# Rational functions and vertical asymptotes

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

This subtopic covers rational function definitions, classification of removable (hole) and non-removable (vertical asymptote) discontinuities, rules for locating vertical asymptotes, and analyzing function behavior near asymptotes for AP Precalculus.

**Prerequisites:** Evaluating one-sided limits of algebraic functions; Factoring polynomials and finding common roots; Basic rules for function continuity

## Learning objectives

- Define rational functions and identify their domain
- Classify discontinuities as removable (holes) or non-removable (vertical asymptotes)
- Locate vertical asymptotes of rational functions using factoring and limit rules
- Analyze function behavior near vertical asymptotes using sign testing

## Classifying Discontinuities: Removable vs Non-Removable

For any rational function $f(x) = \frac{N(x)}{D(x)}$, all points where $D(x) = 0$ are discontinuities, since the function is undefined there. To classify a discontinuity at $x=a$, first factor both numerator and denominator completely.

- If $(x-a)$ is a common factor of both $N(x)$ and $D(x)$, the discontinuity is **removable (a hole)**: the two-sided limit exists and is finite, but the function is undefined at $x=a$.
- If $(x-a)$ is only a factor of $D(x)$ and does not cancel after removing all common factors, the discontinuity is **non-removable**, corresponding to a vertical asymptote at $x=a$.

**Worked example:** Classify all discontinuities of $f(x) = \frac{x^2 - 4x + 3}{x^2 - 2x - 3}$.

1. Factor the numerator and denominator completely:
2. $$x^2 - 4x + 3 = (x-1)(x-3) \quad x^2 - 2x - 3 = (x-3)(x+1)$$
3. Rewrite the function for all $x \neq 3$:
4. $$f(x) = \frac{(x-1)(x-3)}{(x-3)(x+1)}$$
5. Original denominator equals zero at $x=3$ and $x=-1$, so these are the only discontinuities.
6. The common factor $(x-3)$ cancels out, so $x=3$ is a removable discontinuity (hole). After canceling, the reduced function is $\frac{x-1}{x+1}$. $x=-1$ still makes the denominator zero with no remaining common factors, so $x=-1$ is a non-removable discontinuity corresponding to a vertical asymptote.

> **Exam tip:** Always factor out all common factors completely before classifying discontinuities; partial factoring can lead to misidentifying holes as vertical asymptotes.

## The Limit Rule for Locating Vertical Asymptotes

The formal definition of a vertical asymptote relies on one-sided limits: the line $x=a$ is a vertical asymptote if at least one of the following holds:

$$\lim_{x \to a^-} f(x) = \pm\infty \quad \text{or} \quad \lim_{x \to a^+} f(x) = \pm\infty$$

For rational functions, this definition simplifies to a straightforward rule that works for all cases: if $f(x) = \frac{N(x)}{D(x)}$ is written in lowest terms (no common factors between numerator and denominator), then every real zero of $D(x)$ is a vertical asymptote. The lowest terms condition is non-negotiable.

> **info**
>
> Vertical asymptotes are always vertical lines of the form $x=a$, never $y=a$ (which describes horizontal asymptotes).

**Worked example:** Find all vertical asymptotes of $g(x) = \frac{2x^2 + 5x - 3}{x^2 - 4x + 4}$.

1. Factor the numerator and denominator completely:
2. $$2x^2 + 5x - 3 = (2x - 1)(x + 3) \quad x^2 - 4x + 4 = (x-2)^2$$
3. Check for common factors: there are no shared linear factors, so the function is already in lowest terms. Find all real roots of the denominator: $(x-2)^2 = 0$ gives a repeated root at $x=2$.
4. Confirm the numerator is non-zero at $x=2$:
5. $$N(2) = (2(2)-1)(2+3) = 15 \neq 0$$
6. The limit condition for a vertical asymptote is satisfied, so the only vertical asymptote is the line $x=2$.

