# Rational functions and holes

> AP Precalculus · Unit 1: Polynomial and Rational Functions
> Source: https://www.owlsprep.com/study/ap-precalculus-u1-rational-functions-and-holes/

This guide covers identifying removable discontinuities (holes) in rational functions, factoring for common linear factors, calculating hole coordinates, distinguishing holes from vertical asymptotes, and exam-ready graphing rules for AP Precalculus.

**Prerequisites:** Polynomial factoring techniques; Evaluating limits of rational functions; Basic rational function graphing rules

## Learning objectives

- Identify removable discontinuities (holes) in rational functions
- Calculate full (x,y) coordinates of holes
- Distinguish holes from vertical asymptotes
- Classify discontinuities for AP Precalculus exam questions

## Core Concepts: What Is a Hole?

**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.

*Calculator:* allowed

## Finding the x-coordinate of a Hole

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 $x=a$.

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

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

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

*Calculator:* allowed

## Finding the y-coordinate of a Hole

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 $\frac{0}{0}$.

> **tip**
>
> If your simplified function is a polynomial, evaluating the $y$-coordinate is just plugging in $a$ directly; you do not need to compute a limit unnecessarily.

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

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:
3. $$f(x) = \frac{2(x+3)\cancel{(x-2)}}{(x+5)\cancel{(x-2)}} = \frac{2(x+3)}{x+5}, \quad x \neq 2$$
4. Evaluate the simplified function at $x=2$:
5. $$\frac{2(2+3)}{2+5} = \frac{10}{7}$$
6. Conclusion: The hole is at:
7. $$\left(2, \frac{10}{7}\right)$$

*Calculator:* allowed

## Distinguishing Holes from Vertical Asymptotes

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.

> **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.

**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. $$g(x) = \frac{(x-4)^2(x+2)\cancel{(x-1)}}{\cancel{(x-1)}(x+3)\cancel{(x-4)}} = \frac{(x-4)(x+2)}{x+3}, \quad x \neq 1, x \neq 4$$
5. Calculate $y$-coordinates: For $x=1$:
6. $$y = \frac{(1-4)(1+2)}{1+3} = -\frac{9}{4}$$
7. For $x=4$:
8. $$y = \frac{(0)(6)}{7} = 0$$
9. Conclusion: Holes are at $\left(1, -\frac{9}{4}\right)$ and $(4, 0)$. The discontinuity at $x=-3$ is a vertical asymptote.

**Check your understanding**

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

1. Which of the following gives the coordinates of all holes in the function $f(x) = \frac{3x^2 - 10x - 8}{x^2 - 16}$?

   - A) No holes
   - B) Hole at $(4, \frac{7}{4})$ only
   - C) Hole at $(-4, \frac{1}{2})$ only
   - D) Holes at $(4, \frac{7}{4})$ and $(-4, \frac{1}{2})$

   *Why:* 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.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** 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 it fails: 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: Always cancel all common factors between numerator and denominator first, then plug $x=a$ into the simplified function to get the $y$-coordinate.
- **Wrong:** You classify a shared root of the numerator and denominator as a vertical asymptote.
  - Why it fails: Students confuse the fact that all discontinuities come from denominator roots with the type of discontinuity, forgetting that shared roots are removable.
  - Correct: After factoring, cross-check every root of the denominator against the roots of the numerator to confirm if it is shared before classifying.
- **Wrong:** You cancel a quadratic common factor and count one hole for the entire quadratic, rather than one hole per linear root.
  - Why it fails: Students forget that non-linear common factors factor into multiple linear roots, each producing their own hole.
  - Correct: Always factor all polynomials completely into unique linear factors before identifying holes.
- **Wrong:** When graphing, you draw a closed dot at the hole's coordinates instead of an open circle.
  - Why it fails: Students confuse the existence of the limit at $x=a$ with the function being defined at $x=a$.
  - Correct: Always mark a hole with an open circle to indicate the original function is undefined at that point.
- **Wrong:** You count multiple holes at the same $x$-coordinate when $(x-a)$ is raised to a power as a common factor.
  - Why it fails: Students think the exponent of the common factor creates multiple discontinuities at the same $x$-value.
  - Correct: Any power of $(x-a)$ as a common factor creates exactly one hole at $x=a$, regardless of the exponent.

## 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 | $y = f_{\text{simplified}}(a)$, 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 $(a, f_{\text{simplified}}(a))$ | 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 |

## 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.

- [Equivalent representations of polynomial and rational expressions](https://www.owlsprep.com/study/ap-precalculus-u1-equivalent-representations-of-polynomial-and/)
- [Transformations of functions](https://www.owlsprep.com/study/ap-precalculus-u1-transformations-of-functions/)
- [Function model selection and assumption articulation](https://www.owlsprep.com/study/ap-precalculus-u1-function-model-selection-and-assumption/)

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