Rational Functions

Introduction

Many mathematical relationships involve dividing one quantity by another. For example, the time required to complete a journey depends on distance divided by speed, while the average cost of producing an item depends on total cost divided by the number of items produced.

These relationships can often be represented using rational functions.

A rational function is a function that can be written as a fraction in which both the numerator and denominator are polynomials. Rational functions are important because they allow us to model situations involving rates, ratios, inverse relationships, and changing quantities.

However, rational functions have an important restriction: division by zero is undefined.

This means that certain input values must be excluded from the domain of a rational function. Understanding these restrictions is essential when simplifying expressions, interpreting graphs, and solving equations.

In this lesson, we will explore how rational functions are defined, how to recognize rational expressions, how to determine domain restrictions, and how to simplify rational expressions correctly.

What Is a Rational Function?

A rational function is a function that can be expressed in the form:

where:

• P(x) is a polynomial called the numerator.

• Q(x) is a polynomial called the denominator.

• Q(x) cannot equal zero.

  For example:

  \[ f(x)=\frac{x+3}{x-2} \]

  is a rational function because both x + 3 and x − 2 are polynomials.

  Other examples include:

  \[ f(x)=\frac{5}{x} \]

  \[ g(x)=\frac{x^2-9}{x+4} \]

  \[ h(x)=\frac{2x^2+3x-1}{x^2-4} \]

  A rational function can contain linear, quadratic, cubic, or higher-degree polynomials.

  Rational and Non-Rational Expressions

  Not every expression containing a fraction is a rational expression.

  Expression	Rational?	Explanation
  (x + 2)/(x − 5)	Yes	Both parts are polynomials
  7/x²	Yes	Both parts are polynomials
  (x² + 1)/3	Yes	A polynomial divided by a nonzero constant
  √x/(x + 1)	No	√x is not a polynomial
  sin(x)/x	No	sin(x) is not a polynomial
  3x² − 4x + 1	Yes	Can be written over 1

  Important: Every polynomial function is also a rational function because it can be written as a polynomial divided by 1. However, not every rational function is a polynomial function.

  Understanding the Graph of a Rational Function

  Consider the simple rational function:

  \[ f(x)=\frac{1}{x} \]

  The table below shows how the output changes as x changes.

  x	f(x) = 1/x
  −4	−0.25
  −2	−0.5
  −1	−1
  −0.5	−2
  0	Undefined
  0.5	2
  1	1
  2	0.5
  4	0.25

  The function has two separate branches. One branch lies in the first quadrant, while the other lies in the third quadrant.

  As x approaches zero from the positive side, the output increases without bound. As x approaches zero from the negative side, the output decreases without bound.

  The function is not defined at x = 0 because division by zero is impossible.

  Identifying Rational Expressions

  A rational expression is an algebraic expression that can be written as the quotient of two polynomials.

  For example:

  \[ \frac{3x+5}{x-2} \]

  is rational because both the numerator and denominator are polynomials.

  To determine whether an expression is rational, ask:

  1. Can the numerator be written as a polynomial?

  2. Can the denominator be written as a nonzero polynomial?

  3. Are there any variable exponents or operations that prevent the expression from being written as a ratio of polynomials?

  Worked Example 1: Identifying Rational Expressions

  Determine whether each expression is rational.

  a) (x² + 4)/(x − 7)

  Both the numerator and denominator are polynomials.

  Therefore, the expression is rational.

  b) √x/(x + 2)

  The numerator contains √x = x^(1/2), which is not a polynomial.

  Therefore, the expression is not rational.

  c) 5x³ − 2x + 8

  Rewrite the expression as:

  (5x³ − 2x + 8)/1

  Both the numerator and denominator are polynomials.

  Therefore, the expression is rational.

  d) 1/(x² + 9)

  Both the numerator and denominator are polynomials.

  Therefore, the expression is rational.

  Domain Restrictions and Excluded Values

  The domain of a function is the set of all input values for which the function is defined.

  For a rational function, the denominator cannot equal zero.

  Consider:

  \[ f(x)=\frac{4}{x-3} \]

  If x = 3, the denominator becomes:

  3 − 3 = 0

  This would produce:

  4/0

  Since division by zero is undefined, x = 3 cannot be included in the domain.

  We call x = 3 an excluded value.

  The domain is all real numbers except 3.

  Using interval notation:

  \[ (-\infty,3)\cup(3,\infty) \]

  How to Find Domain Restrictions

  Follow these steps:

  1. Identify the denominator.

  2. Set the denominator equal to zero.

  3. Solve the resulting equation.

  4. Exclude those solutions from the domain.

  Worked Example 2: A Linear Denominator

  Find the domain of:

  \[ f(x)=\frac{2x+1}{x-5} \]

  Step 1: Identify the denominator.

  x − 5

  Step 2: Set it equal to zero.

  x − 5 = 0

  Step 3: Solve.

  x = 5

  Step 4: State the domain.