> **Exam tip:** On AP Precalculus free-response questions, you must write vertical asymptotes as full equations of lines (e.g., $x=2$, not just $2$) to earn full credit.

## Analyzing Function Behavior Near Vertical Asymptotes

Once you have identified a vertical asymptote at $x=a$, you will often need to determine whether the function approaches $+\infty$, $-\infty$, or opposite infinities on either side of $a$ for graphing or limit questions. The fastest way to do this is to test the sign of the reduced rational function on each side of $x=a$.

- Repeated root (even exponent on $(x-a)$ in denominator): the term is always positive for $x \neq a$, so the function approaches the same infinity on both sides.
- Distinct root (odd exponent on $(x-a)$): the sign changes across the asymptote, so the function approaches opposite infinities on either side.

**Worked example:** For $h(x) = \frac{x + 2}{(x - 1)(x + 4)}$, identify the vertical asymptotes and describe the behavior of $h(x)$ near each asymptote using one-sided limits.

1. Confirm there are no common factors between numerator and denominator, so the function is in lowest terms. Denominator equals zero at $x=1$ and $x=-4$, so these are the vertical asymptotes.
2. Analyze behavior at $x=1$: Test $x=0.9$ (left of 1): numerator $= 2.9 > 0$, $(0.9 - 1) = -0.1 < 0$, $(0.9 + 4) = 4.9 > 0$. Overall sign is negative, so $\lim_{x \to 1^-} h(x) = -\infty$. Test $x=1.1$ (right of 1): overall sign is positive, so $\lim_{x \to 1^+} h(x) = +\infty$.
3. Analyze behavior at $x=-4$: Test $x=-4.1$ (left of -4): overall sign is negative, so $\lim_{x \to -4^-} h(x) = -\infty$. Test $x=-3.9$ (right of -4): overall sign is positive, so $\lim_{x \to -4^+} h(x) = +\infty$.

> **Exam tip:** When calculating the sign of the function near an asymptote, any even-powered factor can be ignored entirely because it is always positive, cutting down on calculation time.

## AP-Style Practice Worked Examples

**Check your understanding**

Test your understanding with this multiple-choice question:

1. Which of the following gives all vertical asymptotes of the function $f(x) = \frac{x^2 - 9}{x^3 - 4x^2 + 3x}$?

   - $x=0$, $x=1$, $x=3$
   - $x=0$, $x=1$
   - $x=-3$, $x=3$
   - $x=1$ only

   *Answer:* $x=0$, $x=1$

   *Why:* Correct. Factor numerator: $x^2 - 9 = (x-3)(x+3)$, denominator: $x(x-1)(x-3)$. Cancel common $(x-3)$, leaving reduced denominator $x(x-1)$. Only $x=0$ and $x=1$ are vertical asymptotes; $x=3$ is a removable hole.

**Worked example:** Let $f(x) = \frac{3x^2 - 12}{x^2 + 2x - 8}$. (a) Find all discontinuities, classify each as removable or non-removable. (b) Write all vertical asymptote equations. (c) Describe one-sided behavior near each vertical asymptote.

1. Part (a): Factor numerator and denominator:
2. $$3x^2 - 12 = 3(x-2)(x+2) \quad x^2 + 2x - 8 = (x+4)(x-2)$$
3. Discontinuities at $x=2$ and $x=-4$. $(x-2)$ is common, so $x=2$ is removable. $x=-4$ does not cancel, so it is non-removable.
4. Part (b): The only vertical asymptote is $x=-4$.
5. Part (c): Reduced function is $\frac{3(x+2)}{x+4}$ for $x \neq 2$. Left of $x=-4$: ratio is positive, so $\lim_{x \to -4^-} f(x) = +\infty$. Right of $x=-4$: ratio is negative, so $\lim_{x \to -4^+} f(x) = -\infty$.

