Rational Functions

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Textbook Reference

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Try This First

Consider the function $f(x) = \dfrac{x-1}{x-2}$.

Fill in the table below -- predict first, then compute:

Before computing: what do you expect to happen at $x = 2$? What do you expect to happen as $x$ grows very large?

Keep your predictions. Two behaviors drive what you observe: the output near $x = 2$ and the output as $x$ grows large.

Key idea

Division by zero is the central feature of rational functions. At inputs where the denominator is zero, the function has no output -- the domain has a gap. But the interesting question is not just “where is the denominator zero?” It is “what does the function do near those gaps?” The output can either shoot off toward infinity (a vertical asymptote) or quietly cancel to a finite value (a removable hole). These are structurally different behaviors, and telling them apart requires factoring.

Similarly, the long-run behavior of a rational function -- what happens as $x$ grows large -- is controlled by the degrees of the numerator and denominator. Both behaviors come from the same source: where the denominator is small, and what dominates when $x$ is large.

Prerequisite Check

Before this lesson, make sure you can do all of the following:

• Factor polynomials: $x^2 - 4 = (x-2)(x+2)$, $x^2 - x - 6 = (x-3)(x+2)$
• Simplify rational expressions: $\dfrac{x^2 - 4}{x - 2} = x + 2$ when $x \neq 2$
• Identify the degree of a polynomial
• Describe what happens when you divide by a number very close to zero
• Describe end behavior of a polynomial from its leading term

If any of these is uncertain, review Polynomial Functions and Factoring Techniques first.

Quick Reference

Definition. A rational function has the form $$ f(x) = \frac{p(x)}{q(x)} $$ where $p$ and $q$ are polynomials and $q$ is not the zero polynomial.

Horizontal asymptote rules (let $n = \deg p$, $m = \deg q$):

Key Concepts

1. Domain and Where the Denominator Is Zero

The domain of $f(x) = p(x)/q(x)$ is the set of all real numbers where $q(x) \neq 0$. Find the domain by setting $q(x) = 0$ and excluding those values.

Example. Find the domain of $f(x) = \dfrac{3x + 1}{x^2 - 5x + 6}$.

Factor the denominator: $x^2 - 5x + 6 = (x-2)(x-3)$.

Denominator is zero at $x = 2$ and $x = 3$.

Domain: all real numbers except $x = 2$ and $x = 3$.

In interval notation: $(-\infty, 2) \cup (2, 3) \cup (3, +\infty)$.

Two representations, one idea.

Algebraic: $q(x) = (x-2)(x-3) = 0$ at $x = 2$ and $x = 3$.

Graphical: The graph of $f$ has no points with $x$-coordinate $2$ or $x$-coordinate $3$. There are gaps in the curve at those $x$-values.

Translation prompt: Given only the graph with gaps (vertical dashed lines at $x = 2$ and $x = 3$), write the factored denominator. Then reverse: from the factored denominator $(x-2)(x-3)$, describe what you would see on the graph near those $x$-values before you know whether each gap is a vertical asymptote or a hole.

2. Vertical Asymptotes vs. Holes

After finding where $q(a) = 0$, the key question is: does $p(a)$ also equal zero?

Case 1: $q(a) = 0$ and $p(a) \neq 0$. The factor $(x - a)$ is in the denominator but not the numerator. Division by a number approaching zero produces outputs that grow without bound. The graph has a vertical asymptote at $x = a$.

Case 2: $q(a) = 0$ and $p(a) = 0$. Both numerator and denominator share the factor $(x - a)$. After cancellation, the output approaches a finite limit as $x \to a$. The graph has a hole (removable discontinuity) at $x = a$.

Predict-then-check. For $f(x) = \dfrac{x^2 - 1}{x - 1}$, predict: vertical asymptote or hole at $x = 1$?

Prediction: Both numerator and denominator are zero at $x = 1$ (since $1^2 - 1 = 0$). Expect a hole.

Check: Factor: $\dfrac{x^2 - 1}{x - 1} = \dfrac{(x-1)(x+1)}{x-1} = x + 1$ for $x \neq 1$.

As $x \to 1$, the output approaches $1 + 1 = 2$. There is a hole at the point $(1, 2)$, not a vertical asymptote.

Now predict for $g(x) = \dfrac{x^2 - 4}{x - 1}$ at $x = 1$: $g(1) = (1-4)/(0)$, numerator $= -3 \neq 0$. This is a vertical asymptote, not a hole.

3. Behavior Near a Vertical Asymptote

Near a vertical asymptote $x = a$, the function output grows without bound. The direction (positive or negative) depends on the sign of the remaining expression after canceling common factors.

Example. Analyze $f(x) = \dfrac{1}{x - 2}$ near $x = 2$.

For $x$ slightly greater than $2$: $x - 2$ is small and positive, so $\dfrac{1}{x-2}$ is large and positive. $f(x) \to +\infty$.