  All real numbers except x = 5.

  Interval notation:

  \[ (-\infty,5)\cup(5,\infty) \]

  Worked Example 3: A Quadratic Denominator

  Find the domain of:

  \[ g(x)=\frac{x+2}{x^2-9} \]

  Step 1: Set the denominator equal to zero.

  x² − 9 = 0

  Step 2: Factor using the difference of two squares.

  (x − 3)(x + 3) = 0

  Step 3: Solve each factor.

  x − 3 = 0 gives x = 3.

  x + 3 = 0 gives x = −3.

  Step 4: State the domain.

  x ≠ −3 and x ≠ 3.

  Interval notation:

  \[ (-\infty,-3)\cup(-3,3)\cup(3,\infty) \]

  Both values must be excluded because either would make the denominator zero.

  Worked Example 4: A Denominator with No Real Zeros

  Find the domain of:

  \[ h(x)=\frac{x-1}{x^2+4} \]

  Set the denominator equal to zero:

  x² + 4 = 0

  x² = −4

  There are no real solutions because the square of a real number cannot be negative.

  Therefore, the denominator is never zero for real x.

  Domain: all real numbers.

  Interval notation:

  \[ (-\infty,\infty) \]

  Excluded Values and Graphs

  Excluded values can affect the graph of a rational function in different ways.

  Two important features are vertical asymptotes and holes.

  Vertical Asymptotes

  A vertical asymptote is a vertical line that the graph approaches as the function's output increases or decreases without bound.

  Consider:

  \[ f(x)=\frac{1}{x-2} \]

  The denominator becomes zero at x = 2.

  As x approaches 2 from the right, f(x) increases without bound.

  As x approaches 2 from the left, f(x) decreases without bound.

  Therefore, x = 2 is a vertical asymptote.

  Holes in Graphs

  A hole occurs when a value is excluded from the domain but the expression can be simplified by cancelling a common factor.

  Consider:

  \[ g(x)=\frac{x^2-4}{x-2} \]

  Factor the numerator:

  \[ g(x)=\frac{(x-2)(x+2)}{x-2} \]

  Cancel the common factor:

  g(x) = x + 2, where x ≠ 2.

  The graph follows the line y = x + 2, but the point (2, 4) is missing.

  This missing point is called a hole or removable discontinuity.

  The key difference is that the graph approaches a finite value at a hole, whereas it grows without bound near a vertical asymptote.

  Simplifying Rational Expressions

  Rational expressions can often be simplified by factoring the numerator and denominator and cancelling common factors.

  This is similar to simplifying numerical fractions.

  For example:

  12/18 = (6 × 2)/(6 × 3) = 2/3

  In algebra, the same principle applies.

  However, we must remember that only common factors can be cancelled, not individual terms.

  Worked Example 5: Simplifying by Factoring

  Simplify:

  \[ \frac{x^2-9}{x-3} \]

  Step 1: Factor the numerator.

  x² − 9 = (x − 3)(x + 3)

  Step 2: Rewrite the expression.

  \[ \frac{(x-3)(x+3)}{x-3} \]

  Step 3: Cancel the common factor.

  x + 3

  Step 4: State the restriction.

  The original denominator is zero when x = 3.

  Therefore:

  Simplified expression: x + 3, where x ≠ 3.

  Even though the denominator has disappeared, the restriction remains.

  Worked Example 6: Simplifying a Quadratic Expression

  Simplify:

  \[ \frac{x^2+5x+6}{x^2+4x+3} \]

  Step 1: Factor the numerator.

  x² + 5x + 6 = (x + 2)(x + 3)

  Step 2: Factor the denominator.

  x² + 4x + 3 = (x + 1)(x + 3)

  Step 3: Rewrite.

  \[ \frac{(x+2)(x+3)}{(x+1)(x+3)} \]

  Step 4: Cancel the common factor (x + 3).

  \[ \frac{x+2}{x+1} \]

  Step 5: Identify the excluded values from the original denominator.

  (x + 1)(x + 3) = 0

  x = −1 or x = −3

  Final answer:

  \[ \frac{x+2}{x+1},\quad x\ne-1,-3 \]

  Notice that x = −3 is still excluded even though its factor was cancelled.

  Worked Example 7: Cancelling Numerical and Algebraic Factors

  Simplify:

  \[ \frac{6x^2+12x}{9x} \]

  Step 1: Factor the numerator.

