Rational functions and vertical asymptotes — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Definition of rational functions, distinguishing removable and non-removable discontinuities, the limit rule for vertical asymptotes, step-by-step techniques for locating vertical asymptotes, and analyzing function behavior near vertical asymptotes.

You should already know: How to evaluate one-sided limits of algebraic functions. How to factor polynomials and find common roots of numerator and denominator. Basic rules for function continuity.

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 vertical asymptotes?

A rational function is formally defined as the ratio of two polynomials, written in the standard form $f(x)=\frac{N(x)}{D(x)}$, where $N(x)$ (numerator) and $D(x)$ (denominator) are polynomials with real coefficients, and $D(x)$ is not the zero polynomial. This topic is part of AP Precalculus Unit 1: Polynomial and Rational Functions, accounting for approximately 7-10% of the total AP exam weight, and questions about vertical asymptotes appear in both multiple-choice (MCQ) and free-response (FRQ) sections. Vertical asymptotes are vertical lines $x=a$ where a rational function approaches positive or negative infinity as $x$ approaches $a$ from one or both sides, representing non-removable discontinuities of the function. Unlike holes (removable discontinuities), vertical asymptotes cannot be eliminated by redefining the function at a single point, and they divide the domain of a rational function into separate continuous intervals. Mastery of vertical asymptotes is required for graphing rational functions, analyzing limit behavior, and solving applied problems in chemistry, economics, and physics that involve proportional relationships.

2. Classifying Discontinuities: Removable vs Non-Removable

For any rational function $f(x)=\frac{N(x)}{D(x)}$, the domain is all real numbers $x$ such that $D(x)\ne 0$. Any $x=a$ that makes $D(x)=0$ is automatically a point of discontinuity, since the function is undefined at that input. To classify the discontinuity, we first factor both $N(x)$ and $D(x)$ completely into linear factors over the reals. If $(x-a)$ is a common factor of both $N(x)$ and $D(x)$, we can cancel this common factor from the numerator and denominator, meaning the discontinuity at $x=a$ is removable (called a hole). A removable discontinuity means the two-sided limit of $f(x)$ as $x\to a$ exists and is finite, even though $f(a)$ itself is undefined. If $(x-a)$ is a factor of $D(x)$ but not a factor of the reduced numerator (after canceling all common factors), then the discontinuity at $x=a$ is non-removable, and this corresponds to a vertical asymptote at $x=a$. At a non-removable discontinuity, at least one one-sided limit of $f(x)$ is infinite.

Worked Example

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

1. Factor the numerator and denominator completely: $x^2-4x+3=(x-1)(x-3)$, and $x^2-2x-3=(x-3)(x+1)$. We can rewrite the function as $f(x)=\frac{(x-1)(x-3)}{(x-3)(x+1)}$ for all $x\ne 3$.
2. Identify all points where the original denominator equals zero: $x=3$ and $x=-1$, so these are the only discontinuities.
3. Check for common factors: $(x-3)$ is shared by both numerator and denominator, so it cancels out. This means $x=3$ is a removable discontinuity (a hole).
4. After canceling, the reduced function is $\frac{x-1}{x+1}$. The input $x=-1$ still makes the denominator zero, and there is no remaining common factor to cancel, so $x=-1$ is a non-removable discontinuity corresponding to a vertical asymptote.

Exam tip: Always factor out all common factors, including constants, before classifying discontinuities—partial factoring can leave common factors uncanceled, leading you to misidentify a hole as a vertical asymptote.

3. 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 of $f(x)$ if at least one of the following holds: $$\underset{x\to a^{-}}{\lim }f(x)=\pm \infty \text{or}\underset{x\to a^{+}}{\lim }f(x)=\pm \infty$$ For rational functions, this definition simplifies to a straightforward test that works for all cases: if $f(x)=\frac{N(x)}{D(x)}$ is a rational function written in lowest terms (no common factors between numerator and denominator), then $f(x)$ has a vertical asymptote at every real zero of $D(x)$. This rule holds because if $D(a)=0$ and $N(a)\ne 0$, the numerator approaches a non-zero finite value while the denominator approaches 0, so the ratio of the two approaches $\pm \infty$, satisfying the formal limit definition. The "lowest terms" condition is non-negotiable: if a common factor $(x-a)$ exists, the discontinuity at $x=a$ is removable, and there is no vertical asymptote there. It is also important to remember that vertical asymptotes are always vertical lines of the form $x=a$, never $y=a$ (which describes horizontal asymptotes).

Worked Example

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

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

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

4. 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 $x=a$. The sign of a ratio is determined by the product of the signs of the numerator and denominator: positive divided by positive is positive, positive divided by negative is negative, and so on. A positive value for the function near $a$ means the function approaches $+\infty$ as $x\to a$, while a negative value means it approaches $-\infty$. For repeated roots (even exponents on the $(x-a)$ factor in the denominator), the squared term is always positive for $x\ne a$, so it does not change the sign of the function across the asymptote, meaning the function approaches the same infinity on both sides. For distinct roots (odd exponents), the sign changes across the asymptote, so the function approaches opposite infinities on either side.

