Rational functions and end behavior — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: End behavior classification of rational functions, the leading term ratio rule, horizontal, oblique, and curvilinear asymptote identification, limit notation for end behavior, and distinguishing end behavior from local behavior.

You should already know: Limit notation for infinite inputs, polynomial degree and leading coefficient rules, factoring polynomials to find discontinuities.

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 end behavior?

A rational function is defined as any function that can be written as the ratio of two polynomials $f(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials with no common factors (after simplification) and $Q(x)$ is not the zero polynomial. End behavior of a function describes the trend of its output values $f(x)$ as the input $x$ grows without bound, either in the positive direction ($x\to +\infty$) or the negative direction ($x\to -\infty$). According to the official AP Precalculus Course and Exam Description (CED), this topic accounts for approximately 1.5-2% of total exam score weight, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is most commonly tested in questions asking to identify asymptotes, match functions to their graphs, interpret long-term trends in contextual models, and evaluate limits at infinity for rational functions. This topic builds directly on polynomial end behavior and lays the foundation for analyzing all discontinuous functions later in the course.

2. The Leading Term Rule for End Behavior

When analyzing the end behavior of any rational function $f(x)=\frac{P(x)}{Q(x)}$, as $x$ approaches $\pm \infty$, the highest-degree (leading) term of each polynomial dominates all lower-degree terms. Lower-degree terms become negligible compared to the leading term as $|x|$ grows very large. For example, for $P(x)=3x^3+2x^2-5x+1$, as $x\to +\infty$, $3x^3$ is orders of magnitude larger than the sum of all lower-degree terms, so we can approximate $P(x)\approx 3x^3$ for very large $|x|$.

This means the entire end behavior of $f(x)$ is identical to the end behavior of the ratio of the leading terms of $P(x)$ and $Q(x)$: $$f(x)\approx \frac{a_n{x}^n}{b_m{x}^m}=\left(\frac{a_n}{b_m}\right)x^{n-m}$$ where $n$ is the degree of $P(x)$, $m$ is the degree of $Q(x)$, $a_n$ is the leading coefficient of $P(x)$, and $b_m$ is the leading coefficient of $Q(x)$. This core rule allows us to classify all end behavior patterns for rational functions.

Worked Example

Identify the end behavior of $f(x)=\frac{2x^4-3x^2+7x-12}{5x^2+4x-8}$ by writing the limit behavior as $x\to +\infty$ and $x\to -\infty$.

1. First, identify degrees and leading terms: the numerator $P(x)$ has degree $n=4$, leading term $2x^4$; the denominator $Q(x)$ has degree $m=2$, leading term $5x^2$.
2. Apply the leading term rule: for large $|x|$, $f(x)\approx \frac{2x^4}{5x^2}=\frac{2}{5}{x}^2$.
3. Evaluate end behavior for $x\to +\infty$: as $x$ grows positive and large, $\frac{2}{5}{x}^2$ grows without bound to $+\infty$, so ${\lim }_{x\to +\infty }f(x)=+\infty$.
4. Evaluate end behavior for $x\to -\infty$: even for large negative $x$, $x^2$ remains positive and grows without bound, so ${\lim }_{x\to -\infty }f(x)=+\infty$.

Exam tip: When you first start a problem, always write down $n$ (degree of numerator) and $m$ (degree of denominator) explicitly before applying the leading term rule. This avoids mixing up degrees and misclassifying the end behavior, a common MCQ trap.

3. Classifying End Behavior Asymptotes

From the leading term rule, we can classify the end behavior asymptote (the function that $f(x)$ approaches as $x\to \pm \infty$) based on the relationship between $n$ and $m$:

• Case 1: $n<m$: The exponent $n-m$ is negative, so $\frac{a_n}{b_m}{x}^{n-m}=\frac{a_n}{b_m{x}^{m-n}}$, which approaches 0 as $|x|\to \infty$. This gives a horizontal asymptote at $y=0$.
• Case 2: $n=m$: The exponent $n-m=0$, so $x^0=1$, and the ratio approaches the constant $\frac{a_n}{b_m}$. This gives a horizontal asymptote at $y=\frac{a_n}{b_m}$.
• Case 3: $n=m+1$: The exponent $n-m=1$, so the leading term ratio is linear. We use polynomial long division to find the full linear equation $y=mx+b$, which is called an oblique (slant) asymptote.
• Case 4: $n\ge m+2$: The end behavior follows a polynomial of degree $n-m$, called a curvilinear asymptote. This is rarely tested on the AP Precalculus exam.