  6x² + 12x = 6x(x + 2)

  Step 2: Rewrite.

  \[ \frac{6x(x+2)}{9x} \]

  Step 3: Cancel x and reduce 6/9.

  \[ \frac{2(x+2)}{3} \]

  The original denominator requires x ≠ 0.

  Final answer:

  2(x + 2)/3, where x ≠ 0.

  Why Domain Restrictions Must Be Found Before Simplifying

  Consider:

  \[ f(x)=\frac{x^2-1}{x-1} \]

  The original denominator is zero at x = 1.

  After factoring:

  \[ f(x)=\frac{(x-1)(x+1)}{x-1} \]

  Simplifying gives:

  f(x) = x + 1, where x ≠ 1.

  The simplified expression x + 1 is defined at x = 1, but the original rational function is not.

  Therefore, the original function and the unrestricted line y = x + 1 do not have exactly the same domain.

  The correct simplified form must preserve all original excluded values.

  Real-World Applications of Rational Functions

  Rational functions appear in many practical situations involving rates, costs, efficiency, and measurement.

  Application 1: Travel Time

  The relationship between distance, speed, and time is:

  Time = Distance / Speed

  Suppose a car travels 240 km at a constant speed of v km/h.

  The time required is:

  \[ t(v)=\frac{240}{v} \]

  This is a rational function.

  Speed (km/h)	Time (hours)
  40	6
  60	4
  80	3
  100	2.4
  120	2

  As speed increases, travel time decreases.

  Mathematically, v = 0 must be excluded because division by zero is undefined. In the physical model, only positive speeds are meaningful.

  Application 2: Average Production Cost

  A company manufactures reusable water bottles.

  The company has a fixed operating cost of $600 and a production cost of $8 per bottle.

  If x bottles are produced, the total cost is:

  C(x) = 600 + 8x

  The average cost per bottle is:

  \[ A(x)=\frac{600+8x}{x} \]

  This can be rewritten as:

  \[ A(x)=\frac{600}{x}+8 \]

  For 100 bottles:

  A(100) = 600/100 + 8 = $14

  For 300 bottles:

  A(300) = 600/300 + 8 = $10

  For 600 bottles:

  A(600) = 600/600 + 8 = $9

  As production increases, the fixed cost is distributed across more bottles, reducing the average cost.

  The mathematical restriction is x ≠ 0. For this application, x must be a positive whole number.

  Application 3: Electrical Resistance

  For an electrical circuit, Ohm's law gives:

  V = IR

  Rearranging:

  \[ I=\frac{V}{R} \]

  where:

  • I = current in amperes

  • V = voltage in volts

  • R = resistance in ohms

  If the voltage is fixed at 12 V:

  \[ I(R)=\frac{12}{R} \]

  For a resistance of 4 Ω:

  I = 12/4 = 3 A

  For a resistance of 6 Ω:

  I = 12/6 = 2 A

  The current decreases as resistance increases.

  In this simple physical model, resistance must be positive. The expression is mathematically undefined at R = 0.

  Common Mistakes

  Mistake 1: Forgetting to Check the Denominator

  Incorrect:

  For f(x) = 3/(x − 4), the domain is all real numbers.

  Correct:

  The denominator is zero at x = 4.

  Therefore, x ≠ 4.

  Mistake 2: Cancelling Terms Instead of Factors

  Incorrect:

  \[ \frac{x+3}{x+5}=\frac35 \]

  The x terms cannot be cancelled because they are part of sums, not common factors.

  Correct:

  The expression (x + 3)/(x + 5) cannot be simplified by cancellation.

  Mistake 3: Forgetting Restrictions After Simplification

  Incorrect:

  (x² − 16)/(x − 4) = x + 4, with no restrictions.

  Correct:

  (x² − 16)/(x − 4) = x + 4, where x ≠ 4.

  Mistake 4: Assuming Every Excluded Value Is a Vertical Asymptote

  An excluded value may produce a hole rather than a vertical asymptote.

  For example:

  (x² − 1)/(x − 1) has a hole at x = 1.

  By contrast:

  1/(x − 1) has a vertical asymptote at x = 1.

  Mistake 5: Confusing the Mathematical Domain with the Practical Domain

  The function f(x) = 100/x is mathematically defined for every real number except zero.

  However, if x represents the number of workers completing a job, negative values and fractional workers are not meaningful.

  The practical domain may therefore be positive integers.

  Key Terms

  • Rational Function: A function that can be written as the ratio of two polynomials, with a denominator that is not the zero polynomial.

  • Rational Expression: An algebraic expression written as a quotient of two polynomials.

  • Polynomial: An algebraic expression containing terms with nonnegative integer powers of variables.

  • Numerator: The expression above the fraction bar.

  • Denominator: The expression below the fraction bar.

  • Domain: The set of all permitted input values of a function.

