MGU Mathematics Series — Trigonometry and Precalculus
Part III: Advanced Algebraic Functions
Polynomials · Rational Functions · Asymptotes · Limits Preview
Chapter 8 of 14
MGU Mathematics Series  ·  Trigonometry and Precalculus  ·  Library Textbook Edition

Chapter 8:
Polynomial and
Rational Functions

End behavior, zeros and multiplicity, polynomial division, vertical and horizontal asymptotes, holes, rational inequalities, and the bridge to limits in Calculus I

■ Math Level: Precalculus ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapters 1–7 (Functions, Transformations, Trigonometry)  ·  Prepares students for Chapter 9 and Calculus I

Chapter 8 — Front Matter

Purpose and Overview

Chapter 8 reviews and deepens the study of polynomial and rational functions that students began in Algebra II. This chapter moves beyond computation and toward graph reasoning. Students will study how a polynomial's degree, leading coefficient, zeros, and their multiplicities control the shape of its graph. They will study how rational functions behave near domain restrictions, vertical asymptotes, horizontal asymptotes, slant asymptotes, and holes. They will practice polynomial long division, synthetic division, and rational inequalities. They will connect polynomial and rational behavior to function modeling and to the limit ideas that define Calculus I.

This chapter should feel less like a review of algebra rules and more like a serious study of how function structure controls graph behavior. By the end of this chapter, students should be able to look at a polynomial or rational function and describe, before graphing, what the function will do on both ends, near its zeros, and near any discontinuities. That kind of algebraic insight is the foundation of everything that follows in calculus, engineering, science, and applied mathematics.

Opening Question

A biologist studying a population of fish in a small lake models the population using a rational function. When she examines the graph, she notices something unexpected: the function has a hole at a particular input value, and it approaches a constant level far in the future even though the fish population never actually stops changing. Why would a rational function have a hole? What does a horizontal asymptote mean in a biological model? How does the algebra of numerators and denominators control the long-run behavior of the fish population? By the end of this chapter, you will be able to answer these questions precisely.

14Sections
12Objectives
19Key Terms
15+Worked Examples

Learning Objectives

  1. 1Identify polynomial functions by degree, leading term, and coefficient.
  2. 2Describe and predict end behavior using degree and leading coefficient.
  3. 3Identify zeros of polynomial functions and describe graph behavior at each zero using multiplicity.
  4. 4Sketch graphs of polynomial functions using end behavior, zeros, and multiplicities.
  5. 5Perform polynomial long division and synthetic division correctly.
  6. 6Apply the Remainder Theorem and Factor Theorem.
  7. 7Identify the domain of a rational function by finding values that cause division by zero.
  8. 8Identify vertical asymptotes, holes, horizontal asymptotes, and slant asymptotes from a rational function's equation.
  9. 9Sketch graphs of rational functions using asymptotes, holes, intercepts, and end behavior.
  10. 10Solve rational inequalities using sign analysis.
  11. 11Build and interpret polynomial and rational models in applied situations.
  12. 12Connect polynomial and rational graph behavior to the intuitive idea of limits.
Chapter 8 — Contents

Sections at a Glance

Sections 8.1 – 8.4
Polynomial Functions: Structure, Behavior, and Graphs
Review · End Behavior · Zeros and Multiplicity · Graphing
Section 8.1

Polynomial Functions Review

A polynomial function is a smooth, continuous function built entirely from non-negative integer powers of x. You have worked with polynomial functions throughout Algebra I and Algebra II — linear functions, quadratic functions, and cubic functions are all polynomials. Chapter 8 extends that work to polynomials of higher degree and builds the tools needed to read their graphs precisely.

The standard form of a polynomial function is f(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀, where n is the degree and aₙ is the leading coefficient. A polynomial function has no fractions with variables in the denominator, no radicals containing variables, and no negative or fractional exponents. It is a single, smooth expression made of powers of x.

DegreeNameExample
0Constantf(x) = 5
1Linearf(x) = 3x − 2
2Quadraticf(x) = x² − 4x + 3
3Cubicf(x) = x³ − 2x² − x + 2
4Quarticf(x) = x⁴ − 5x² + 4
5Quinticf(x) = x⁵ − x³ + x
Identifying Polynomial Functions

Check every term for non-integer exponents, negative exponents, or variable denominators. f(x) = x^(1/2) is not a polynomial (fractional exponent). f(x) = x⁻³ is not a polynomial (negative exponent). f(x) = 3x⁴ − 2x² + 7 is a polynomial. Every polynomial function has domain all real numbers and is continuous and smooth everywhere.

Example 8.1.1
Identifying polynomial functions
  • ProblemClassify each as a polynomial or not. If polynomial, state degree and leading coefficient.
    (a) f(x) = 4x³ − 7x + 1   (b) g(x) = 3/x² + 5x − 2   (c) h(x) = x⁵ − 2x^(3/2) + 6   (d) k(x) = −6x⁴ + 3x³ − x² + 8
  • ThinkCheck each function for non-integer exponents, negative exponents, or variable denominators.
  • Solve (a) Every term has a non-negative integer exponent. Polynomial. Degree 3, leading coefficient 4.
    (b) 3/x² = 3x⁻². Negative exponent. Not a polynomial.
    (c) x^(3/2) has a fractional exponent. Not a polynomial.
    (d) Every term has a non-negative integer exponent. Polynomial. Degree 4, leading coefficient −6.
Answer: (a) polynomial, degree 3, leading coeff. 4   (b) not a polynomial   (c) not a polynomial   (d) polynomial, degree 4, leading coeff. −6
Section 8.2

Degree, Leading Coefficient, and End Behavior

The most powerful observation you can make about a polynomial without graphing it is its end behavior — the direction the graph moves as x moves far to the right or far to the left. End behavior depends entirely on the leading term aₙxⁿ. As x becomes extremely large or extremely negative, the highest-degree term overwhelms all others.

