MGU Mathematics Series — Algebra II
Part III: Rational, Radical, and Power Functions
Restrictions · Asymptotes · Rational Equations · Inverse Variation
Chapter 7 of 14
MGU Mathematics Series  ·  Algebra II  ·  Library Textbook Edition

Chapter 7:
Rational Expressions
and Functions

Domain restrictions, simplification, operations, solving rational equations, asymptotes, holes, and inverse variation — the full language of polynomial ratios

■ Math Level: Algebra II ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapter 6: Polynomial Equations and Theorems  ·  Prepares for Chapter 8: Radical Functions and Rational Exponents

Chapter 7 — Front Matter

Purpose and Overview

Division is not just an operation. It is a question: what happens when you share, split, or scale? Rational expressions ask that question with polynomials — and the answer always depends on what the denominator is allowed to be.

You have spent considerable time working with polynomials — adding them, multiplying them, factoring them. Now you take the next step: dividing them. When one polynomial is placed over another in a fraction, the result is called a rational expression. These objects are central to algebra, precalculus, and a wide range of applied mathematics.

But rational expressions are not simply "algebraic fractions" in a mechanical sense. They carry a hidden layer of meaning: every rational expression comes with conditions on what values its variable is allowed to take. A denominator equal to zero does not produce a very large number, a very small number, or infinity — it produces nothing at all, a value that is simply undefined. Understanding, identifying, and respecting these restrictions is the central skill of this chapter.

In this chapter you will learn to simplify rational expressions without losing their domain restrictions, perform all four arithmetic operations on them, solve equations that contain them (while catching solutions that secretly violate the restrictions), and read the graphs of rational functions — a rich family of curves featuring asymptotes, holes, and behavior that polynomial graphs cannot produce.

Opening Question

A company spends $8,000 in fixed costs each month and $12 in variable costs for every unit it produces. If it produces $x$ units, the average cost per unit is $A(x) = \frac{8000 + 12x}{x}$. What happens to the average cost as production grows very large? Can the average cost ever reach exactly $12? Why does the formula break down at $x = 0$? By the end of this chapter, you will be able to answer all of these questions — and understand the graph that makes the answers visible.

14Sections
6Objectives
18Key Terms
20+Worked Examples

Learning Objectives

  1. 1Define rational expressions and identify their domains by finding excluded values.
  2. 2Simplify, multiply, divide, add, and subtract rational expressions correctly, preserving all restrictions.
  3. 3Simplify complex fractions using two reliable methods.
  4. 4Solve rational equations by clearing denominators, and identify and discard extraneous solutions.
  5. 5Describe and sketch rational functions, including vertical and horizontal asymptotes and holes.
  6. 6Model real-world situations using rational functions and inverse variation.
Chapter 7 — Contents

Sections at a Glance

Sections 7.1 – 7.3
What Rational Expressions Are and How to Simplify Them
Definition · Restrictions and Domain · Simplification in Lowest Terms
Section 7.1

What Rational Expressions Are

A rational expression is any expression of the form $\dfrac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \neq 0$. The word rational comes from ratio — a rational expression is a ratio of polynomials, in exactly the same way that a rational number is a ratio of integers.

Recall that a polynomial is a sum of terms of the form $ax^n$ where $a$ is a real number and $n$ is a non-negative integer. This means rational expressions exclude variables in exponents, variables under radicals, or variables in their own sub-denominators.

ExpressionNumeratorDenominatorRational?
$\dfrac{5}{x^2+1}$$5$$x^2+1$Yes
$\dfrac{x^2-1}{x-1}$$x^2-1$$x-1$Yes
$\dfrac{x+3}{4}$$x+3$$4$ (nonzero constant)Yes
$\dfrac{\sqrt{x}}{x+1}$$\sqrt{x}$ — not a polynomial$x+1$No

Rational expressions appear throughout mathematics and its applications. Rate problems use distance divided by time. Physics describes gravitational force and electric field intensity with rational models. Economics uses average cost and price elasticity. Understanding rational expressions gives you a language to describe how quantities scale, diminish, and interact — and to interpret the conditions under which those descriptions break down.

Looking at Values

Consider $\dfrac{x+1}{x-3}$. When $x = 5$, the value is $\frac{6}{2} = 3$. When $x = 0$, the value is $\frac{1}{-3} = -\frac{1}{3}$. But when $x = 3$, the denominator equals zero — the expression is undefined. This is not a technicality; it is the fundamental fact that shapes everything in this chapter.

Practice 7.1Concept Check
  1. Which of the following are rational expressions? Explain your reasoning for each. (a) $\dfrac{x^2-3x+1}{x+4}$    (b) $\dfrac{\sqrt{x+1}}{x-2}$    (c) $\dfrac{7}{x^3-1}$    (d) $\dfrac{x+2}{x^{-1}+1}$
  2. Evaluate $\dfrac{2x+3}{x-1}$ at $x = 0$, $x = 4$, and $x = -2$. Then explain what happens at $x = 1$.
  3. Give two real-world situations that could be modeled by a rational expression. Identify what each variable represents.
Section 7.2

Restrictions and Excluded Values

Core Principle

A rational expression $\dfrac{P(x)}{Q(x)}$ is undefined at any value of $x$ for which $Q(x) = 0$. These values are called excluded values or restrictions. The domain is all real numbers except those excluded values.

To find the domain: set the denominator equal to zero and solve. Every solution is excluded.