  • Domain Restriction: A condition limiting the values that a variable can take.

  • Excluded Value: An input value that makes a function undefined.

  • Factor: An expression multiplied by another expression to form a product.

  • Simplification: Rewriting an expression in an equivalent, usually less complicated form while preserving restrictions.

  • Vertical Asymptote: A vertical line approached by a graph as its function values increase or decrease without bound.

  • Hole: A missing point on a graph caused by a removable discontinuity.

  • Interval Notation: A method of representing sets of real numbers using intervals.

  Key Takeaways

  1. A rational function can be written as f(x) = P(x)/Q(x), where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial.

  2. Rational expressions include fractions with polynomial numerators and denominators, as well as ordinary polynomials written over 1.

  3. The denominator of a rational function must never equal zero.

  4. Domain restrictions are found by setting the original denominator equal to zero and solving.

  5. Excluded values must remain excluded even after simplifying an expression.

  6. Rational expressions are simplified by factoring and cancelling common factors.

  7. Terms cannot be cancelled unless they are part of a common factor.

  8. Excluded values can produce holes or vertical asymptotes.

  9. Rational functions are useful for modelling rates, costs, travel times, and many scientific relationships.

  Check Your Understanding

  Question 1: Identifying Rational Expressions

  Determine whether each expression is rational.

  a) (x + 4)/(x² − 1)

  b) √x/(x − 2)

  c) 4x³ − 2x + 7

  d) 1/(x² + 5)

  Question 2: Domain Restrictions

  Find the domain of:

  \[ f(x)=\frac{5x+2}{x+6} \]

  Question 3: Multiple Excluded Values

  Find all excluded values of:

  \[ g(x)=\frac{x-4}{x^2-16} \]

  Question 4: Simplifying Rational Expressions

  Simplify:

  \[ \frac{x^2-25}{x-5} \]

  State any restrictions.

  Question 5: Factoring Quadratics

  Simplify:

  \[ \frac{x^2+7x+12}{x^2+5x+6} \]

  State the excluded values.

  Question 6: Real-World Application

  A company has a fixed cost of $400 and a variable cost of $12 per item.

  a) Write a rational function for the average cost per item.

  b) Calculate the average cost when 50 items are produced.

  c) Explain why x = 0 is excluded.

  d) Describe what happens to the average cost as production increases.

   

  Fully Worked Answers

  Answer 1

  a) Rational. Both the numerator and denominator are polynomials.

  b) Not rational. √x is not a polynomial.

  c) Rational. The expression can be written over 1.

  d) Rational. Both the numerator and denominator are polynomials.

  Answer 2

  Given:

  f(x) = (5x + 2)/(x + 6)

  Set the denominator equal to zero:

  x + 6 = 0

  x = −6

  Therefore:

  Domain: all real numbers except −6.

  Interval notation:

  \[ (-\infty,-6)\cup(-6,\infty) \]

  Answer 3

  Given:

  g(x) = (x − 4)/(x² − 16)

  Set the denominator equal to zero:

  x² − 16 = 0

  Factor:

  (x − 4)(x + 4) = 0

  Solve:

  x = 4 or x = −4

  Excluded values: x = −4 and x = 4.

  Even though the factor (x − 4) could be cancelled, x = 4 must remain excluded.

  Answer 4

  Given:

  (x² − 25)/(x − 5)

  Factor the numerator:

  (x − 5)(x + 5)/(x − 5)

  Cancel the common factor:

  x + 5

  The original denominator is zero at x = 5.

  Final answer: x + 5, where x ≠ 5.

  Answer 5

  Given:

  (x² + 7x + 12)/(x² + 5x + 6)

  Factor the numerator:

  x² + 7x + 12 = (x + 3)(x + 4)

  Factor the denominator:

  x² + 5x + 6 = (x + 2)(x + 3)

  Rewrite:

  \[ \frac{(x+3)(x+4)}{(x+2)(x+3)} \]

  Cancel (x + 3):

  \[ \frac{x+4}{x+2} \]

  The original denominator equals zero at x = −2 and x = −3.

  Final answer: (x + 4)/(x + 2), where x ≠ −2 and x ≠ −3.

  Answer 6

  a) Write the function.

  Total cost:

  C(x) = 400 + 12x

  Average cost:

  \[ A(x)=\frac{400+12x}{x} \]

  Or:

  A(x) = 400/x + 12

  b) Calculate for 50 items.

  A(50) = 400/50 + 12

  A(50) = 8 + 12

  Average cost = $20 per item.

  c) Explain the restriction.

  At x = 0, the denominator becomes zero.

  It is also impossible to calculate an average cost per item when no items are produced.

  d) Describe the trend.

  As x increases, 400/x becomes smaller.