Think of it this way. An even-degree polynomial with a positive leading coefficient looks like a smile — both ends go up. An even-degree polynomial with a negative leading coefficient looks like a frown — both ends go down. An odd-degree polynomial with a positive leading coefficient falls on the left and rises on the right. An odd-degree polynomial with a negative leading coefficient rises on the left and falls on the right.

DegreeLeading Coefficientx → −∞x → +∞Shape
EvenPositivef(x) → +∞f(x) → +∞Both ends rise (smile)
EvenNegativef(x) → −∞f(x) → −∞Both ends fall (frown)
OddPositivef(x) → −∞f(x) → +∞Falls left, rises right
OddNegativef(x) → +∞f(x) → −∞Rises left, falls right
Tip — End Behavior vs. Middle Behavior

"End behavior" refers to the two tails of the graph, not the middle. No matter how complicated a polynomial's middle section is, its tails are always determined by the leading term alone. The inner terms matter for finding zeros and turning points, but not for what happens as x races toward ±∞.

Example 8.2.1
Describing end behavior
  • ProblemDescribe the end behavior of:
    (a) f(x) = 3x⁴ − 5x³ + x − 7   (b) g(x) = −2x⁵ + 4x³ − 1   (c) h(x) = −x⁶ + 3x⁴ − 2
  • ThinkIdentify the degree and the sign of the leading coefficient. Apply the four rules above.
  • Solve (a) Degree 4 (even), leading coeff. 3 (positive). Both ends rise.
    As x → −∞, f(x) → +∞. As x → +∞, f(x) → +∞.

    (b) Degree 5 (odd), leading coeff. −2 (negative). Left rises, right falls.
    As x → −∞, g(x) → +∞. As x → +∞, g(x) → −∞.

    (c) Degree 6 (even), leading coeff. −1 (negative). Both ends fall.
    As x → −∞, h(x) → −∞. As x → +∞, h(x) → −∞.
Answer: (a) both ends rise   (b) left rises, right falls   (c) both ends fall
Section 8.3

Zeros and Multiplicity

A zero of a polynomial function is a value of x where the output equals zero. Geometrically, zeros correspond to x-intercepts — places where the graph crosses or touches the x-axis. The behavior of the graph at a zero is determined by the multiplicity of the corresponding factor.

If a polynomial f(x) has a factor (x − c)ᵏ, then x = c is a zero of multiplicity k. The key distinction is simple: odd multiplicity → graph crosses; even multiplicity → graph bounces.

Multiplicity Rules

k = 1 (odd): The graph crosses the x-axis at c in a standard transversal manner.

k = 2 (even): The graph touches the x-axis at c and turns back, like a parabola vertex.

k odd, k ≥ 3: The graph crosses the x-axis at c with a flattened S-shape.

k even, k ≥ 4: The graph bounces at c with a flatter curve.

Note: The sum of all multiplicities of a polynomial equals its degree.

Example 8.3.1
Identifying zeros and multiplicities
  • ProblemFind the zeros of f(x) = (x − 2)³(x + 1)²(x − 5) and describe the graph behavior at each zero.
  • ThinkRead the multiplicities directly from the factored form. The degree is 3 + 2 + 1 = 6.
  • Solve x − 2 = 0 → x = 2, multiplicity 3 (odd). Graph crosses at x = 2 with a flattened S-shape.
    x + 1 = 0 → x = −1, multiplicity 2 (even). Graph bounces at x = −1.
    x − 5 = 0 → x = 5, multiplicity 1 (odd). Graph crosses at x = 5 in a standard manner.
  • CheckSum of multiplicities: 3 + 2 + 1 = 6. Degree is 6. Consistent.
Answer: Zeros at x = −1 (bounce), x = 2 (cross, flattened), x = 5 (cross, standard).
Example 8.3.2
Writing a polynomial from zeros
  • ProblemWrite a polynomial of least degree with zeros at x = −3 (mult. 1), x = 0 (mult. 2), and x = 4 (mult. 1), with a positive leading coefficient.
  • ThinkBuild the polynomial by writing a factor for each zero, including its multiplicity. Use a = 1 for positive leading coefficient.
  • Solvef(x) = (x + 3)¹ · x² · (x − 4)¹ = x²(x + 3)(x − 4)
  • CheckDegree: 2 + 1 + 1 = 4. Positive leading coefficient (from x² · x · x = x⁴, coefficient 1). Zeros: x = 0 (mult 2), x = −3 (mult 1), x = 4 (mult 1). ✓
Answer: f(x) = x²(x + 3)(x − 4)
Section 8.4

Graphing Polynomial Functions

To sketch a polynomial function without technology, use a five-step approach that assembles information from end behavior, zeros, and multiplicities into a coherent graph. A smooth polynomial graph never crosses itself, and its number of turning points is at most n − 1 where n is the degree.

Five-Step Process for Sketching Polynomials

Step 1: Determine the degree and leading coefficient. Describe end behavior.

Step 2: Find the y-intercept by evaluating f(0).

Step 3: Find the zeros by factoring completely. List each zero with its multiplicity.

Step 4: Determine behavior at each zero using the odd/even multiplicity rule.

Step 5: Sketch the graph. Start from the left tail, move through the zeros correctly, end with the right tail.

Example 8.4.1
Graphing a polynomial function
  • ProblemSketch the graph of f(x) = −(x − 1)²(x + 2).
  • ThinkThis is a degree-3 polynomial. Identify degree, leading term, zeros, and behaviors using the five steps.
  • Step 1Degree 3, leading coefficient −1 (negative). Left end rises, right end falls.
  • Step 2f(0) = −(0 − 1)²(0 + 2) = −(1)(2) = −2. The y-intercept is (0, −2).
  • Step 3Zeros: x = 1 (multiplicity 2), x = −2 (multiplicity 1).
  • Step 4At x = 1: even multiplicity → graph bounces. At x = −2: odd multiplicity → graph crosses.
  • Step 5Starting from upper left, the graph rises from −∞. It crosses the x-axis at x = −2 coming down. It passes through (0, −2). It bounces off the x-axis at x = 1 from below, then falls to −∞ on the right.
Graph description: Degree-3 polynomial that rises left, crosses at x = −2, bounces at x = 1, falls right. Y-intercept at (0, −2). Mark intercepts at (−2, 0), (0, −2), and (1, 0).
Section 8.5
Polynomial Division: Long Division, Synthetic Division, and Key Theorems
Long Division · Synthetic Division · Remainder Theorem · Factor Theorem
Section 8.5

Polynomial Division

When working with polynomials, it is often useful to divide one polynomial by another — especially when factoring is not immediately obvious. Polynomial division is the algebraic analog of long division with integers. The two main methods are polynomial long division and synthetic division.