Worked Example

Find the equation of the end behavior asymptote for $f(x)=\frac{4x^2-5x+9}{2x-1}$, and classify the type of asymptote.

1. Identify degrees: $n=2$ (numerator), $m=1$ (denominator), so $n=m+1$, meaning we expect an oblique asymptote.
2. Perform polynomial long division of the numerator by the denominator: Dividing $4x^2-5x+9$ by $2x-1$ gives $f(x)=2x-\frac{3}{2}+\frac{\frac{15}{2}}{2x-1}$, with $\frac{15}{2}$ as the remainder.
3. As $x\to \pm \infty$, the remainder term $\frac{\frac{15}{2}}{2x-1}$ approaches 0, so $f(x)$ approaches $y=2x-\frac{3}{2}$.
4. Classification: This is a linear end behavior asymptote, so it is an oblique (slant) asymptote.

Exam tip: Always remember that an oblique asymptote only exists when the numerator degree is exactly one greater than the denominator degree. If it is two or more higher, there is no oblique asymptote, which is a common MCQ distractor.

4. End Behavior vs Local Behavior

A core distinction that is often tested is the difference between end behavior (behavior for very large $|x|$, i.e., as $x\to \pm \infty$) and local behavior (behavior near a finite input $x=c$, such as near a vertical asymptote or hole). End behavior is driven entirely by the relative degrees of the numerator and denominator, while local behavior near a discontinuity is driven by the roots of the denominator.

One common misconception is that a rational function can never cross any asymptote. In reality, a function can never cross a vertical asymptote (it is undefined at the vertical asymptote, and blows up to $\pm \infty$ near it), but it can cross a horizontal or oblique end behavior asymptote at any finite value of $x$. Crossing at a finite $x$ does not change the end behavior, because the asymptote only describes the trend for very large $|x|$.

Worked Example

For $f(x)=\frac{3x(x-2)}{(x+1)(x-2)}$, identify (a) its horizontal end behavior asymptote, and (b) explain whether the discontinuity at $x=2$ affects the end behavior.

1. First, simplify the function: the common $(x-2)$ factor cancels, so the simplified function is $f(x)=\frac{3x}{x+1}$ for $x\ne 2$, with a hole at $x=2$.
2. Find the end behavior: the degrees of the numerator and denominator are both 1 ($n=m=1$), so the horizontal asymptote is $y=\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}=\frac{3}{1}=3$.
3. The discontinuity at $x=2$ is a local discontinuity at a finite input. It does not change the leading terms of the numerator or denominator, so it has no impact on the end behavior as $x\to \pm \infty$.
4. Even though the function is undefined at $x=2$, this does not change the fact that as $x$ grows very large, the function approaches $y=3$.

Exam tip: Any time you see a common factor that creates a hole, remember that holes are local discontinuities and never change the end behavior or end behavior asymptote of the rational function.

5. Common Pitfalls (and how to avoid them)

• Wrong move: When $n=m$, reverse the leading coefficient ratio, getting $y=\frac{b_m}{a_n}$ instead of $y=\frac{a_n}{b_m}$. Why: Students mix up numerator and denominator when memorizing the rule, instead of writing the ratio explicitly. Correct move: Always write the ratio explicitly as $\frac{\text{leading coefficient of numerator}}{\text{leading coefficient of denominator}}$ before simplifying, do not rely on memorized wording alone.
• Wrong move: Claim an oblique asymptote exists when the numerator degree is 2 or more higher than the denominator. Why: Students incorrectly generalize that any higher numerator degree means an oblique asymptote, forgetting the "exactly one higher" requirement. Correct move: Always write down $n$ and $m$, then check if $n=m+1$ explicitly before concluding an oblique asymptote exists.
• Wrong move: Eliminate an answer choice just because the graph crosses a horizontal asymptote at a finite $x$. Why: Students confuse the "no crossing" rule for vertical asymptotes with the rule for end behavior asymptotes. Correct move: Only apply the "no crossing" rule to vertical asymptotes; crossing a horizontal/oblique asymptote at finite $x$ is allowed and does not affect end behavior.
• Wrong move: For $f(x)=\frac{-3x^3}{2x^2}$, state that ${\lim }_{x\to -\infty }f(x)=-\infty$. Why: Students forget to check the sign of the power when $x$ is negative, only looking at the leading coefficient sign. Correct move: After finding the leading term ratio, explicitly evaluate the sign for $x\to +\infty$ and $x\to -\infty$ separately when the exponent $n-m$ is odd.
• Wrong move: Claim the end behavior limit does not exist because the function has a hole at a finite $x$. Why: Students confuse local undefined points with end behavior, mixing up discontinuity location. Correct move: End behavior depends only on behavior for very large $|x|$, so any discontinuity at a finite $x$ never changes end behavior.