Example 7.2.1
Finding the domain — linear denominator
  • ProblemFind the domain of $\dfrac{x+5}{x-7}$.
  • ThinkSet the denominator equal to zero and solve for the excluded value.
  • Solvex - 7 = 0 → x = 7
  • CheckSubstituting $x = 7$ gives denominator $0$. ✓ All other values are defined.
Answer: Domain = all real numbers except $x = 7$. Restriction: $x \neq 7$.
Example 7.2.2
Finding the domain — factorable quadratic denominator
  • ProblemFind the domain of $\dfrac{3x}{x^2-9}$.
  • ThinkFactor the denominator first, then set each factor equal to zero.
  • Solve x² - 9 = (x - 3)(x + 3) = 0 x - 3 = 0 → x = 3 x + 3 = 0 → x = -3
Answer: Domain = all real numbers except $x = 3$ and $x = -3$. In interval notation: $(-\infty,-3) \cup (-3,3) \cup (3,\infty)$.
Example 7.2.3
Finding the domain — denominator with no real zeros
  • ProblemFind the domain of $\dfrac{x^2-1}{x^2+1}$.
  • ThinkSet $x^2 + 1 = 0$. Can this equal zero for any real number?
  • Solvex² + 1 = 0 → x² = -1 (no real solutions)
Answer: Domain = all real numbers. No restrictions. The expression is defined for every real $x$.
Restrictions Belong to the Expression — Not Just the Denominator You See

When you simplify a rational expression, you may cancel a factor from the denominator, making its zero invisible in the result. But that restriction does not disappear. Simplification changes the form of an expression, not its domain. This principle is the most important idea in Section 7.3.

Practice 7.2Restriction Practice
  1. $\dfrac{x}{x+6}$
  2. $\dfrac{3x-1}{x^2-16}$
  3. $\dfrac{5}{x^2+4}$
  4. $\dfrac{x+3}{x^2-x-6}$
  5. $\dfrac{2x^2}{x^3-8}$
  6. $\dfrac{x^2-4x}{x^2+4x}$
  7. $\dfrac{1}{x(x-1)(x+5)}$
  8. $\dfrac{x+2}{x^4-x^2}$

For each: state all excluded values and write the domain in interval notation.

Section 7.3

Simplifying Rational Expressions

A rational expression is in simplified form (lowest terms) when the numerator and denominator share no common polynomial factors other than constants. To simplify, factor both numerator and denominator completely, then cancel any factors that appear in both. The rule underlying cancellation is: $\dfrac{a \cdot c}{b \cdot c} = \dfrac{a}{b}$, provided $c \neq 0$ — which in our context corresponds to an actual restriction that must be recorded.

Step-by-Step Process for Simplification

Step 1 — State all restrictions from the original denominator.

Step 2 — Factor the numerator and denominator completely.

Step 3 — Cancel common factors.

Step 4 — Write the result with the restrictions attached.

Example 7.3.1
Simplifying — difference of squares
  • ProblemSimplify $\dfrac{x^2-4}{x-2}$.
  • Restrict$x - 2 = 0 \Rightarrow x \neq 2$.
  • Factorx² - 4 = (x - 2)(x + 2)
  • Cancel[(x - 2)(x + 2)] / (x - 2) = x + 2
  • CheckThe simplified form $x + 2$ appears defined at $x = 2$, but the original expression is not. The restriction $x \neq 2$ must carry forward.
Answer: $x + 2$, where $x \neq 2$.
Example 7.3.2
Simplifying — trinomial numerator and denominator
  • ProblemSimplify $\dfrac{x^2+3x+2}{x^2-1}$.
  • Restrict$x^2 - 1 = (x-1)(x+1) = 0 \Rightarrow x \neq 1,\, x \neq -1$.
  • FactorNumerator: (x + 1)(x + 2) Denominator: (x - 1)(x + 1)
  • Cancel[(x + 1)(x + 2)] / [(x - 1)(x + 1)] = (x + 2) / (x - 1)
Answer: $\dfrac{x+2}{x-1}$, where $x \neq 1$ and $x \neq -1$.
Example 7.3.3
Simplifying — recognizing opposite factors
  • ProblemSimplify $\dfrac{x-5}{5-x}$.
  • ThinkNotice that $5 - x = -(x - 5)$. So the denominator is the opposite of the numerator.
  • Solve(x - 5) / [-(x - 5)] = -1
Answer: $-1$, where $x \neq 5$.
What Cannot Be Cancelled — Terms vs. Factors

Wrong: $\dfrac{x^2+4}{x} = x + 4$. Here $x^2$ and $x$ are terms in the numerator, not factors of the whole numerator.

Correct: $\dfrac{x(x+4)}{x} = x + 4$. Here $x$ is a factor of the entire numerator. Only factors cancel — never terms within a sum.

Practice 7.3Simplification Practice
  1. $\dfrac{6x^2}{3x}$
  2. $\dfrac{x^2-25}{x+5}$
  3. $\dfrac{x^2-x-12}{x^2-16}$
  4. $\dfrac{2x^2+6x}{x^2+3x}$
  5. $\dfrac{x^2-4x+4}{x^2-4}$
  6. $\dfrac{x^3-8}{x^2-4}$
  7. $\dfrac{x^2+6x+9}{x^2+x-6}$
  8. $\dfrac{6-x}{x^2-36}$

Simplify completely. State all restrictions.

Sections 7.4 – 7.7
Arithmetic Operations on Rational Expressions
Multiplication · Division · Addition and Subtraction · Complex Fractions
Section 7.4

Multiplying Rational Expressions

Multiplication Rule

$\dfrac{P}{Q} \cdot \dfrac{R}{S} = \dfrac{P \cdot R}{Q \cdot S}$

In practice: factor first, cancel common factors across any numerator and any denominator, then multiply the remaining factors. This avoids working with large polynomials.