Polynomial Long Division

Polynomial long division follows the same divide-multiply-subtract-bring-down cycle as integer long division. To divide f(x) by d(x): divide the leading term, multiply, subtract, bring down, and repeat until the remainder has lower degree than the divisor. The result is written as:

f(x) / d(x) = q(x) + r(x)/d(x)
Example 8.5.1
Polynomial long division
  • ProblemDivide 2x³ − 5x² + 3x − 7 by x − 2.
  • ThinkUse polynomial long division, dividing leading terms at each step.
  • SolveDivide 2x³ by x: get 2x² Multiply: 2x²(x − 2) = 2x³ − 4x² Subtract: (2x³ − 5x²) − (2x³ − 4x²) = −x² Bring down 3x. New dividend: −x² + 3x Divide −x² by x: get −x Multiply: −x(x − 2) = −x² + 2x Subtract: (−x² + 3x) − (−x² + 2x) = x Bring down −7. New dividend: x − 7 Divide x by x: get 1 Multiply: 1(x − 2) = x − 2 Subtract: (x − 7) − (x − 2) = −5 ← remainder
  • CheckRemainder Theorem: f(2) = 2(8) − 5(4) + 3(2) − 7 = 16 − 20 + 6 − 7 = −5. The remainder −5 matches. ✓
Answer: 2x² − x + 1 − 5/(x − 2)

Synthetic Division

Synthetic division is a streamlined method for dividing a polynomial by a linear factor (x − c). It works only for linear divisors. The method records only the coefficients and performs arithmetic in a compact format. Always include a 0 coefficient for any missing degree in the polynomial before setting up the table.

Example 8.5.2
Synthetic division
  • ProblemDivide 3x⁴ − 2x³ + 0x² − x + 5 by x − 3.
  • ThinkUse synthetic division with c = 3. Include 0 for the missing x² term.
  • SolveCoefficients: 3 −2 0 −1 5 (c = 3) 3 | 3 −2 0 −1 5 | 9 21 63 186 ───────────────────────── 3 7 21 62 191 ← remainder Quotient coefficients: 3, 7, 21, 62 Remainder: 191
  • Checkf(3) = 3(81) − 2(27) + 0 − 3 + 5 = 243 − 54 − 3 + 5 = 191. Matches. ✓
Answer: 3x³ + 7x² + 21x + 62 + 191/(x − 3)

The Remainder Theorem and Factor Theorem

Two Fundamental Theorems

Remainder Theorem: If f(x) is divided by (x − c), the remainder equals f(c). You can evaluate f(c) by dividing and reading the remainder — no substitution required.

Factor Theorem: (x − c) is a factor of f(x) if and only if f(c) = 0. Finding a zero and finding a factor are the same thing.

Example 8.5.3
Using the Factor Theorem
  • ProblemDetermine whether (x + 2) is a factor of f(x) = x³ − x² − 4x + 4.
  • ThinkApply the Factor Theorem. (x + 2) is a factor if and only if f(−2) = 0.
  • Solvef(−2) = (−2)³ − (−2)² − 4(−2) + 4 = −8 − 4 + 8 + 4 = 0 ✓
  • FactorSince f(−2) = 0, divide out (x + 2) using synthetic division:
    −2 | 1 −1 −4 4 | −2 6 −4 ───────────────────── 1 −3 2 0 Quotient: x² − 3x + 2 = (x − 1)(x − 2) Full factorization: f(x) = (x + 2)(x − 1)(x − 2)
Answer: Yes, (x + 2) is a factor. Fully factored: f(x) = (x + 2)(x − 1)(x − 2).
Student Warning — Missing Terms in Synthetic Division

If a polynomial is missing a term — for example, x³ + 2x − 5 has no x² term — the coefficient table must include 0 for the missing degree. Write all coefficients in descending order, inserting 0 for any missing degree, before setting up the synthetic division array.

Sections 8.6 – 8.10
Rational Functions: Domain, Asymptotes, Holes, and Graphs
Review · Domain Restrictions · Vertical Asymptotes · Horizontal and Slant Asymptotes · Holes
Section 8.6

Rational Functions Review

A rational function is a ratio of two polynomials:

f(x) = p(x) / q(x), where q(x) ≠ 0

Unlike polynomial functions, rational functions are not always defined on all real numbers, and their graphs may include breaks, asymptotes, and holes. The graph of a rational function can have vertical asymptotes (where the denominator is zero and the factor does not cancel), holes (where a factor of the denominator cancels with a matching factor in the numerator), horizontal or slant asymptotes (describing long-run behavior), x-intercepts (where the numerator is zero), and a y-intercept (where x = 0, if in the domain).

Section 8.7

Domain Restrictions

The domain of a rational function consists of all real numbers except values where the denominator equals zero. To find domain restrictions, set the denominator equal to zero and solve. Every excluded value corresponds to either a vertical asymptote or a hole in the graph.