  Therefore, the average cost decreases and approaches $12 per item.

  The average cost does not fall below $12 in this model because each item has a variable production cost of $12.

   

  This completes Understanding Rational Functions, covering the definition, identification of rational expressions, domain restrictions, excluded values, simplification, graphical features, and practical applications.

Introduction

Fractions are not limited to numbers. In algebra, fractions can contain variables, expressions, and polynomials. These are called rational expressions, and simplifying them is an essential skill in mathematics.

Consider the numerical fraction:

12/18 = 2/3

We simplify this fraction by dividing the numerator and denominator by their common factor of 6.

The same principle applies to algebraic fractions. For example:

(x² − 9)/(x − 3)

By factoring the numerator, we can identify a common factor and simplify the expression.

However, simplifying rational expressions requires special care. Unlike ordinary numerical fractions, rational expressions may contain values that make the denominator equal to zero. These values must be identified and excluded, even if the factors responsible are cancelled during simplification.

In this lesson, we will learn how to factor numerators and denominators, simplify rational expressions correctly, identify restrictions, recognize equivalent expressions, and verify our results using algebraic and numerical methods.

What Is a Rational Expression?

A rational expression is a fraction whose numerator and denominator are polynomials.

Its general form is:

where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial.

Examples include:

• (x + 3)/(x − 5)

• (x² − 16)/(x + 4)

• (2x² + 6x)/(4x)

• (x² + 5x + 6)/(x² − 9)

The numerator is the expression above the fraction bar, while the denominator is the expression below it.

The fundamental rule is that the denominator cannot equal zero.

Factoring Numerators and Denominators

Factoring is the process of rewriting an expression as a product of simpler expressions.

For example:

x² + 5x + 6 = (x + 2)(x + 3)

Factoring is especially useful when simplifying rational expressions because it reveals common factors that can be cancelled.

Common Factoring Techniques

Factoring Out the Greatest Common Factor

Consider:

6x² + 12x

Both terms contain the common factor 6x.

Therefore:

6x² + 12x = 6x(x + 2)

This technique should generally be checked first when factoring a polynomial.

Factoring a Difference of Squares

The identity is:

For example:

x² − 49 = (x − 7)(x + 7)

And:

4x² − 25 = (2x − 5)(2x + 5)

Factoring Quadratic Trinomials

Consider:

x² + 8x + 15

We need two numbers that:

• Multiply to give 15.

• Add to give 8.

The numbers are 3 and 5.

Therefore:

x² + 8x + 15 = (x + 3)(x + 5)

Factoring Quadratics with Leading Coefficients

Consider:

2x² + 7x + 3

We seek two binomial factors whose product gives the original quadratic.

2x² + 7x + 3 = (2x + 1)(x + 3)

Check by expanding:

(2x + 1)(x + 3)

= 2x² + 6x + x + 3

= 2x² + 7x + 3

This confirms the factorization.

The Fundamental Rule of Simplifying Rational Expressions

To simplify a rational expression, divide the numerator and denominator by their common nonzero factors.

For example:

When x ≠ −2, the common factor (x + 2) is nonzero and can be cancelled.

The simplified expression is:

3/5, where x ≠ −2.

The restriction is important because the original expression is undefined at x = −2.

Only factors can be cancelled, not individual terms.

For example:

(x + 3)/(x + 5)

The x terms cannot be cancelled because they are part of separate sums.

By contrast:

x(x + 3)/[x(x + 5)]

contains a common factor of x that can be cancelled, provided x ≠ 0.

A Reliable Four-Step Simplification Method

1. Find restrictions: Set the original denominator equal to zero and solve.

2. Factor completely: Factor both the numerator and denominator wherever possible.

3. Cancel common factors: Divide out factors appearing in both the numerator and denominator.

4. Verify: Check the simplified result by algebraic reasoning or substituting permitted values.

This order prevents the common mistake of losing excluded values during cancellation.

Worked Example 1: Simplifying with a Common Monomial Factor

Simplify:

Step 1: Identify restrictions.

18x = 0

x = 0

Therefore, x ≠ 0.

Step 2: Factor numerator and denominator.

12x² = 6x(2x)

18x = 6x(3)

Rewrite:

Step 3: Cancel 6x.

Step 4: Verify with x = 2.

Original:

12(2²)/[18(2)] = 48/36 = 4/3

Simplified:

2(2)/3 = 4/3

Both expressions give the same result.

Final answer: 2x/3, where x ≠ 0.

Worked Example 2: Simplifying Using the Difference of Squares

Simplify:

Step 1: Find restrictions.

x − 4 = 0

x = 4

Therefore, x ≠ 4.