Example 7.4.1
Multiplying — factor and cancel across both fractions
  • ProblemMultiply $\dfrac{x^2-9}{x+4} \cdot \dfrac{x+4}{x+3}$.
  • Restrict$x \neq -4$ (from first denominator); $x \neq -3$ (from second denominator).
  • Factor[(x-3)(x+3)] / (x+4) · (x+4) / (x+3)
  • Cancel(x-3)·[x+3]·[x+4] / [x+4]·[x+3] = x - 3
  • CheckLet $x = 0$: original gives $\frac{-9}{4} \cdot \frac{4}{3} = \frac{-9}{3} = -3$. Simplified: $0 - 3 = -3$. ✓
Answer: $x - 3$, where $x \neq -4$ and $x \neq -3$.
Example 7.4.2
Multiplying — factoring trinomials before cancelling
  • ProblemMultiply $\dfrac{2x^2+x-1}{x^2-4} \cdot \dfrac{x^2-2x}{4x-2}$.
  • Restrict$x \neq 2,\, x \neq -2$ (from $x^2-4$); $x \neq \frac{1}{2}$ (from $4x - 2 = 0$).
  • Factor Num 1: (2x - 1)(x + 1) Den 1: (x - 2)(x + 2) Num 2: x(x - 2) Den 2: 2(2x - 1)
  • CancelCancel (2x-1) and (x-2). Result: x(x+1) / [2(x+2)]
Answer: $\dfrac{x(x+1)}{2(x+2)}$, where $x \neq 2,\, x \neq -2,\, x \neq \frac{1}{2}$.
Section 7.5

Dividing Rational Expressions

Division Rule

$\dfrac{P}{Q} \div \dfrac{R}{S} = \dfrac{P}{Q} \cdot \dfrac{S}{R}$

Dividing equals multiplying by the reciprocal. Important: the numerator of the divisor ($R$) becomes a denominator after flipping — so $R \neq 0$ becomes an additional restriction.

Example 7.5.1
Dividing rational expressions
  • ProblemDivide $\dfrac{x^2-25}{x+3} \div \dfrac{x-5}{x^2-9}$.
  • Restrict$x \neq -3,\, x \neq 3$ (denominators); $x \neq 5$ (divisor's numerator, since it becomes a denominator).
  • Rewrite[x² - 25] / (x + 3) · (x² - 9) / (x - 5)
  • Factor[(x-5)(x+5)] / (x+3) · [(x-3)(x+3)] / (x-5)
  • CancelCancel (x-5) and (x+3). Result: (x+5)(x-3)
Answer: $(x+5)(x-3)$, where $x \neq \pm 3,\, x \neq 5$.
Section 7.6

Adding and Subtracting Rational Expressions

To add or subtract rational expressions, they must share a common denominator. When denominators differ, find the Least Common Denominator (LCD) by factoring each denominator and taking each distinct factor to its highest power.

Example 7.6.1
Adding — same denominator
  • ProblemAdd $\dfrac{3x}{x+2} + \dfrac{6}{x+2}$.
  • ThinkSame denominator — combine numerators directly, then simplify.
  • Solve (3x + 6) / (x + 2) = 3(x + 2) / (x + 2) = 3
  • CheckEven though the result is just $3$, the restriction $x \neq -2$ from the original denominator remains.
Answer: $3$, where $x \neq -2$.
Example 7.6.2
Adding — different denominators
  • ProblemAdd $\dfrac{2}{x} + \dfrac{3}{x+1}$.
  • Restrict$x \neq 0,\, x \neq -1$. LCD $= x(x+1)$.
  • Rewrite 2(x+1) / [x(x+1)] + 3x / [x(x+1)]
  • Solve(2x + 2 + 3x) / [x(x+1)] = (5x + 2) / [x(x+1)]
Answer: $\dfrac{5x+2}{x(x+1)}$, where $x \neq 0,\, x \neq -1$.
Example 7.6.3
Subtracting — different denominators
  • ProblemSubtract $\dfrac{5}{x^2-4} - \dfrac{1}{x+2}$.
  • Factor$x^2 - 4 = (x-2)(x+2)$. Restrictions: $x \neq 2,\, x \neq -2$. LCD $= (x-2)(x+2)$.
  • Rewrite5/[(x-2)(x+2)] - (x-2)/[(x-2)(x+2)]
  • Solve[5 - (x - 2)] / [(x-2)(x+2)] = (7 - x) / [(x-2)(x+2)]
Answer: $\dfrac{7-x}{(x-2)(x+2)}$, where $x \neq 2,\, x \neq -2$.
Distribute the Negative Over the Entire Numerator

When subtracting, the negative sign applies to every term in the numerator of the second fraction. $\dfrac{A}{Q} - \dfrac{B-C}{Q} = \dfrac{A-B+C}{Q}$, not $\dfrac{A-B-C}{Q}$. Missing this is one of the most common errors in the chapter. See Section 7.14, Mistake 4.

Section 7.7

Complex Fractions

A complex fraction is a fraction in which the numerator, the denominator, or both contain fractions themselves. Two reliable methods handle them: combine the numerator and denominator into single fractions then divide (Method 1), or multiply the entire expression through by the LCD of all inner fractions (Method 2).

Example 7.7.1
Complex fraction — Method 2 (multiply by LCD)
  • ProblemSimplify $\dfrac{\dfrac{3}{x}+2}{\dfrac{5}{x}-1}$.
  • ThinkThe LCD of all inner fractions is $x$. Multiply every term in numerator and denominator by $x$.
  • Solve x · (3/x) + x · 2 = 3 + 2x (numerator) x · (5/x) - x · 1 = 5 - x (denominator) Result: (3 + 2x) / (5 - x)
Answer: $\dfrac{3+2x}{5-x}$, where $x \neq 0,\, x \neq 5$.
Example 7.7.2
Complex fraction — Method 2 with cancellation
  • ProblemSimplify $\dfrac{1-\dfrac{1}{x^2}}{1+\dfrac{1}{x}}$.
  • ThinkLCD of inner fractions is $x^2$. Multiply every term by $x^2$.
  • Solve Numerator: x² - 1 = (x-1)(x+1) Denominator: x² + x = x(x+1) Result: (x-1)(x+1) / [x(x+1)] = (x-1)/x
Answer: $\dfrac{x-1}{x}$, where $x \neq 0,\, x \neq -1$.
Practice 7.4 – 7.7Operations Practice

Perform the indicated operation and simplify. State all restrictions.