Example 8.7.1
Finding the domain of rational functions
  • ProblemFind the domain of: (a) f(x) = (3x − 1)/(x − 5)   (b) g(x) = (x² + 4)/(x² − 9)   (c) h(x) = (x + 2)/(x² + 1)
  • Solve (a) x − 5 = 0 → x = 5. Domain: (−∞, 5) ∪ (5, +∞).
    (b) x² − 9 = 0 → (x−3)(x+3) = 0 → x = 3 or x = −3. Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, +∞).
    (c) x² + 1 = 0 has no real solutions since x² + 1 ≥ 1 > 0 always. Domain: all real numbers, (−∞, +∞).
Answer: (a) (−∞, 5) ∪ (5, +∞)   (b) (−∞, −3) ∪ (−3, 3) ∪ (3, +∞)   (c) all real numbers
Section 8.8

Vertical Asymptotes

A vertical asymptote occurs at a value x = a where the denominator of a rational function equals zero and the factor (x − a) does not cancel from the numerator. Near a vertical asymptote, the function values grow without bound. The distinction between asymptotes and holes is critical: a common factor creates a hole, not an asymptote.

Finding Vertical Asymptotes — Three Steps

Step 1: Factor both the numerator and denominator completely.

Step 2: Cancel any common factors. (Common factors create holes, not asymptotes.)

Step 3: Set the remaining denominator equal to zero. Each solution gives a vertical asymptote.

Example 8.8.1
Finding vertical asymptotes
  • ProblemFind the vertical asymptotes of f(x) = (2x − 6)/(x² − x − 6).
  • Factor Numerator: 2x − 6 = 2(x − 3)
    Denominator: x² − x − 6 = (x − 3)(x + 2)
  • Cancelf(x) = 2(x − 3) / [(x − 3)(x + 2)]The factor (x − 3) cancels → creates a hole at x = 3.
    Simplified: f(x) = 2/(x + 2), x ≠ 3
  • AsymptoteRemaining denominator: x + 2 = 0 → x = −2. Vertical asymptote: x = −2.
  • CheckAt x = −2, the denominator factor (x + 2) does not cancel. At x = 3, the factor cancels, giving a hole rather than an asymptote. ✓
Answer: Vertical asymptote at x = −2. Hole at x = 3.
Behavior Near a Vertical Asymptote

For f(x) = 2/(x + 2): as x → −2⁻ (from the left), x + 2 approaches 0 from the negative side, so f(x) → −∞. As x → −2⁺ (from the right), x + 2 approaches 0 from the positive side, so f(x) → +∞. The graph falls on the left of x = −2 and rises on the right.

Section 8.9

Horizontal and Slant Asymptotes

While vertical asymptotes describe what happens near domain restrictions, horizontal and slant asymptotes describe what happens at the far ends of the graph. The horizontal asymptote of a rational function is determined by comparing the degree of the numerator p(x) to the degree of the denominator q(x). Let n = degree of numerator, m = degree of denominator.

CaseConditionHorizontal Asymptote
Case 1n < m (numerator degree less than denominator)y = 0 (the x-axis)
Case 2n = m (equal degrees)y = aₙ/bₘ (ratio of leading coefficients)
Case 3n > m (numerator degree exceeds denominator)No horizontal asymptote; if n = m + 1, slant asymptote exists
Student Warning — Graphs Can Cross Horizontal Asymptotes

Horizontal asymptotes describe long-run behavior only — what happens as x → ±∞. The graph of a rational function can cross its horizontal asymptote in the middle section. Do not assume the graph must stay on one side of the asymptote for all finite values of x.

Example 8.9.1
Finding horizontal asymptotes
  • ProblemFind the horizontal asymptote of:
    (a) f(x) = (3x − 1)/(2x² + 5)   (b) g(x) = (5x³ − 2x)/(x³ + 4)   (c) h(x) = (x² − 3)/(x − 1)
  • ThinkCompare degrees of numerator and denominator for each.
  • Solve (a) Numerator degree 1 < denominator degree 2. Horizontal asymptote: y = 0.
    (b) Equal degrees (both 3). Leading coefficients 5 and 1. Horizontal asymptote: y = 5/1 = 5.
    (c) Numerator degree 2 > denominator degree 1. No horizontal asymptote. (Degree exceeds by exactly 1 → slant asymptote exists.)
Answer: (a) y = 0   (b) y = 5   (c) no horizontal asymptote

Slant Asymptotes

A slant (oblique) asymptote occurs when the degree of the numerator exceeds the degree of the denominator by exactly 1. Find it by performing polynomial long division. As x → ±∞, the remainder fraction approaches 0 and the graph approaches the quotient line.

Example 8.9.2
Finding a slant asymptote
  • ProblemFind the slant asymptote of h(x) = (x² − 3)/(x − 1).
  • ThinkDegree of numerator (2) exceeds degree of denominator (1) by exactly 1. Perform polynomial long division.
  • SolveDivide x² − 3 by x − 1: Divide x² by x: get x Multiply: x(x − 1) = x² − x Subtract: (x² − 3) − (x² − x) = x − 3 Divide x by x: get 1 Multiply: 1(x − 1) = x − 1 Subtract: (x − 3) − (x − 1) = −2 ← remainder h(x) = x + 1 + (−2)/(x − 1)
  • CheckAs x → ±∞, −2/(x − 1) → 0. The graph approaches the line y = x + 1. ✓
Answer: Slant asymptote: y = x + 1.
Section 8.10

Holes in Rational Graphs

A hole — formally called a removable discontinuity — occurs at a value x = c where a factor (x − c) cancels from both the numerator and denominator. The function is undefined at x = c, but the graph approaches a specific finite y-value as x approaches c. Holes appear as open circles in the graph.

Finding Holes — Four Steps

Step 1: Factor numerator and denominator completely.

Step 2: Identify all common factors that cancel.

Step 3: The x-values of the common factors give the x-coordinates of the holes.

Step 4: Substitute the x-value into the simplified form to find the y-coordinate of each hole.