Step 2: Factor the numerator.

x² − 16 = (x − 4)(x + 4)

Step 3: Rewrite and cancel.

Step 4: Verify with x = 6.

Original:

(36 − 16)/(6 − 4) = 20/2 = 10

Simplified:

6 + 4 = 10

The results agree.

Final answer: x + 4, where x ≠ 4.

Worked Example 3: Factoring Both Numerator and Denominator

Simplify:

Step 1: Find restrictions from the original denominator.

x² + x − 6 = (x + 3)(x − 2)

Therefore:

x ≠ −3 and x ≠ 2.

Step 2: Factor the numerator.

x² + 5x + 6 = (x + 2)(x + 3)

Step 3: Rewrite the expression.

Step 4: Cancel the common factor (x + 3).

Step 5: Verify with x = 0.

Original:

(0 + 0 + 6)/(0 + 0 − 6) = −1

Simplified:

(0 + 2)/(0 − 2) = −1

Both results agree.

Final answer: (x + 2)/(x − 2), where x ≠ −3 and x ≠ 2.

Worked Example 4: Simplifying with Numerical Coefficients

Simplify:

Step 1: Find restrictions.

9x = 0

x = 0

Therefore, x ≠ 0.

Step 2: Factor the numerator.

6x² − 18x = 6x(x − 3)

Step 3: Rewrite.

Step 4: Cancel x and simplify 6/9.

Final answer: 2(x − 3)/3, where x ≠ 0.

Worked Example 5: More Than One Common Factor

Simplify:

Step 1: Find restrictions.

x² − 4 = (x − 2)(x + 2)

Therefore:

x ≠ 2 and x ≠ −2.

Step 2: Factor the numerator.

x² − 5x + 6 = (x − 2)(x − 3)

Step 3: Rewrite.

Step 4: Cancel (x − 2).

Final answer: (x − 3)/(x + 2), where x ≠ 2 and x ≠ −2.

Notice that although only one factor was cancelled, both original restrictions remain.

Worked Example 6: Recognizing When an Expression Cannot Be Simplified

Consider:

Factor both parts:

There are no common factors.

Therefore, the expression cannot be simplified by cancellation.

The original denominator is zero at x = 2 and x = −2.

Final answer: (x² − 16)/(x² − 4), where x ≠ −2 and x ≠ 2.

An expression does not always become simpler just because its numerator and denominator can be factored.

Recognizing Equivalent Rational Expressions

Two rational expressions are equivalent on a specified domain if they produce the same output for every input in that domain.

For example:

is true whenever x ≠ 3.

The expression on the left is undefined at x = 3, while x + 3 is defined there.

Therefore, the expressions agree for all x ≠ 3, but they are not identical as functions with their natural domains.

Worked Example 7: Determining Equivalence

Determine whether the following expressions are equivalent:

A(x) = (x² + 4x + 3)/(x + 1)

B(x) = x + 3

Factor A(x):

A(x) = (x + 1)(x + 3)/(x + 1)

Cancel the common factor:

A(x) = x + 3, where x ≠ −1.

Therefore, A(x) and B(x) give the same output whenever x ≠ −1.

However, B(x) is also defined at x = −1.

Conclusion: They are equivalent on the restricted domain x ≠ −1, but they do not have the same natural domain.

Using Cross-Multiplication to Check Equivalence

Suppose we want to compare:

A(x) = (x + 2)/(x − 1)

B(x) = (x² + x − 2)/(x² − 2x + 1)

Factor B(x):

Numerator:

x² + x − 2 = (x + 2)(x − 1)

Denominator:

x² − 2x + 1 = (x − 1)²

Therefore:

Cancel one factor of (x − 1):

B(x) = (x + 2)/(x − 1)

Both expressions have the restriction x ≠ 1.

Therefore, A(x) and B(x) are equivalent rational functions with the same domain.

Cross-multiplication offers another method:

For A/B and C/D, check whether:

AD = BC

This establishes equality wherever both original denominators are nonzero.

Verifying Simplified Results

Simplifying an expression is not complete until we are confident that the factorization and cancellation were performed correctly.

There are three useful verification methods.

Method 1: Expanding the Factors

Suppose:

Expanding the numerator gives:

(x + 2)(x + 3) = x² + 5x + 6

Therefore, the original expression is:

(x² + 5x + 6)/(x + 3)

The factorization is correct, and cancellation is valid for x ≠ −3.

Method 2: Substituting Numerical Values

Consider:

(x² − 9)/(x − 3) = x + 3, where x ≠ 3.

Choose x = 5.

Original:

(25 − 9)/(5 − 3) = 16/2 = 8

Simplified:

5 + 3 = 8

The results agree.

A numerical check can detect errors, but testing one or two values does not prove that two expressions are equivalent for every permitted input.