Multiplication

  1. $\dfrac{x+3}{x-1} \cdot \dfrac{x^2-1}{x^2-9}$
  2. $\dfrac{x^2-4}{x^2+5x+6} \cdot \dfrac{x+3}{x-2}$
  3. $\dfrac{3x}{x+2} \cdot \dfrac{x^2-4}{6x^2}$

Division

  1. $\dfrac{x^2-9}{x+4} \div \dfrac{x-3}{x^2-16}$
  2. $\dfrac{x^2-1}{x+3} \div \dfrac{x-1}{x^2+6x+9}$
  3. $\dfrac{4x^2}{x-5} \div \dfrac{8x^3}{x^2-25}$

Addition and Subtraction

  1. $\dfrac{3}{x+1} + \dfrac{2}{x-1}$
  2. $\dfrac{x}{x-3} - \dfrac{9}{x^2-9}$
  3. $\dfrac{2}{x^2-4} + \dfrac{3}{x+2}$
  4. $\dfrac{x+1}{x^2-x-6} - \dfrac{x-2}{x^2+x-2}$

Complex Fractions

  1. $\dfrac{x/y}{x^2/y^2}$
  2. $\dfrac{\dfrac{1}{x}+1}{1-\dfrac{1}{x}}$
  3. $\dfrac{\dfrac{3}{x^2}-\dfrac{1}{x}}{\dfrac{2}{x}+1}$
Sections 7.8 – 7.9
Solving Rational Equations
Clearing Denominators · Extraneous Solutions · Checking is Mandatory
Section 7.8

Solving Rational Equations

A rational equation is an equation containing at least one rational expression. The standard strategy is to clear all fractions at once by multiplying both sides by the LCD of every denominator in the equation. This converts the rational equation into a polynomial equation that is easier to solve.

Six-Step Process for Rational Equations

Step 1 — State the restrictions from all denominators.

Step 2 — Find the LCD of all denominators.

Step 3 — Multiply both sides by the LCD. This eliminates all fractions.

Step 4 — Solve the resulting polynomial equation.

Step 5 — Check every solution against the restrictions. Discard extraneous ones.

Step 6 — State valid solutions only.

Example 7.8.1
Solving a rational equation — one variable term
  • ProblemSolve $\dfrac{2}{x} + \dfrac{1}{3} = \dfrac{5}{x}$.
  • Restrict$x \neq 0$. LCD $= 3x$.
  • Multiply3x·(2/x) + 3x·(1/3) = 3x·(5/x) 6 + x = 15 x = 9
  • Check$9 \neq 0$ ✓. Substitute: $\frac{2}{9} + \frac{1}{3} = \frac{2}{9} + \frac{3}{9} = \frac{5}{9}$ ✓.
Answer: $x = 9$.
Example 7.8.2
Solving — equation reduces to a contradiction
  • ProblemSolve $\dfrac{x}{x-2} = \dfrac{3}{x-2} + 1$.
  • Restrict$x \neq 2$. LCD $= x - 2$.
  • Multiplyx = 3 + (x - 2) = x + 1 0 = 1 (contradiction)
  • InterpretA contradiction means the equation has no solution — not an algebra error.
Answer: No solution.
Example 7.8.3
Solving — quadratic after clearing
  • ProblemSolve $\dfrac{3}{x+2} + \dfrac{1}{x} = \dfrac{5}{2x}$.
  • Restrict$x \neq -2,\, x \neq 0$. LCD $= 2x(x+2)$.
  • Multiply 6x + 2(x+2) = 5(x+2) 6x + 2x + 4 = 5x + 10 8x + 4 = 5x + 10 3x = 6 → x = 2
  • Check$2 \neq -2$ and $2 \neq 0$ ✓. Valid solution.
Answer: $x = 2$.
Section 7.9

Extraneous Solutions

When we multiply both sides of an equation by the LCD — which may contain a variable — we assume that the LCD is nonzero. But what if a solution makes the LCD equal zero? That value was introduced by the multiplication step itself, not by the original equation. It is called an extraneous solution.

An extraneous solution does not arise from an algebra error. It is a structural feature of the solving method. Checking solutions is not optional — it is part of the solution process.

Example 7.9.1
Identifying an extraneous solution
  • ProblemSolve $\dfrac{x+2}{x-2} + \dfrac{x}{x+2} = \dfrac{8}{x^2-4}$.
  • Restrict$x^2 - 4 = (x-2)(x+2)$. Restrictions: $x \neq 2,\, x \neq -2$. LCD $= (x-2)(x+2)$.
  • Multiply (x+2)² + x(x-2) = 8 x²+4x+4 + x²-2x = 8 2x²+2x+4 = 8 → x²+x-2 = 0 (x+2)(x-1) = 0 → x = -2 or x = 1
  • Check$x = -2$: violates restriction $x \neq -2$. Extraneous — discard. $x = 1$: valid. Verify: $\frac{3}{-1} + \frac{1}{3} = \frac{8}{-3}$ → $-3 + \frac{1}{3} = -\frac{8}{3}$ ✓.
Answer: $x = 1$ only. (Note: $x = -2$ is extraneous.)
SituationWhat Happens
Solution equals a restrictionExtraneous — discard
Solution does not equal any restrictionValid — keep
All candidate solutions are extraneousNo solution
Equation reduces to a true tautologyAll non-restricted values are solutions
Practice 7.8 – 7.9Equation Practice

Solve each equation. Identify any extraneous solutions. State restrictions first.

  1. $\dfrac{3}{x} = \dfrac{1}{4}$
  2. $\dfrac{x+1}{x-2} = \dfrac{3}{x-2}$
  3. $\dfrac{2}{x-1}+\dfrac{3}{x+1}=\dfrac{10}{x^2-1}$
  4. $\dfrac{x}{x+3}-\dfrac{1}{x-3}=\dfrac{18}{x^2-9}$
  5. $\dfrac{5}{x-2}=\dfrac{x+3}{x^2-4}$
  6. $\dfrac{1}{x}+\dfrac{1}{x-1}=\dfrac{1}{x(x-1)}$
  7. $\dfrac{x+2}{x}=\dfrac{3x}{x+1}$
  8. $\dfrac{x}{x-4}=\dfrac{4}{x-4}+1$
Sections 7.10 – 7.13
Rational Functions, Graphs, and Models
Rational Functions · Asymptotes · Holes · Inverse Variation · Applications
Section 7.10

Rational Functions

A rational function is a function of the form $f(x) = \dfrac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials and $Q(x) \not\equiv 0$. The domain is all real numbers except the zeros of $Q(x)$.