Example 8.10.1
Complete analysis — holes, asymptotes, and intercepts
  • ProblemFind all holes, asymptotes, and intercepts of f(x) = (x² − 4)/(x² − x − 2).
  • Factor Numerator: x² − 4 = (x − 2)(x + 2)
    Denominator: x² − x − 2 = (x − 2)(x + 1)
  • Cancelf(x) = (x − 2)(x + 2) / [(x − 2)(x + 1)] (x − 2) cancels → Simplified: f(x) = (x + 2)/(x + 1), x ≠ 2
  • HoleAt x = 2. Substitute into simplified form: (2 + 2)/(2 + 1) = 4/3. Hole at (2, 4/3).
  • AsymptotesVertical: x + 1 = 0 → x = −1.
    Horizontal: equal degrees, leading coefficients both 1 → y = 1.
  • Interceptsy-intercept: f(0) = 2/1 = 2 → (0, 2).
    x-intercept: x + 2 = 0 → x = −2 → (−2, 0).
Answer: Hole at (2, 4/3). Vertical asymptote x = −1. Horizontal asymptote y = 1. Intercepts at (−2, 0) and (0, 2).
Graph description: Looks like (x+2)/(x+1) with an open circle at (2, 4/3), a vertical asymptote at x = −1 (dashed), and a horizontal asymptote at y = 1 (dashed).
Section 8.11
Rational Inequalities and Sign Analysis
Critical Values · Sign Charts · Solution Intervals
Section 8.11

Rational Inequalities

A rational inequality asks where a rational expression is positive, negative, or zero. Solving it requires sign analysis — tracking whether the expression is positive or negative across intervals defined by critical values. Values where the denominator equals zero are never included in the solution, even when the inequality is non-strict (≤ or ≥).

Solving Rational Inequalities — Seven Steps

Step 1: Move all terms to one side so the inequality has 0 on one side.

Step 2: Factor numerator and denominator completely.

Step 3: Find all critical values: zeros of the numerator and zeros of the denominator.

Step 4: Plot critical values on a number line, dividing it into intervals.

Step 5: Test a value in each interval.

Step 6: Determine the sign of the expression in each interval.

Step 7: Write the solution. Include numerator zeros (where expression = 0) only for non-strict inequalities. Never include denominator zeros.

Example 8.11.1
Solving a rational inequality
  • ProblemSolve (x − 1)/(x + 3) ≥ 0.
  • ThinkFind critical values from numerator and denominator zeros. Build a sign chart.
  • Criticalx − 1 = 0 → x = 1 (numerator zero)    x + 3 = 0 → x = −3 (denominator zero)
    Number line divided into three intervals: (−∞, −3), (−3, 1), (1, +∞)
  • Sign Chart
    IntervalTest Value(x − 1)(x + 3)Quotient
    (−∞, −3)x = −5−−+
    (−3, 1)x = 0−+−
    (1, +∞)x = 2+++
  • SolvePositive: (−∞, −3) and (1, +∞). Zero at x = 1 (include since ≥ 0). Excluded: x = −3 (denominator zero, never include).
Answer: x ∈ (−∞, −3) ∪ [1, +∞)
Example 8.11.2
A more complex rational inequality
  • ProblemSolve (x² − x − 6)/(x − 4) < 0.
  • FactorNumerator: x² − x − 6 = (x − 3)(x + 2). Denominator: x − 4.
    Critical values: x = −2, x = 3 (numerator zeros), x = 4 (denominator zero).
  • Sign Chart
    Interval(x + 2)(x − 3)(x − 4)Expression
    (−∞, −2)−−−−
    (−2, 3)+−−+
    (3, 4)++−−
    (4, +∞)++++
  • SolveNegative intervals: (−∞, −2) and (3, 4). Strict inequality, so x = −2 and x = 3 (numerator zeros) are not included. x = 4 (denominator zero) is never included.
Answer: x ∈ (−∞, −2) ∪ (3, 4)
Section 8.12

Polynomial and Rational Models

Polynomial functions model situations where behavior is smooth, continuous, and influenced by multiple powers: motion and position, revenue and cost analysis, engineering stress and deflection, curve fitting. Rational functions model situations where a quantity is bounded, inversely related to another, or approaches a limiting value: population models with carrying capacity, drug concentration in the bloodstream, electrical resistance, average cost per unit.

A rational model often has a horizontal asymptote that represents a long-run limiting value — the level the quantity can approach but never quite reach.

Example 8.12.1
Rational model for average cost
  • SituationA factory has daily fixed costs of $800 and variable cost of $12 per part. Write a rational function for the average cost per part when x parts are produced. Analyze the long-run behavior.
  • BuildTotal daily cost: C(x) = 800 + 12x Average cost per part: C̄(x) = (12x + 800) / x, x > 0
  • Interpret At x = 1: C̄(1) = 812/1 = $812 per part (fixed cost dominates).
    At x = 100: C̄(100) = 2000/100 = $20 per part.
    Long-run: equal degrees; leading coefficients 12 and 1. Horizontal asymptote: y = 12.
  • CheckAs production grows, average cost approaches $12 per part — the variable cost only. The fixed cost per part shrinks toward zero. The company can never produce parts for less than $12 each, no matter how much they produce. ✓
Answer: C̄(x) = (12x + 800)/x. Horizontal asymptote at y = 12. Long-run average cost approaches $12 per part.
Sections 8.13 – 8.14
Common Mistakes and the Bridge to Calculus
Eight Recurring Errors · End Behavior as Limits · Asymptotes and Continuity
Section 8.13

Common Polynomial and Rational Function Mistakes

Mistake 1 — Confusing multiplicity 1 with multiplicity 2 at x-intercepts
Sketching a bounce where a cross belongs, or vice versa.

Correction: Always count the multiplicity of each factor. Odd multiplicity → cross; even multiplicity → bounce.

Mistake 2 — Stating wrong end behavior by ignoring the sign of the leading coefficient
Claiming a degree-4 polynomial with negative leading coefficient has both ends rising.

Correction: Check both degree (even or odd) and the sign of the leading coefficient before describing end behavior.

Mistake 3 — Canceling factors without noting holes
Simplifying a rational function and declaring it fully analyzed without noting the hole created by the canceled factor.

Correction: When a factor cancels, record the hole before simplifying. The hole occurs at the zero of the canceled factor; its y-coordinate comes from the simplified expression.