Method 3: Checking Restrictions

After simplification, return to the original denominator.

For example:

The original denominator requires x ≠ 2.

Therefore, the correct statement is:

(x² − 4)/(x − 2) = x + 2, where x ≠ 2.

This confirms that the simplified form has preserved the original restriction.

Graphical Interpretation of Simplification

Simplification can reveal important information about the graph of a rational function.

Consider:

Factor:

f(x) = (x − 1)(x + 1)/(x − 1)

Simplify:

f(x) = x + 1, where x ≠ 1.

The graph is the straight line y = x + 1, except for the missing point (1, 2).

This missing point is called a hole or removable discontinuity.

The simplified equation describes the same line everywhere except at the excluded value.

A graphing calculator may display the complete line without making the hole obvious, so algebraic restrictions must always be checked.

Real-World Applications

Application 1: Simplifying Area Ratios

Suppose the area of a rectangular garden is:

A = x² + 5x + 6

and its width is:

W = x + 2

The length is:

L = A/W

Therefore:

Factor the numerator:

L = (x + 2)(x + 3)/(x + 2)

Simplify:

L = x + 3, where x ≠ −2.

For a physical garden, the dimensions must also be positive, so x must be greater than −2.

Application 2: Average Production Cost

A factory has a total production cost:

C(x) = 5x² + 20x

where x is the number of items produced.

The average cost per item is:

Factor the numerator:

A(x) = 5x(x + 4)/x

Simplify:

A(x) = 5(x + 4), where x ≠ 0.

For 10 items:

A(10) = 5(14) = $70 per item.

The simplified form makes calculations easier.

In the real-world context, x must be a positive whole number.

Application 3: Simplifying Physics Formulae

Suppose an object moves a distance:

d = 3t² + 6t

over a time interval t, with t greater than zero.

Its average speed over the interval is:

v = d/t

Therefore:

Factor:

v = 3t(t + 2)/t

Simplify:

v = 3(t + 2), where t ≠ 0.

For t = 4 seconds:

v = 3(6) = 18 m/s.

This illustrates how simplifying rational expressions can make scientific formulae easier to use.

Common Mistakes

Mistake 1: Cancelling Terms Instead of Factors

Incorrect:

(x + 4)/(x + 6) = 4/6

The x terms cannot be cancelled because they are not common factors.

Correct:

The expression (x + 4)/(x + 6) cannot be simplified by cancellation.

Mistake 2: Forgetting to Factor Completely

Consider:

(x² + 5x + 6)/(x² + x − 6)

Without factoring, common factors may not be obvious.

Correct factorization:

(x + 2)(x + 3)/[(x + 3)(x − 2)]

Simplified:

(x + 2)/(x − 2), where x ≠ −3 and x ≠ 2.

Mistake 3: Losing Excluded Values

Incorrect:

(x² − 9)/(x − 3) = x + 3, with no restriction.

Correct:

x + 3, where x ≠ 3.

Mistake 4: Incorrect Factorization

Incorrect:

x² + 7x + 12 = (x + 2)(x + 6)

Expanding gives:

x² + 8x + 12

Correct:

x² + 7x + 12 = (x + 3)(x + 4)

Mistake 5: Assuming Every Expression Can Be Simplified

Not every rational expression contains common factors.

For example:

(x + 2)/(x + 5)

cannot be simplified by cancellation.

Mistake 6: Relying Only on Numerical Verification

Two different expressions might give the same answer for a particular input.

Numerical substitution is useful for checking, but algebraic verification is needed to establish equivalence for all permitted values.

Key Terms

• Rational Expression: A quotient of two polynomials.

• Numerator: The polynomial above the fraction bar.

• Denominator: The polynomial below the fraction bar.

• Factor: An expression that is multiplied by another expression.

• Common Factor: A factor shared by two or more expressions.

• Greatest Common Factor: The largest factor shared by all terms.

• Factoring: Rewriting an expression as a product of factors.

• Difference of Squares: A factoring pattern of the form a² − b² = (a − b)(a + b).

• Simplification: Rewriting an expression in a less complicated equivalent form.

• Excluded Value: A value that makes the original denominator zero.

• Domain Restriction: A condition excluding inputs that make an expression undefined.

• Equivalent Rational Expressions: Expressions that give the same values over a specified common domain.

• Verification: Checking whether a mathematical result is correct.

• Removable Discontinuity: A missing point in a graph caused by a cancelled denominator factor.

• Hole: The missing point associated with a removable discontinuity.