For $x$ slightly less than $2$: $x - 2$ is small and negative, so $\dfrac{1}{x-2}$ is large and negative. $f(x) \to -\infty$.

Numerical check:

The values confirm the prediction: left side goes to $-\infty$, right side goes to $+\infty$.

Ask why: Does the graph ever actually reach $x = 2$? No -- $x = 2$ is not in the domain. The graph approaches the vertical line $x = 2$ but never touches it. The asymptote is a line that describes the graph’s trend, not a wall that the graph bounces off of (see the misconception below).

4. Horizontal Asymptotes and End Behavior

As $x \to \pm\infty$, the leading terms of numerator and denominator dominate all other terms. Divide numerator and denominator by the highest power of $x$ that appears.

Example: $n < m$ (numerator degree less than denominator degree).

$f(x) = \dfrac{3x + 1}{x^2 - 5x + 6}$.

As $x \to \infty$, numerator grows like $3x$ and denominator grows like $x^2$. The ratio $3x / x^2 = 3/x \to 0$.

Horizontal asymptote: $y = 0$.

Example: $n = m$ (equal degrees).

$f(x) = \dfrac{4x^2 - 3}{2x^2 + 7}$.

Divide numerator and denominator by $x^2$: $$ f(x) = \frac{4 - 3/x^2}{2 + 7/x^2} \to \frac{4}{2} = 2 \text{ as } x \to \pm\infty. $$

Horizontal asymptote: $y = 2$.

Two representations, one idea.

Algebraic: The ratio of leading coefficients $4/2 = 2$.

Graphical/numerical: Compute $f(100) = (40000 - 3)/(20000 + 7) = 39997/20007 \approx 1.9992$, and $f(1000) \approx 1.999992$. The output approaches 2.

Translation prompt: Explain in one sentence why the lower-degree terms become negligible as $x$ grows large. Then use that explanation to determine the horizontal asymptote of $\dfrac{5x^3 - 2x}{3x^3 + x^2 + 1}$ without computing anything beyond the leading terms.

Example: $n > m$ (numerator degree greater).

$f(x) = \dfrac{x^3}{x^2 + 1}$.

The output grows like $x^3/x^2 = x$ for large $x$. There is no horizontal asymptote. (There is an oblique asymptote; see the extension below.)

Named Misconception

Misconception ID: asymptote-as-wall

A vertical asymptote at $x = a$ is sometimes described as a “wall” that the graph cannot cross. This image is misleading for two reasons.

First, a vertical asymptote is a line that describes the graph’s behavior near $x = a$: the output grows without bound as $x$ approaches $a$ from one or both sides. It is a trend description, not a physical barrier.

Second, a horizontal asymptote can be crossed. The function $f(x) = \dfrac{\sin x}{x}$ has horizontal asymptote $y = 0$ but crosses it infinitely many times (at every $x = k\pi$ for nonzero integer $k$). Even simpler: $f(x) = \dfrac{x}{x^2+1}$ has horizontal asymptote $y = 0$ and equals $0$ at $x = 0$.

Why the wall image breaks: The asymptote $y = L$ means only that $f(x) \to L$ as $x \to \infty$. It says nothing about whether $f(x) = L$ is possible for finite $x$.

Correct image: Think of the asymptote as a trend line that the graph approaches but (for vertical asymptotes) cannot reach because that $x$-value is outside the domain. For horizontal asymptotes, the graph may or may not cross the asymptote for finite $x$; what matters is the long-run behavior.

Worked Example

Problem. Analyze $f(x) = \dfrac{x^2 - x - 6}{x^2 - 4}$.

Find: (a) domain, (b) holes, (c) vertical asymptotes, (d) horizontal asymptote, (e) zeros.

Step 1: Predict. Both numerator and denominator are degree 2. Expect a horizontal asymptote at $y = 1/1 = 1$ (ratio of leading coefficients).

Step 2: Factor. $$ \text{Numerator: } x^2 - x - 6 = (x-3)(x+2) $$ $$ \text{Denominator: } x^2 - 4 = (x-2)(x+2) $$

Step 3: Domain. Denominator is zero at $x = 2$ and $x = -2$. Domain: all reals except $x = 2$ and $x = -2$.

Step 4: Holes vs. vertical asymptotes.

At $x = -2$: both numerator and denominator are zero ($(x+2)$ is a common factor). Cancel: $$ f(x) = \frac{(x-3)(x+2)}{(x-2)(x+2)} = \frac{x-3}{x-2} \quad \text{for } x \neq -2. $$ As $x \to -2$: output $\to \dfrac{-2-3}{-2-2} = \dfrac{-5}{-4} = \dfrac{5}{4}$. Hole at $\left(-2, \dfrac{5}{4}\right)$.

At $x = 2$: denominator is zero, numerator (after cancellation) $= 2 - 3 = -1 \neq 0$. Vertical asymptote at $x = 2$.