Mistake 4 — Including a vertical asymptote where there is actually a hole
f(x) = (x² − 4)/(x − 2) has a vertical asymptote at x = 2.

Correction: Factor completely and cancel common factors before identifying asymptotes. The factor (x − 2) cancels with (x − 2) in the numerator, creating a hole at x = 2, not an asymptote.

Mistake 5 — Misidentifying the horizontal asymptote when degrees are equal
Stating y = 0 for all rational functions regardless of degree comparison.

Correction: Compare degrees first. When degrees are equal, the horizontal asymptote is y = aₙ/bₘ (ratio of leading coefficients), not y = 0.

Mistake 6 — Including denominator zeros in rational inequality solutions
Writing x ∈ [−3, 1] when the denominator zero at x = −3 should be excluded.

Correction: Values where the denominator equals zero are never in the domain. Use open circles and open intervals for all denominator zeros regardless of the inequality type.

Mistake 7 — Forgetting placeholder zeros in synthetic division
Setting up synthetic division for x³ + 2x − 5 using coefficients 1, 2, −5 instead of 1, 0, 2, −5.

Correction: Write all coefficients in descending order, inserting 0 for any missing degree before setting up the array.

Mistake 8 — Believing a graph cannot cross its horizontal asymptote
Assuming the graph of a rational function stays entirely above or below its horizontal asymptote.

Correction: Horizontal asymptotes describe long-run behavior only. The graph can cross the asymptote in the middle section. Find intercepts and other features separately.

Section 8.14

Preparing for Limits and Calculus

The ideas in Chapter 8 are the direct precursors of the central concepts in Calculus I. This section draws the explicit connections.

End behavior and limits at infinity. When you wrote "as x → +∞, f(x) → L," you were already using the notation of limits. In calculus, a horizontal asymptote means limx→+∞ f(x) = L. Students who understand horizontal asymptotes already understand limits at infinity in their intuitive form.

Vertical asymptotes and one-sided limits. When you described a graph approaching +∞ from one side and −∞ from the other near x = a, you were describing one-sided limits. Calculus I will formalize the notation limx→a⁻ and limx→a⁺ and connect them to the definition of continuity.

Holes and removable discontinuities. A hole at x = c means the function is undefined there, but the limit from both sides exists and equals the y-coordinate of the hole. Calculus begins with this idea. Continuity requires that the limit at every point equals the function value there. Holes are the simplest concrete examples of discontinuities.

Polynomial structure and the derivative. The derivative of a function at a point is the limit of the average rate of change as the interval shrinks toward zero: f′(x) = limh→0 [(f(x+h) − f(x)) / h]. Polynomials are the first functions for which this limit is computed. Students who understand polynomial structure will find differentiation rules natural.

Bridge to Calculus I

Every step of Calculus I builds directly on Chapter 8. You will compute limits precisely for polynomial and rational functions, learn the formal definition of continuity, and differentiate polynomial and rational functions using power, product, quotient, and chain rules. The asymptote, hole, and end-behavior analysis you practiced here is the informal calculus you already know.

Chapter 8 — Back Matter

Chapter Summary

Key Terms Review

polynomial function
A function f(x) = aₙxⁿ + … + a₀ where n is a non-negative integer and aₙ ≠ 0. Smooth and continuous on all real numbers.
degree
The highest power of x with a nonzero coefficient in a polynomial. Controls end behavior together with the leading coefficient.
leading coefficient
The coefficient of the highest-degree term. Determines whether the ends of the graph rise or fall.
end behavior
The direction a function moves as x → +∞ and as x → −∞. Determined solely by the leading term.
zero of a function
A value c such that f(c) = 0. Also called a root or x-intercept. Corresponds geometrically to a point on the x-axis.
multiplicity
The number of times a factor (x − c) appears in a polynomial. Odd multiplicity → graph crosses; even multiplicity → graph bounces.
polynomial division
A method for dividing one polynomial by another, analogous to long division with integers.
synthetic division
A shortcut method for dividing a polynomial by a linear factor (x − c). Works only for linear divisors.
Remainder Theorem
If f(x) is divided by (x − c), the remainder equals f(c). Lets you evaluate f(c) via division.
Factor Theorem
(x − c) is a factor of f(x) if and only if f(c) = 0. Connects zeros to factors.
rational function
A function f(x) = p(x)/q(x) where p(x) and q(x) are polynomials and q(x) ≠ 0. May have asymptotes and holes.
domain restriction
A value of x excluded from the domain because it makes the denominator equal to zero.
vertical asymptote
A vertical line x = a that the graph approaches but never reaches, occurring where a denominator factor does not cancel.
hole
A removable discontinuity — a single missing point where a denominator factor cancels with a numerator factor. Shown as an open circle.
horizontal asymptote
A horizontal line y = L that the graph approaches as x → ±∞. Determined by comparing degrees of numerator and denominator.
slant asymptote
A non-horizontal line y = mx + b that the graph approaches as x → ±∞. Occurs when numerator degree exceeds denominator degree by exactly 1.
rational inequality
An inequality involving a rational expression. Solved by finding sign changes across intervals defined by critical values.
sign chart
A tool for tracking the sign of a rational or polynomial expression across intervals defined by its critical values.
removable discontinuity
A point where a function is undefined, but the limit exists from both sides. Corresponds to a hole in the graph of a rational function.

Concept Review Questions

Answer each question in complete sentences without performing detailed computation.

  1. What is the difference between a polynomial function and a rational function? Give one example of each.
  2. How does the leading term of a polynomial determine its end behavior? Why don't the other terms matter for end behavior?
  3. Explain in your own words what multiplicity tells you about the behavior of a polynomial graph at an intercept.
  4. What is the difference between a vertical asymptote and a hole in the graph of a rational function? What algebraic event produces each?
  5. A rational function has a horizontal asymptote at y = 4. What does this mean about the long-run behavior? Can the graph cross y = 4 for finite values of x?
  6. When does a rational function have a slant asymptote instead of a horizontal asymptote?
  7. Why are the zeros of the denominator excluded from the solution of a rational inequality, even when the inequality is non-strict (≤)?
  8. Explain the Remainder Theorem in your own words. How does it connect division to function evaluation?
  9. What does the Factor Theorem tell you about the relationship between zeros and factors of a polynomial?
  10. How does understanding horizontal asymptotes prepare you for understanding limits at infinity in calculus?