The most fundamental rational function is the reciprocal function $f(x) = \dfrac{1}{x}$. It is undefined at $x = 0$. For large positive $x$, $f(x)$ is small and positive. For $x$ just greater than 0, $f(x)$ is very large. For negative $x$, $f(x)$ is negative. The graph has two branches — one in the first quadrant and one in the third — and never touches the coordinate axes. The axes are asymptotes of this graph.

Behaviors Unique to Rational Functions

Rational functions can exhibit behaviors that polynomial functions cannot: vertical asymptotes (function approaches $\pm\infty$), horizontal asymptotes (function approaches a constant), holes (removable discontinuities), and oblique asymptotes. Each of these is explained fully in Sections 7.11 and 7.12.

Standard Analysis for $f(x) = P(x)/Q(x)$

1. Domain — find where $Q(x) = 0$.

2. Simplify — factor and cancel; identify holes at cancelled factors.

3. Vertical asymptotes — zeros of $Q$ remaining after simplification.

4. Horizontal/oblique asymptote — compare degrees of $P$ and $Q$.

5. $x$-intercepts — zeros of simplified numerator.

6. $y$-intercept — evaluate $f(0)$ if defined.

7. Sign analysis — determine where $f$ is positive or negative.

Section 7.11

Vertical and Horizontal Asymptotes

A vertical asymptote occurs at $x = a$ if $f(x) \to \pm\infty$ as $x \to a$ from either side. This happens when $x = a$ is a zero of the denominator that does not cancel with a zero of the numerator. After simplifying a rational function, the vertical asymptotes occur at values where the simplified denominator equals zero.

Example 7.11.1
Asymptotic behavior near a vertical asymptote
  • Function$f(x) = \dfrac{1}{x-2}$. Vertical asymptote at $x = 2$.
  • From rightAs $x \to 2^+$: denominator $\to 0^+$, so $f(x) \to +\infty$.
  • From leftAs $x \to 2^-$: denominator $\to 0^-$, so $f(x) \to -\infty$.
Vertical asymptote $x = 2$; function changes sign across the asymptote.

A horizontal asymptote describes the end behavior — what $f(x)$ approaches as $x \to \pm\infty$. Let $n$ be the degree of the numerator and $m$ the degree of the denominator.

Degree ComparisonHorizontal AsymptoteExample
$n < m$ (numerator lower)$y = 0$ (the $x$-axis)$\dfrac{3x}{x^2+1}$: $y = 0$
$n = m$ (same degree)$y = \dfrac{\text{leading coeff of }P}{\text{leading coeff of }Q}$$\dfrac{4x^2-1}{2x^2+3}$: $y = 2$
$n = m+1$ (numerator one higher)Oblique (slant) asymptote — use long division$\dfrac{x^2-1}{x+3}$: $y = x-3$
$n > m+1$No horizontal or simple oblique asymptote
Crossing Asymptotes

Vertical asymptotes are never crossed by the function. Horizontal asymptotes can be crossed for finite values of $x$ — they only describe end behavior as $x \to \pm\infty$. Finding where $f(x) = $ (the horizontal asymptote value) tells you any crossing point.

Section 7.12

Holes in Graphs

When a factor cancels from both numerator and denominator during simplification, the zero of that factor becomes a hole — formally called a removable discontinuity. The function is still undefined at that point, but unlike a vertical asymptote, the function approaches a finite limit there. On a graph, a hole is shown as an open circle.

Holes vs. Vertical Asymptotes

Hole: Factor cancels from both numerator and denominator. $f(x)$ approaches a finite limit. Shown as an open circle. $y$-coordinate found by substituting into the simplified expression.

Vertical Asymptote: Factor remains in denominator after simplification. $f(x) \to \pm\infty$. Shown as a dashed vertical line.

Example 7.12.1
Finding a hole and a vertical asymptote in the same function
  • ProblemAnalyze $f(x) = \dfrac{x^2+x-6}{x^2-x-2}$.
  • FactorNumerator: (x+3)(x-2) Denominator: (x-2)(x+1)
  • SimplifyCancel (x-2): simplified form = (x+3)/(x+1)
  • Hole at x=2Substitute into simplified form: $\frac{2+3}{2+1} = \frac{5}{3}$. Hole at $\left(2,\frac{5}{3}\right)$.
  • V. Asymptote$x + 1 = 0 \Rightarrow x = -1$. Remains in denominator after simplification. Vertical asymptote at $x = -1$.
  • H. AsymptoteDegrees equal (both 1 in simplified form); ratio of leading coefficients $= 1$. Horizontal asymptote $y = 1$.
Hole at $\left(2,\tfrac{5}{3}\right)$; vertical asymptote $x = -1$; horizontal asymptote $y = 1$.
Diagram Instruction — Rational Function Graph

Draw a coordinate plane with dashed vertical line $x = -1$ and dashed horizontal line $y = 1$. Sketch two branches of the simplified function $(x+3)/(x+1)$: one in the region $x > -1$, one in the region $x < -1$. Each branch approaches the asymptotes but never crosses the vertical one. Mark an open circle at $\left(2,\frac{5}{3}\right)$ on the right branch to indicate the hole.

Section 7.13

Rational Models and Inverse Variation

Two quantities $x$ and $y$ exhibit inverse variation if $y = \dfrac{k}{x}$ for some nonzero constant $k$, called the constant of variation. As $x$ increases, $y$ decreases proportionally — they vary inversely. Equivalently, the product $xy = k$ is constant.

