatsheet

Category	Rule / Formula	Notes
Rational Function	$f (x) = \frac{N ( x )}{D ( x )}$ , $N (x), D (x)$ 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

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
Find hole coordinates multiple choice
2023 · AP Precalculus
Classify discontinuities free response

What's Next

Mastering holes in rational functions is a critical prerequisite for upcoming topics in Unit 1, including 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 free-response 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.