## Common pitfalls

- **Wrong:** Immediately conclude $x=a$ is a vertical asymptote after finding $D(a)=0$, without checking for common numerator factors.
  - Why it fails: Students rush after finding a root of the denominator and forget to check for removable discontinuities.
  - Correct: Always factor both numerator and denominator and cancel all common factors before identifying vertical asymptotes.
- **Wrong:** Write a vertical asymptote as $y=a$ instead of $x=a$.
  - Why it fails: Confuses vertical asymptotes (vertical lines, constant x-value) with horizontal asymptotes (horizontal lines, constant y-value).
  - Correct: Remember vertical lines have constant x-values, so all vertical asymptote equations are of the form $x=a$.
- **Wrong:** Claim a function can cross a vertical asymptote, incorrectly extending the rule for horizontal asymptotes.
  - Why it fails: Students mix up properties of horizontal and vertical asymptotes.
  - Correct: Vertical asymptotes are outside the domain of the function, so the function can never have a point on the asymptote or cross it.
- **Wrong:** Assume the sign of the function changes across a vertical asymptote at a repeated root.
  - Why it fails: Students assume all roots change sign, forgetting that even powers are always positive.
  - Correct: Check the exponent of the $(x-a)$ factor in the denominator—if even, the sign does not change across $x=a$.
- **Wrong:** Leave extraneous common factors in the denominator when locating asymptotes, leading to extra false vertical asymptotes.
  - Why it fails: Students do not cancel all common factors after factoring.
  - Correct: Cancel every common linear factor, leaving only non-removable roots in the denominator before identifying asymptotes.

## Cheatsheet

| Category | Rule / Formula | Notes |
| --- | --- | --- |
| Rational function definition | $f(x) = \frac{N(x)}{D(x)}$, $N(x), D(x)$ polynomials, $D(x)$ non-zero | Domain is all $x$ with $D(x) \neq 0$ |
| Removable discontinuity (hole) | $D(a)=0$, $(x-a)$ common factor of $N(x), D(x)$ | $\lim_{x \to a} f(x)$ exists, no vertical asymptote |
| Non-removable discontinuity (vertical asymptote) | $D(a)=0$, $(x-a)$ not a factor of reduced $N(x)$ | At least one one-sided limit is $\pm\infty$ |
| Vertical asymptote rule (lowest terms) | $x=a$ is vertical asymptote if $D(a)=0$ and $N(a) \neq 0$ | Only applies after canceling all common factors |
| Formal limit definition | $x=a$ is vertical asymptote if $\lim_{x \to a^-} f(x) = \pm\infty$ or $\lim_{x \to a^+} f(x) = \pm\infty$ | Only one side needs to be infinite |
| Sign behavior: repeated root | $(x-a)^n$, $n$ even: always positive for $x \neq a$ | No sign change, $f(x)$ approaches same infinity both sides |
| Sign behavior: distinct root | $(x-a)^n$, $n$ odd: sign changes across $x=a$ | $f(x)$ approaches opposite infinities on each side |
| Vertical asymptote equation form | $x=a$ (constant $a$) | Never written as $y=a$ (that is a horizontal asymptote) |

## What's next

Mastering vertical asymptotes is a critical prerequisite for upcoming topics in AP Precalculus Unit 1: horizontal and slant asymptotes, and full graphing of rational functions. Without correctly identifying vertical asymptotes, you cannot correctly sketch rational function graphs or analyze their full domain behavior, which is a common free-response task on the AP exam. This topic also lays the foundation for analyzing discontinuities in other types of non-polynomial functions later in the course, including logarithmic and reciprocal trigonometric functions, where vertical asymptotes also appear. The distinction between removable and non-removable discontinuities also connects directly to the study of continuity and limits in calculus, which AP Precalculus prepares you for.

- [Rational functions and holes](https://www.owlsprep.com/study/ap-precalculus-u1-rational-functions-and-holes/)
- [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/)

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