Variation TypeEquation FormExample
Inverse$y = k/x$Speed and travel time for fixed distance
Joint$y = kxz$Area of rectangle: $A = lw$
Combined$y = kx/z$Pressure: $P = kT/V$ (ideal gas)
Example 7.13.1
Inverse variation — finding the constant and a missing value
  • ProblemThe time $t$ to complete a job varies inversely with the number of workers $n$. With 4 workers the job takes 9 hours. How long with 6 workers?
  • Set upt = k/n → 9 = k/4 → k = 36
  • Solvet = 36/6 = 6 hours
  • CheckMore workers should mean less time: $6 < 9$ ✓.
Answer: 6 hours with 6 workers.
Example 7.13.2
Average cost model — horizontal asymptote interpretation
  • ProblemA company has fixed costs of $\$8{,}000$ and variable costs of $\$12$ per unit. Write the average cost function $A(x)$ and find its horizontal asymptote. Interpret the result.
  • ModelA(x) = (8000 + 12x) / x = 12 + 8000/x
  • AsymptoteAs $x \to \infty$, $8000/x \to 0$, so $A(x) \to 12$. Horizontal asymptote: $y = 12$.
  • InterpretThe average cost can never reach exactly $\$12$ — no matter how many units are produced, the fixed costs are still being spread across them. But as production grows, the average cost approaches $\$12$ (the variable cost per unit).
Answer: $A(x) = 12 + \dfrac{8000}{x}$; horizontal asymptote $y = 12$; average cost approaches but never reaches $\$12$ per unit.
Example 7.13.3
Work-rate problem
  • ProblemPipe A fills a tank in 6 hours; pipe B in 4 hours. How long do they take together?
  • Set up1/6 + 1/4 = 1/t LCD = 12: 2/12 + 3/12 = 1/t → 5/12 = 1/t t = 12/5 = 2.4 hours
  • Check$2.4 < 4$ (faster than either pipe alone) ✓.
Answer: $t = 2.4$ hours (2 hours 24 minutes) working together.
Practice 7.10 – 7.13Graphing Rational Functions and Applications

For each function: (a) state the domain, (b) find all asymptotes and holes with coordinates, (c) find intercepts, (d) describe end behavior.

  1. $f(x) = \dfrac{1}{x+3}$
  2. $f(x) = \dfrac{2x}{x-1}$
  3. $f(x) = \dfrac{x^2-4}{x-2}$
  4. $f(x) = \dfrac{x+1}{x^2-x-6}$
  5. $f(x) = \dfrac{x^2}{x^2-4}$
  6. $f(x) = \dfrac{x^2+2x-3}{x^2-1}$

Applications — solve and interpret each answer in context.

  1. Work Rate. Maria can paint a room in 5 hours; José in 8 hours. How long do they take together?
  2. Average Cost. A factory has fixed monthly costs of $\$12{,}000$ and variable costs of $\$20$ per unit. Write and interpret the average cost function. What is the horizontal asymptote?
  3. Inverse Variation. The resistance $R$ of a wire varies inversely with its cross-sectional area $A$. A wire with area $0.01\,\text{cm}^2$ has resistance $50\,\Omega$. Find the resistance when the area is $0.025\,\text{cm}^2$.
  4. Optics. The thin lens equation is $\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}$. Solve for $d_i$ as a function of $d_o$ and $f$. Then find $d_i$ when $f = 20\,\text{cm}$ and $d_o = 60\,\text{cm}$.
Section 7.14
Common Mistakes with Rational Expressions
Eight Recurring Errors · Explanations and Corrections
Section 7.14

Common Rational Expression Mistakes

Mistake 1 — Canceling Terms Instead of Factors
(x² + 5) / x = x + 5 Correct: x² + 5 cannot be simplified this way. x does not divide the entire numerator as a factor. Compare: x(x + 5)/x = x + 5 ✓ — here x IS a factor of the whole numerator.
Mistake 2 — Canceling Across Addition
(x + 3)/(x + 7) = 3/7 Correct: x cannot be cancelled from a sum. Only factors of the entire numerator and denominator cancel. (x + 3)/(x + 7) is already in simplest form.
Mistake 3 — Dropping Restrictions After Simplification
(x² - 9)/(x - 3) = x + 3 (no restriction stated) Correct: Write x + 3, where x ≠ 3. The original expression was undefined at x = 3. Simplification changes the form, not the domain.
Mistake 4 — Mis-distributing the Negative in Subtraction
x/(x+2) − (x−4)/(x+2) = (x − x − 4)/(x+2) = −4/(x+2) Correct: −(x − 4) = −x + 4, so the numerator is x − x + 4 = 4. Answer: 4/(x+2). The negative sign distributes over every term in the second numerator.
Mistake 5 — Adding by Adding Numerators and Denominators
1/x + 1/(x+2) = 2/(2x+2) Correct: Find the LCD = x(x+2), rewrite, then add numerators: (x+2+x)/[x(x+2)] = (2x+2)/[x(x+2)]. Fraction addition requires a common denominator — you cannot add numerators and denominators separately.
Mistake 6 — Accepting Extraneous Solutions
Solving and getting x = 3, then accepting it without checking — when x = 3 makes a denominator zero. Correct: Always substitute every candidate solution into the original equation. Any value that makes a denominator zero must be discarded as extraneous.
Mistake 7 — Treating "No Solution" as an Algebra Error
After clearing, the equation gives 0 = 3. Student re-solves repeatedly looking for an answer. Correct: A contradiction after correct algebra means the equation genuinely has no solution. This is a valid conclusion, not an error.
Mistake 8 — Confusing Holes and Vertical Asymptotes
Both occur at zeros of the original denominator, so the student treats them as the same. Correct: A HOLE occurs when a factor cancels — the function approaches a finite limit. A VERTICAL ASYMPTOTE occurs when a factor does NOT cancel — the function approaches ±∞. Always simplify first, then classify.
Chapter 7 — Back Matter

