Rational functions and vertical asymptotes

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

1. Classifying Discontinuities: Removable vs Non-Removable★★☆☆☆⏱ 4 min

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 $f (x) = \frac{x ^{2} - 4 x + 3}{x ^{2} - 2 x - 3}$ .

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.

2. The Limit Rule for Locating Vertical Asymptotes★★★☆☆⏱ 4 min

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.

📐 Worked Example

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

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.

3. Analyzing Function Behavior Near Vertical Asymptotes★★★☆☆⏱ 4 min

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.

4. AP-Style Practice Worked Examples★★★★☆⏱ 3 min

✓ Quick check

Test your understanding with this multiple-choice question:

Which of the following gives all vertical asymptotes of the function $f (x) = \frac{x ^{2} - 9}{x ^{3} - 4 x ^{2} + 3 x}$ ?
- $x = 0$ , $x = 1$ , $x = 3$
- $x = 0$ , $x = 1$
- $x = - 3$ , $x = 3$
- $x = 1$ only
Reveal answer
1 —
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 $f (x) = \frac{3 x ^{2} - 12}{x ^{2} + 2 x - 8}$ . (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 $lim_{x \to - 4^{+}} f (x) = - \infty$ .

5. Common Pitfalls

Wrong move:

Immediately conclude $x = a$ is a vertical asymptote after finding $D (a) = 0$ , without checking for common numerator factors.

Why:

Students rush after finding a root of the denominator and forget to check for removable discontinuities.

Correct move:

Always factor both numerator and denominator and cancel all common factors before identifying vertical asymptotes.

Wrong move:

Write a vertical asymptote as $y = a$ instead of $x = a$ .

Why:

Confuses vertical asymptotes (vertical lines, constant x-value) with horizontal asymptotes (horizontal lines, constant y-value).

Correct move:

Remember vertical lines have constant x-values, so all vertical asymptote equations are of the form $x = a$ .

Wrong move:

Claim a function can cross a vertical asymptote, incorrectly extending the rule for horizontal asymptotes.

Why:

Students mix up properties of horizontal and vertical asymptotes.

Correct move:

Vertical asymptotes are outside the domain of the function, so the function can never have a point on the asymptote or cross it.

Wrong move:

Assume the sign of the function changes across a vertical asymptote at a repeated root.

Why:

Students assume all roots change sign, forgetting that even powers are always positive.

Correct move:

Check the exponent of the $(x - a)$ factor in the denominator—if even, the sign does not change across $x = a$ .

Wrong move:

Leave extraneous common factors in the denominator when locating asymptotes, leading to extra false vertical asymptotes.

Why:

Students do not cancel all common factors after factoring.

Correct move:

Cancel every common linear factor, leaving only non-removable roots in the denominator before identifying asymptotes.

6. Quick Reference 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)

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
MCQ: Find all vertical asymptotes
2023 · AP Precalculus
FRQ: Classify discontinuities

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.