Polynomial Graphing Practice

For problems 1–8, state the degree, leading coefficient, end behavior, and identify all zeros with multiplicities. Describe the graph behavior at each zero.

  1. f(x) = (x − 4)²(x + 1)
  2. f(x) = −(x + 2)(x − 3)(x − 1)
  3. f(x) = x³(x − 5)²
  4. f(x) = −2(x + 1)⁴
  5. f(x) = (x − 2)³(x + 3)²(x − 1)
  6. f(x) = 3x(x² − 4)(x + 5)
  7. f(x) = −(x² − 1)²
  8. f(x) = x²(x − 3)²(x + 2)²

For problems 9–12, sketch a rough graph. Label all intercepts.

  1. f(x) = (x − 1)²(x + 3)
  2. f(x) = −x(x − 4)(x + 2)²
  3. f(x) = (x + 1)(x − 2)³
  4. f(x) = −2(x² − 9)(x + 1)²

For problems 13–16, write a polynomial satisfying the given conditions.

  1. Degree 3; zeros at x = −1, 2, 5; positive leading coefficient.
  2. Degree 4; zeros at x = 0 (mult 2) and x = 3 (mult 2); negative leading coefficient.
  3. Degree 5; zeros at x = −2 (mult 1), x = 0 (mult 2), x = 4 (mult 2).
  4. Minimum degree; bounces at x = 1 and x = −1, crosses at x = 3.
Part BPolynomial Division Practice

Problems 17–22: use polynomial long division.

  1. (x³ − 2x² + 5x − 6) ÷ (x − 1)
  2. (2x³ + 3x² − x + 4) ÷ (x + 2)
  3. (x⁴ − 5x² + 2) ÷ (x² − 1)
  4. (3x³ − 7x + 1) ÷ (x − 3)
  5. (x³ + 8) ÷ (x + 2)
  6. (x⁴ − 1) ÷ (x − 1)

Problems 23–28: use synthetic division.

  1. (2x³ − x² + 4x − 5) ÷ (x − 2)
  2. (x⁴ − 3x³ + x − 2) ÷ (x + 1)
  3. (5x³ − 10x) ÷ (x − 2)
  4. (x³ + 27) ÷ (x + 3)
  5. (3x⁴ − 2x³ + x − 7) ÷ (x − 1)
  6. (x⁵ − 1) ÷ (x − 1)

Problems 29–32: use the Remainder Theorem to evaluate.

  1. f(x) = x³ − 4x + 1; find f(3) via division.
  2. g(x) = 2x⁴ − x³ + 3x − 5; find g(−2) via division.
  3. h(x) = x³ + 5x² − 2x + 8; find h(1) via division.
  4. k(x) = 4x³ − 3x² + 2x − 1; find k(−1) via division.

Problems 33–36: use the Factor Theorem.

  1. Is (x − 2) a factor of f(x) = x³ − 3x² + 4?
  2. Is (x + 1) a factor of g(x) = 2x³ + x² − 3x − 2?
  3. Is (x − 3) a factor of h(x) = x⁴ − 5x³ + 6x² + 4x − 8?
  4. Is (x + 5) a factor of k(x) = x³ + 4x² − 3x − 10?
Part CRational Function Practice

Problems 37–44: find the domain in interval notation.

  1. f(x) = (x + 3)/(x − 7)
  2. g(x) = 4/(x² − 16)
  3. h(x) = (x² + 1)/(x² + 4)
  4. k(x) = 2x/(x² + x − 6)
  5. f(x) = (x − 2)/(x³ − x)
  6. g(x) = (3x + 1)/(x² + 2x + 5)
  7. h(x) = (x² − 9)/(x² − 6x + 9)
  8. k(x) = (x + 1)/(x³ − 8)

Problems 45–52: find all vertical asymptotes and holes.

  1. f(x) = (x + 4)/(x² − 16)
  2. g(x) = (x² − x − 6)/(x − 3)
  3. h(x) = 2x/(x² + x)
  4. k(x) = (x² − 9)/(x² + 5x + 6)
  5. f(x) = (x² − 4)/(x² + x − 2)
  6. g(x) = (x³ − x)/(x² − 1)
  7. h(x) = 3(x−1)(x+2)/[(x+2)(x−4)]
  8. k(x) = (x² + x)/(x³ + x²)

Problems 53–60: find all horizontal and slant asymptotes.

  1. f(x) = (5x − 3)/(2x + 1)
  2. g(x) = (x² + 1)/(x³ − 4)
  3. h(x) = 3x³/(x³ + 1)
  4. k(x) = (x² − 4)/(x + 1)
  5. f(x) = (2x² + 3)/(x² − x)
  6. g(x) = (x³ + 2x)/(x² − 1)
  7. h(x) = 4x/(x² + 9)
  8. k(x) = (x² − x − 6)/(x − 4)
Part DInequality Practice

Solve each rational inequality. Write the solution in interval notation.

  1. (x − 2)/(x + 5) > 0
  2. 3/(x − 1) ≤ 0
  3. (x + 4)/(x − 3) < 0
  4. (x² − 4)/(x + 1) ≥ 0
  5. (x−1)(x+2)/(x − 4) < 0
  6. 2x/(x² − 9) > 0
  7. (x² + x − 6)/(x + 4) ≤ 0
  8. (x − 3)/[(x+1)(x−5)] ≥ 0
  9. (x² − 5x + 6)/(x² − 1) < 0
  10. (x + 3)²/(x − 2) > 0

Modeling Problems

Write a model, find and interpret all relevant features, and answer the specific question in context.