Chapter Summary

Key Terms Review

rational expression
A fraction of the form $P(x)/Q(x)$ where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial.
excluded value
A value of the variable for which the denominator of a rational expression equals zero; the expression is undefined there.
domain
The set of all input values for which a rational expression or function is defined — all reals except the excluded values.
restriction
A condition $x \neq a$ that excludes a value making some denominator zero. Restrictions persist through all simplification steps.
simplified form
A rational expression in lowest terms — numerator and denominator share no common polynomial factors other than constants.
LCD
Least Common Denominator — the polynomial of lowest degree divisible by all denominators; built from all distinct factors at their highest powers.
complex fraction
A fraction whose numerator, denominator, or both contain fractions.
rational equation
An equation containing one or more rational expressions. Solved by multiplying through by the LCD to clear denominators.
extraneous solution
A value obtained algebraically during solving that does not satisfy the original equation, usually because it violates a domain restriction.
rational function
A function $f(x) = P(x)/Q(x)$ where $P$ and $Q$ are polynomials. Its graph can have asymptotes and holes that polynomial graphs cannot.
vertical asymptote
A vertical line $x = a$ the graph approaches but never reaches; occurs at zeros of the simplified denominator.
horizontal asymptote
A horizontal line $y = b$ the graph approaches as $x \to \pm\infty$; determined by comparing degrees of numerator and denominator.
oblique asymptote
A non-horizontal, non-vertical line the graph approaches as $x \to \pm\infty$; occurs when numerator's degree exceeds denominator's degree by exactly 1.
hole
A missing point on the graph of a rational function where a common factor cancels; the function approaches a finite limit but is undefined there. Also called a removable discontinuity.
inverse variation
A relationship $y = k/x$ where $k$ is a nonzero constant. As one variable increases, the other decreases proportionally. Equivalently, $xy = k$.
constant of variation
The constant $k$ in a variation equation that determines the scale of the relationship.
joint variation
A relationship $y = kxz$; $y$ varies jointly as both $x$ and $z$.
combined variation
A relationship involving both direct and inverse variation, such as $y = kx/z$.

Concept Review Questions

These questions test understanding, not computation.

  1. Explain in your own words why the value $x = 3$ must be excluded from the domain of $\dfrac{x-3}{x-3}$, even though the numerator and denominator are the same.
  2. A student simplifies $\dfrac{x^2-4}{x-2}$ to $x+2$ and concludes both expressions are identical in every way. Is the student correct? Explain fully.
  3. Why does dividing by a rational expression introduce an additional restriction beyond those in the original problem?
  4. Describe the difference between a hole and a vertical asymptote. What algebraic condition causes each?
  5. A rational equation yields two candidate solutions, $x = 1$ and $x = -3$, but the restriction states $x \neq 1$. What is the solution set?
  6. Under what condition does $f(x) = P(x)/Q(x)$ have a horizontal asymptote of $y = 0$?
  7. Can a rational function's graph cross a horizontal asymptote? Can it cross a vertical asymptote? Explain each.
  8. What is the difference between $xy = 5$ and $y = 5x$ in terms of variation type?
  9. Why can you not simplify $\dfrac{x^2+4}{x^2}$ by canceling $x^2$?
  10. A rational equation is cleared of fractions and the resulting polynomial has three solutions. After checking, all three are extraneous. What is the solution set? Is this possible even without an algebra error?

Skill Practice

Part ADomain and Restrictions
  1. $\dfrac{3x-1}{x^2-16}$
  2. $\dfrac{x^2+2x+1}{x^2-9}$
  3. $\dfrac{x^3-1}{x^2-5x+6}$
  4. $\dfrac{5}{x^2+4}$
Part BSimplification
  1. $\dfrac{6x^2-6}{3x^2+3x}$
  2. $\dfrac{3x^2-12}{x^2+4x+4}$
  3. $\dfrac{x^3-8}{x^2-4}$
  4. $\dfrac{2x^3-2x}{x^3+x^2-x-1}$
Part CMultiplication and Division
  1. $\dfrac{2x^2+x-1}{x^2-4} \cdot \dfrac{x^2-2x}{4x-2}$
  2. $\dfrac{3x}{x^2-1} \div \dfrac{9x^2}{x+1}$
  3. $\dfrac{x^2-25}{x+3} \div \dfrac{x-5}{x^2-9}$
  4. $\dfrac{x^2-4}{x^2+5x+6} \cdot \dfrac{x^2+6x+9}{x^2-2x}$
Part DAddition and Subtraction
  1. $\dfrac{2}{x} + \dfrac{3}{x+1}$
  2. $\dfrac{x}{x-3} - \dfrac{2}{x+1}$
  3. $\dfrac{5}{x^2-4} - \dfrac{1}{x+2}$
  4. $\dfrac{x+1}{x^2-x-6} - \dfrac{x-2}{x^2+x-2}$
Part ESolving Rational Equations
  1. $\dfrac{3}{x+2}+\dfrac{1}{x}=\dfrac{5}{2x}$
  2. $\dfrac{2}{x-1}+\dfrac{3}{x+1}=\dfrac{6}{x^2-1}$
  3. $\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=\dfrac{8}{x^2-4}$
  4. $\dfrac{3}{x^2-5x+6}+\dfrac{2}{x-2}=\dfrac{1}{x-3}$

Word Problems

For each: define variables, write the rational model, and interpret the answer in context.

  1. Travel Time. Chen drives from city A to city B at speed $r$ mph. The return trip is 20 mph faster. The distance each way is 180 miles. Write an expression for total travel time $T(r)$ and find $T$ when $r = 60$.
  2. Dilution. A 100-gallon tank contains 5 pounds of dissolved salt. A solution of 2 lb/gallon is pumped in at 3 gal/min, while the tank drains at the same rate. Write and evaluate the concentration after 10 minutes.
  3. Population. A population model gives $P(t) = \dfrac{1200t}{t+10}$, where $t$ is years. Find the horizontal asymptote and interpret it. When does the population first reach 900?
  4. Shared Work. Pump A can drain a pool in 8 hours; pump B in 12 hours. How long does it take both pumps working together? Set up and solve a rational equation.
  5. Combined Variation. The flow rate through a pipe varies jointly with the square of the radius and inversely with the length. A pipe of radius 3 cm and length 100 m has flow rate 27 L/min. Find the flow rate for a pipe of radius 4 cm and length 150 m.
  6. Boyle's Law. For a fixed amount of gas at constant temperature, pressure $P$ varies inversely with volume $V$. A gas at pressure 120 kPa occupies $3\,\text{m}^3$. What pressure corresponds to a volume of $5\,\text{m}^3$?