Step 5: Horizontal asymptote. Degrees are equal; ratio of leading coefficients: $1/1 = 1$. Horizontal asymptote $y = 1$.

Check prediction: matches the prediction from Step 1.

Step 6: Zeros. Zeros of the simplified function $\dfrac{x-3}{x-2}$: numerator $= 0$ when $x = 3$. Check: $x = 3$ is in the domain (denominator at $x = 3$ is $3 - 2 = 1 \neq 0$). Zero at $x = 3$.

Step 7: Verify with a value. $f(10) = (10-3)/(10-2) = 7/8 = 0.875$. The horizontal asymptote $y = 1$ says $f(x) \to 1$ from below for large $x$. $0.875$ is below $1$, consistent.

Common Errors

Common Misconceptions

Leveled Practice

Level 1 -- Direct Application

Problem 1. Find the domain of $f(x) = \dfrac{2x + 3}{x^2 - 9}$ and identify all values excluded.

Problem 2. Find any holes and vertical asymptotes of $g(x) = \dfrac{x^2 - 4}{x^2 - x - 2}$.

Problem 3. Determine the horizontal asymptote (if any) of $h(x) = \dfrac{6x^3 - 1}{2x^3 + x - 5}$.

Level 2 -- Multiple Representations

Problem 4. For $f(x) = \dfrac{x^2 + x - 2}{x^2 - 1}$, find the domain, holes, vertical asymptotes, horizontal asymptote, and zeros. Then make a rough sketch consistent with your findings.

Problem 5. For large positive $x$, is $f(x) = \dfrac{3x^2 - 2x}{x^2 + 1}$ approaching $y = 3$ from above or below? Explain algebraically and confirm numerically.

Level 3 -- Extension

Problem 6 (Low Floor, High Ceiling). Determine the oblique (slant) asymptote of $f(x) = \dfrac{x^2 + 3x - 1}{x + 1}$ and explain why it is an asymptote.

Mastery Checklist

You have mastered this skill when you can do all of the following without referring to notes:

• Find the domain of a rational function by factoring and solving $q(x) = 0$
• Distinguish between a hole and a vertical asymptote by checking for common factors
• Find the coordinates of any hole in the graph
• Determine the horizontal asymptote (or confirm none exists) by comparing degrees
• Find zeros of a rational function by setting the numerator equal to zero and checking the domain
• Describe the sign of $f(x)$ on each side of a vertical asymptote
• Explain why a horizontal asymptote can be crossed for finite $x$

Mental Model

Think of a rational function as a polynomial that has been “divided in half” by another polynomial. Where the denominator is zero, the function has a gap. The gap is either a hole (if the numerator also vanishes there, and the factors cancel) or a vertical asymptote (if the numerator does not vanish there). Near the asymptote, the output surges toward $\pm\infty$ as the denominator shrinks toward zero.

At the far ends (large $|x|$), the function behaves like the ratio of its leading terms. A smaller numerator degree means the denominator wins and the output flattens to zero. Equal degrees mean the outputs balance at the ratio of leading coefficients. A larger numerator degree means the numerator wins and the output grows without bound (giving an oblique asymptote when the degrees differ by exactly 1).

The graph of a rational function is like a polynomial graph that has been stretched, compressed, and cut wherever the denominator fails.

Connections

Within Chapter 1

• Polynomial functions: Rational functions generalize polynomials. A polynomial $p(x) = p(x)/1$ is a rational function with denominator 1. All polynomials are rational functions; not all rational functions are polynomials.
• Domain restrictions: Rational functions introduce the idea that domain matters -- not every rule can be evaluated at every input. This theme continues throughout calculus with square roots, logarithms, and inverse trig functions.

Toward Later Calculus

• Limits (Chapter 2): The calculation of $\displaystyle\lim_{x \to a} f(x)$ when $f(a)$ is undefined (a hole or asymptote) is the central problem of Chapter 2. The tools for analyzing holes and asymptotes here are exactly the tools used there.
• Derivatives of rational functions (Chapter 3): The quotient rule produces rational functions from polynomial inputs. Analyzing the resulting rational functions uses every skill from this lesson.
• Partial fractions (Chapter 5, integration): Integrating rational functions requires decomposing them into simpler pieces. The denominator factoring done here is the first step of that process.

Audience Notes

For students who find the hole vs. asymptote distinction tricky: The key is always to factor first. After full cancellation, check the simplified expression at the excluded $x$-value. If the simplified expression gives a finite number, it is a hole. If the denominator is still zero in the simplified expression, it is a vertical asymptote.

For students who want to go further: A rational function over the real numbers is completely determined (up to a constant) by its zeros, poles (vertical asymptotes), and the degree comparison. This idea generalizes to complex analysis, where rational functions on the complex plane (called rational maps) are classified by their poles and zeros, and the study of their global behavior is a rich area of mathematics.

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