Worked Example

Problem: 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. Solution:

1. Factor the numerator and denominator and confirm no common factors: numerator is $(x+2)$, denominator is $(x-1)(x+4)$, so the function is in lowest terms. The denominator equals zero at $x=1$ and $x=-4$, so vertical asymptotes are at $x=1$ and $x=-4$.
2. Analyze behavior at $x=1$: Test $x=0.9$ (left of 1): numerator $0.9+2=2.9>0$, $(0.9-1)=-0.1<0$, $(0.9+4)=4.9>0$, so overall sign is $\frac{(+)}{(-)(+)}=-$. Thus ${\lim }_{x\to 1^{-}}h(x)=-\infty$. Test $x=1.1$ (right of 1): all terms except $(x-1)$ stay positive, so overall sign is positive, giving ${\lim }_{x\to 1^{+}}h(x)=+\infty$.
3. Analyze behavior at $x=-4$: Test $x=-4.1$ (left of -4): numerator $-4.1+2=-2.1<0$, $(-4.1-1)=-5.1<0$, $(-4.1+4)=-0.1<0$, so overall sign is $\frac{(-)}{(-)(-)}=-$, giving ${\lim }_{x\to -4^{-}}h(x)=-\infty$. Test $x=-3.9$ (right of -4): denominator product becomes positive, numerator stays negative, so overall sign is positive, giving ${\lim }_{x\to -4^{+}}h(x)=+\infty$.

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.

5. Common Pitfalls (and how to avoid them)

• Wrong move: After finding that $D(a)=0$, immediately conclude $x=a$ is a vertical asymptote without checking for a common factor in the numerator. 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: Writes a vertical asymptote as $y=2$ instead of $x=2$. Why: Confuses vertical asymptotes (vertical lines, constant x-value) with horizontal asymptotes (horizontal lines, constant y-value). Correct move: Remember vertical lines have a constant x-value, so all vertical asymptote equations are of the form $x=a$.
• Wrong move: Claims a function can cross a vertical asymptote, extending the rule for horizontal asymptotes incorrectly. Why: Students mix up properties of horizontal and vertical asymptotes. Correct move: Know that 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: Assumes 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: Leaves 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. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives all vertical asymptotes of the function $f(x)=\frac{x^2-9}{x^3-4x^2+3x}$? A) $x=0$, $x=1$, $x=3$ B) $x=0$, $x=1$ C) $x=-3$, $x=3$ D) $x=1$ only

Worked Solution: First, factor both numerator and denominator completely. The numerator is a difference of squares: $x^2-9=(x-3)(x+3)$. The denominator factors to $x(x^2-4x+3)=x(x-1)(x-3)$. Next, cancel the common factor of $(x-3)$ from numerator and denominator, giving the reduced function $\frac{x+3}{x(x-1)}$ for $x\ne 3$. The only roots of the reduced denominator are $x=0$ and $x=1$, neither of which are roots of the reduced numerator. $x=3$ is a removable discontinuity (hole), not a vertical asymptote. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{3x^2-12}{x^2+2x-8}$. (a) Find all discontinuities of $f(x)$, and classify each as removable or non-removable. (b) Write the equations for all vertical asymptotes of $f(x)$. (c) Evaluate the one-sided limits of $f(x)$ as $x$ approaches each vertical asymptote from the left and right, to describe the function's behavior near the asymptotes.

Worked Solution: (a) Factor the numerator and denominator: $3x^2-12=3(x^2-4)=3(x-2)(x+2)$ $x^2+2x-8=(x+4)(x-2)$ Discontinuities occur where the original denominator is zero: $x=2$ and $x=-4$. $(x-2)$ is a common factor, so $x=2$ is removable. $x=-4$ is a root of the denominator that does not cancel with the reduced numerator, so $x=-4$ is non-removable. (b) Vertical asymptotes exist only at non-removable discontinuities where the denominator is zero and the reduced numerator is non-zero. The only vertical asymptote is $\boxed{x=-4}$. (c) The reduced function is $\frac{3(x+2)}{x+4}$ for $x\ne 2$. Testing left of $x=-4$ at $x=-4.1$: numerator is $3(-4.1+2)=-6.3<0$, denominator is $-4.1+4=-0.1<0$, so the ratio is positive, giving ${\lim }_{x\to -4^{-}}f(x)=+\infty$. Testing right of $x=-4$ at $x=-3.9$: numerator is $3(-3.9+2)=-5.7<0$, denominator is $-3.9+4=0.1>0$, so the ratio is negative, giving ${\lim }_{x\to -4^{+}}f(x)=-\infty$.

Question 3 (Application / Real-World Style)

A population biologist models the growth rate of a bacteria colony $r(N)$ (in cells per hour) as a function of the current population size $N$ (in thousands of cells), given by $r(N)=\frac{2N+10}{N^2-9N+18}$ for $N>0$. Find all vertical asymptotes of $r(N)$ in the domain $N>0$, and explain what the asymptote implies about the model's behavior.

Worked Solution: First, factor the denominator: $N^2-9N+18=(N-3)(N-6)$. There are no common factors between the numerator $(2N+10)$ and denominator. The positive roots of the denominator are $N=3$ and $N=6$, so these are the vertical asymptotes. Testing the behavior for $0<N<3$: both $(N-3)$ and $(N-6)$ are negative, numerator is positive, so $r(N)=\frac{(+)}{(-)(-)}=+$, meaning ${\lim }_{N\to 3^{-}}r(N)=+\infty$. In context, this model predicts the growth rate of the colony becomes unbounded as the population approaches 3000 cells, which means the model only applies to population sizes outside the interval $(3,6)$.

7. Quick Reference Cheatsheet

8. What's Next

Mastering vertical asymptotes is a critical prerequisite for the next topics in Unit 1 of AP Precalculus: horizontal and slant asymptotes, and full graphing of rational functions. Without correctly identifying vertical asymptotes, you cannot correctly sketch the graph of a rational function or analyze its behavior over its entire domain, which is a common FRQ 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 functions 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. Next topics to study: Horizontal and slant asymptotes Graphing rational functions Continuity of algebraic functions