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Which of the following gives the equation of the end behavior asymptote of $f(x)=\frac{(2x^2-5)(3x+4)}{6x^3-2x^2+11}$ and the correct limit as $x\to +\infty$? A) Asymptote $y=x$, ${\lim }_{x\to +\infty }f(x)=+\infty$ B) Asymptote $y=1$, ${\lim }_{x\to +\infty }f(x)=1$ C) Asymptote $y=6x$, ${\lim }_{x\to +\infty }f(x)=+\infty$ D) Asymptote $y=0$, ${\lim }_{x\to +\infty }f(x)=0$

Worked Solution: First, expand the numerator to find its degree and leading term: $(2x^2-5)(3x+4)=6x^3+8x^2-15x-20$, so the numerator has degree $n=3$ and leading coefficient 6. The denominator has degree $m=3$ and leading coefficient 6. Since $n=m$, the horizontal asymptote is $y=\frac{6}{6}=1$, and the limit as $x\to +\infty$ equals 1. Distractor A is for students who miscount the numerator degree, distractor C for students who multiply instead of divide leading coefficients, distractor D for students who think $n<m$. The correct answer is B.

Question 2 (Free Response)

Consider the rational function $f(x)=\frac{3x^3-6x^2+5x-10}{x^2-3}$. (a) Find the degree of the numerator and denominator, then classify the type of end behavior asymptote. (b) Find the exact equation of the end behavior asymptote. (c) Evaluate ${\lim }_{x\to -\infty }f(x)$ and describe the end behavior in words.

Worked Solution: (a) The numerator $3x^3-6x^2+5x-10$ has degree $n=3$, and the denominator $x^2-3$ has degree $m=2$. Since $n=m+1$, the end behavior asymptote is an oblique (slant) asymptote. (b) Performing polynomial long division gives: $f(x)=3x-6+\frac{14x-28}{x^2-3}$. As $x\to \pm \infty$, the remainder term $\frac{14x-28}{x^2-3}$ approaches 0, so the oblique asymptote is $\boxed{y=3x-6}$. (c) The leading term ratio is $\frac{3x^3}{x^2}=3x$. As $x\to -\infty$, $3x$ approaches $-\infty$, so $\boxed{\underset{x\to -\infty }{\lim }f(x)=-\infty }$. In words: as $x$ decreases without bound, the output of $f(x)$ also decreases without bound, approaching the oblique asymptote $y=3x-6$.

Question 3 (Application / Real-World Style)

A city planner is modeling the total cost (in millions of dollars) to remove $p$ percent of industrial pollution from a local river, given by the model $C(p)=\frac{12p}{100-p}$ for $0\le p<100$. Describe the end behavior of $C(p)$ as $p\to 100^{-}$ and explain what this means in the context of the problem.

Worked Solution: First, rewrite the function to analyze its behavior as $p$ approaches 100 from below (the practical "end" of the input range for this problem). Factor the leading terms: $C(p)=\frac{12p}{-p+100}$. For $p$ near 100, the leading term of the denominator is $-p$, so $C(p)\approx \frac{12p}{-p}=-12$, but as $p\to 100^{-}$, the denominator approaches 0 from the positive side, so $C(p)=\frac{12(100^{-})}{\text{small positive number}}\to +\infty$. In context: as the city tries to remove closer and closer to 100% of the river pollution, the total cost grows without bound, meaning it is practically infeasible to remove 100% of the pollution with current technology and budget.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for analyzing all rational function features, coming next in Unit 1. You will use the end behavior classification you learned here to sketch complete graphs of rational functions, match functions to their graphs, and interpret long-term behavior in applied modeling problems. Without mastering the leading term rule and asymptote classification, you will struggle to sort through MCQ distractors that mix up different asymptote types and correctly answer FRQ questions asking for end behavior interpretation. This topic also lays the groundwork for limits at infinity that you will use in AP Calculus if you continue after precalculus.

Vertical asymptotes and discontinuities Graphing rational functions Polynomial end behavior Limits at infinity