  1. Average Cost. A restaurant spends $2,400 per month on fixed overhead and $8 in ingredients per meal. Write a rational function for the average cost per meal. Find and interpret the horizontal asymptote. How many meals must be served to bring average cost below $14 per meal?
  2. Drug Concentration. The concentration (mg/L) of a drug after t hours is modeled by C(t) = 8t/(t² + 4) for t ≥ 0. Find concentration at t = 1, t = 2, and t = 4. What happens to concentration as t → ∞? What does the horizontal asymptote mean?
  3. Projectile Height. A ball is thrown upward from 6 feet with initial velocity 48 ft/s: h(t) = −16t² + 48t + 6. Find the t-intercepts and interpret them. Determine when the ball is above 30 feet. Find the maximum height.
  4. Population Model. A deer population is modeled by P(t) = 1200t/(t + 3) where t is years after reintroduction. Find P(1), P(5), and P(20). Find and interpret the horizontal asymptote. What does the model predict about long-term population?
  5. Engineering Deflection. A beam deflection model is given by y(x) = −0.001x(x − 10)(x² − 50). Identify all zeros within the physical span 0 ≤ x ≤ 10. Describe the degree, end behavior, and what the zeros mean about deflection along the beam.
  6. Monthly Profit. A company's monthly revenue model is R(x) = −2x³ + 30x² − 50x thousand dollars, where x is the number of advertising campaigns (0 ≤ x ≤ 12). Find the zeros, describe the behavior on the physical interval, and estimate the number of campaigns that maximizes revenue.

Error Analysis

Identify the error in each solution and provide the correct answer.

Problem 1
f(x) = (x² − 9)/(x − 3) has a vertical asymptote at x = 3.

Find the error and correct it. Does this function have an asymptote at x = 3, a hole, or something else?

Problem 2
g(x) = (4x³ − 2)/(x³ + 5x) has horizontal asymptote y = 0 because the denominator has more terms.

Find the error and state the correct horizontal asymptote.

Problem 3
Solution to (x−2)/(x+1) ≤ 0 is [−1, 2] because the expression is zero at both endpoints and negative between them.

Find the error. Write the correct solution in interval notation.

Problem 4
Synthetic division shows the remainder when f(x) = x³ + 2x − 5 is divided by (x − 2) is 7. Therefore f(2) = −7.

Find the error. What does the Remainder Theorem actually say about f(2)?

Problem 5
f(x) = x³(x − 4)² is graphed with the curve crossing the x-axis at x = 4 because 4 is a zero.

Find the error. What is the correct behavior at x = 4?

Problem 6
The end behavior of f(x) = −3x⁴ + 5x³ − 2 is "falls left, rises right" because the degree is 4 (even).

Find the error. What is the correct end behavior?

Chapter 8 Checkpoint

Chapter 8 Checkpoint
Complete all problems without assistance — assesses all key skills and concepts from Chapter 8
Part A — Polynomial Functions
  1. For f(x) = 2x⁴ − 5x³ + x² − 3, state the degree, leading coefficient, and describe both end behaviors.
  2. Factor completely: f(x) = x⁴ − 5x² + 4. Then list all zeros with multiplicities.
  3. Use polynomial long division to find the quotient and remainder when 3x³ − x² + 2x − 5 is divided by x − 1.
  4. Use the Factor Theorem to determine whether (x + 2) is a factor of g(x) = 2x³ + 3x² − 2x − 3.
  5. Write a degree-4 polynomial with zeros at x = −2 (mult 1), x = 0 (mult 2), and x = 3 (mult 1), with a negative leading coefficient.
  6. Sketch the graph of f(x) = −(x + 1)(x − 2)²(x − 4). Label all intercepts and describe end behavior.
Part B — Rational Functions
  1. Find the domain of f(x) = (x² − 25)/(x² − x − 20) in interval notation.
  2. Find all holes and vertical asymptotes of f(x) = (x² − x − 6)/(x² − 4).
  3. Find the horizontal asymptote of g(x) = (3x² − 7)/(x² + 2x + 1).
  4. Find the slant asymptote of h(x) = (x² + 5x − 2)/(x − 2).
  5. Find all intercepts, asymptotes, and holes of f(x) = (x² − 1)/(x² − x − 2), then sketch the graph.
Part C — Rational Inequalities
  1. Solve (x + 5)/(x − 2) < 0 and write the solution in interval notation.
  2. Solve (x² − x − 6)/(x + 4) ≥ 0 and write the solution in interval notation.
Part D — Modeling
  1. A manufacturer's total production cost is C(x) = 500 + 15x dollars. Write the average cost function, find its horizontal asymptote, and interpret what the asymptote means in this production context.
  2. A concentration model is C(t) = 12t/(t² + 9) mg/L after t minutes. Find C(0), C(3), and the long-run concentration. What does the horizontal asymptote tell you about the substance in the tank over time?
Looking Ahead
Chapter 9: Exponential and Logarithmic Functions

Chapter 8 deepened your ability to read polynomial and rational graphs and connected those ideas to the language of limits. Chapter 9 returns to a function family that grows in a fundamentally different way — not by powers of x, but by powers of a constant base. Exponential functions model multiplicative change: growth that compounds, decay that halves, interest that accumulates. Logarithmic functions are their inverses, turning exponential questions into additive ones. Together, these two families model finance, population, physics, medicine, and the scale of nearly everything in the natural world. The calculus of exponential and logarithmic functions will be among the most important applications you encounter in Calculus I.

MGU Library Connection
  • Appendix J: Polynomial and Rational Function Reference — this textbook
  • Appendix P: Limit and Continuity Preview Reference — this textbook
  • Algebra II, Chapter 6: Polynomial Functions and Factoring
  • Algebra II, Chapter 8: Rational Functions and Expressions
  • Calculus I, Chapter 1: Limits and Continuity
  • Calculus I, Chapter 2: Derivatives of Polynomial and Rational Functions
  • MGU Dictionary entries: polynomial, rational function, asymptote, limit, continuity, multiplicity