Error Analysis

Each problem contains an error. Identify the error precisely and provide the correct solution.

Problem 1
(x² + 3x)/x = x + 3x = 4x

What is wrong? What is the correct simplified form?

Problem 2
(x² − 9)/(x² − 3) = (x−3)(x+3) / [x(x−3)] = (x+3)/x

Identify the factoring error. What is $x^2 - 3$? Can it be factored as $x(x-3)$? Show the correct work.

Problem 3
Student solves 5/(x−5) = x/(x−5) + 1, gets x = 5, and accepts it as valid.

What went wrong? What is the correct conclusion?

Problem 4
3/(x+1) − (x−2)/(x+1) = (3 − x − 2)/(x+1) = (1−x)/(x+1)

Is this correct? Identify any sign error and show the correct numerator.

Problem 5
Student finds domain of f(x) = (x+4)/(x²−16), simplifies to 1/(x−4), and states the only excluded value is x = 4.

What important step did the student skip? What are all the excluded values?

Problem 6
Student claims y = 2/x and y = x/2 "both show inverse variation because one goes up when the other goes down."

Evaluate this claim. Which equation shows inverse variation? What does the other show?

Chapter 7 Checkpoint

Chapter 7 Checkpoint
Complete without assistance — assesses all major skills and concepts from Chapter 7
Part A — Short Answer
  1. What are the excluded values of $\dfrac{x^2-5x}{x^2-5x-14}$?
  2. True or False: The expressions $\dfrac{x^2-1}{x-1}$ and $x+1$ have the same domain. Explain.
  3. What is the horizontal asymptote of $f(x) = \dfrac{7x^3-2}{3x^3+x}$?
  4. If $(x+2)$ is a common factor in the numerator and denominator of a rational function, what feature does this create on the graph?
Part B — Simplification and Operations
  1. Simplify: $\dfrac{x^2+x-12}{x^2-6x+9}$. State all restrictions.
  2. Perform and simplify: $\dfrac{x^2-1}{x+3} \cdot \dfrac{x^2+6x+9}{x^2-x}$. State all restrictions.
  3. Add: $\dfrac{4}{x-2} + \dfrac{x}{x^2-4}$.
  4. Simplify: $\dfrac{\dfrac{1}{x+1}+\dfrac{1}{x-1}}{\dfrac{2}{x^2-1}}$.
Part C — Equations
  1. Solve: $\dfrac{x}{x-4}+\dfrac{1}{x+2}=\dfrac{12}{x^2-2x-8}$. State restrictions and identify any extraneous solutions.
  2. Solve: $\dfrac{2x}{x+1}+\dfrac{1}{x-1}=\dfrac{4}{x^2-1}$. Check all solutions.
Part D — Graphing Analysis
  1. For $f(x) = \dfrac{x^2-x-2}{x^2-4}$:
    (a) Identify the domain.
    (b) Simplify completely and state all restrictions.
    (c) Find all asymptotes and holes with coordinates.
    (d) Find the $x$- and $y$-intercepts.
    (e) Describe the end behavior.
Part E — Variation and Application
  1. $y$ varies inversely as $x$, and $y = 8$ when $x = 3$. (a) Find the constant of variation. (b) Find $y$ when $x = 12$. (c) Find $x$ when $y = 2$.
  2. A company has fixed monthly costs of $\$6{,}000$ and variable costs of $\$18$ per unit. Write and simplify the average cost function $A(x)$. State and interpret the horizontal asymptote. For what production level is the average cost $\$20$ per unit?
Part F — Error Analysis
  1. A student simplifies $5(2x-3)$ — wait, wrong chapter. A student simplifies $\dfrac{x^2+9}{x} = x+9$. What is the error? What is correct?
  2. A student solves a rational equation and gets $x = 2$ and $x = -5$. The denominators in the original equation include $x-2$ and $x+3$. Which solutions, if any, are extraneous?
Bridge Note to Chapter 8
Chapter 8: Radical Functions and Rational Exponents

Chapter 7 placed polynomials in the denominator and asked what happens when that denominator is zero. Chapter 8 places variables under radical signs and asks what happens when the radicand is negative — a different restriction, but the same discipline. In both cases, the central question is identical: where is this expression defined?

The tools you built here — domain tracking before any operation, simplification with preserved restrictions, systematic checking of solutions — transfer directly into radical expressions. The concepts of simplification, extraneous solutions, and restricted domains reappear in Chapter 8 in a new setting. Rational and radical expressions also combine frequently in higher mathematics: rationalizing a denominator (clearing a radical from a denominator) connects both worlds and prepares you for techniques used throughout precalculus and calculus. The habit of always asking where an expression is defined is one of the most valuable disciplines in all of mathematics.

MGU Library Connection
  • Rational Function Reference — Appendix G of this textbook
  • Polynomial Factoring and Division Reference — Appendix F of this textbook
  • Chapter 6: Polynomial Equations and Theorems — prerequisite for rational expressions
  • Chapter 8: Radical Functions and Rational Exponents — the next function family
  • Precalculus: Limits and continuity use the concepts of holes and asymptotes developed here
  • Calculus: Partial fractions (a key integration technique) decompose rational functions; asymptotic analysis extends this chapter's end-behavior concepts
  • Physics: Gravitational force, electric field intensity, and lens equations are modeled by rational functions
  • MGU Dictionary entries: rational expression, excluded value, asymptote, hole, inverse variation