Key Takeaways

1. Rational expressions are simplified by factoring numerators and denominators and cancelling common factors.

2. Always identify excluded values from the original denominator before simplifying.

3. Only common factors can be cancelled; individual terms cannot.

4. The greatest common factor should be checked before using other factoring techniques.

5. Difference-of-squares and quadratic factoring are especially useful for simplifying rational expressions.

6. Cancelled denominator factors still produce restrictions in the original expression.

7. Equivalent rational expressions produce the same outputs over their common permitted domain.

8. Algebraic expansion, substitution, and checking restrictions can help verify simplified results.

9. Simplification can reveal holes and other important features of rational-function graphs.

10. Rational-expression simplification is useful in geometry, economics, physics, and other mathematical applications.

Check Your Understanding

Question 1: Common Factors

Simplify and state restrictions:

Question 2: Difference of Squares

Simplify:

State any excluded values.

Question 3: Factoring Quadratics

Simplify:

State all restrictions.

Question 4: Numerical and Algebraic Factors

Simplify:

State any restrictions.

Question 5: Recognizing Equivalence

Determine whether the following expressions are equivalent over their common domain:

A(x) = (x² − 25)/(x − 5)

B(x) = x + 5

Explain your reasoning and identify any domain differences.

Question 6: Verification

A student simplifies:

a) Verify the factorization.

b) State the excluded value.

c) Check the result using x = 3.

Question 7: Multi-Step Simplification

Simplify completely:

State all restrictions.

Question 8: Real-World Application

A rectangular field has area:

A = x² + 9x + 20

Its width is:

W = x + 4

a) Write an expression for its length.

b) Simplify the expression.

c) State the mathematical restriction.

d) Calculate the length when x = 6.

Fully Worked Answers

Answer 1

Given:

15x²/(20x)

Step 1: Find restrictions.

20x = 0 gives x = 0.

Step 2: Simplify.

15x²/(20x) = 3x/4

Final answer: 3x/4, where x ≠ 0.

Answer 2

Given:

(x² − 49)/(x − 7)

Factor:

(x − 7)(x + 7)/(x − 7)

Cancel:

x + 7

The original denominator is zero at x = 7.

Final answer: x + 7, where x ≠ 7.

Answer 3

Given:

(x² + 6x + 8)/(x² + 5x + 6)

Factor the numerator:

x² + 6x + 8 = (x + 2)(x + 4)

Factor the denominator:

x² + 5x + 6 = (x + 2)(x + 3)

Rewrite:

Cancel (x + 2):

Restrictions from the original denominator:

x ≠ −2 and x ≠ −3.

Final answer: (x + 4)/(x + 3), where x ≠ −2 and x ≠ −3.

Answer 4

Given:

(8x² + 16x)/(12x)

Factor:

8x(x + 2)/(12x)

Cancel x and simplify 8/12:

2(x + 2)/3

The original denominator requires x ≠ 0.

Final answer: 2(x + 2)/3, where x ≠ 0.

Answer 5

A(x) = (x² − 25)/(x − 5)

Factor:

A(x) = (x − 5)(x + 5)/(x − 5)

Simplify:

A(x) = x + 5, where x ≠ 5.

B(x) = x + 5 is defined for all real numbers.

Therefore, A(x) and B(x) are equivalent for x ≠ 5, but their natural domains differ.

Answer 6

Given:

(x² + 3x + 2)/(x + 1)

a) Factor the numerator.

x² + 3x + 2 = (x + 1)(x + 2)

Therefore:

(x + 1)(x + 2)/(x + 1) = x + 2

b) Restriction.

x + 1 = 0

x = −1

Therefore, x ≠ −1.

c) Verify with x = 3.

Original:

(9 + 9 + 2)/(3 + 1) = 20/4 = 5

Simplified:

3 + 2 = 5

Both results agree.

Answer 7

Given:

(2x² − 8x)/(x² − 16)

Step 1: Find restrictions.

x² − 16 = (x − 4)(x + 4)

Therefore:

x ≠ 4 and x ≠ −4.

Step 2: Factor the numerator.

2x² − 8x = 2x(x − 4)

Step 3: Rewrite.

Step 4: Cancel (x − 4).

Final answer: 2x/(x + 4), where x ≠ −4 and x ≠ 4.

Answer 8

a) Write the length expression.

Length = Area / Width

L = (x² + 9x + 20)/(x + 4)

b) Simplify.

Factor the numerator:

x² + 9x + 20 = (x + 4)(x + 5)

Therefore:

L = (x + 4)(x + 5)/(x + 4)

L = x + 5

c) Restriction.

x + 4 ≠ 0

x ≠ −4

For the physical field, both dimensions must be positive, so x greater than −4 is required.

d) Substitute x = 6.

L = 6 + 5

Length = 11 units.

This completes Simplifying Rational Expressions, with detailed coverage of factoring, cancellation, restrictions, equivalent expressions, verification, and practical applications.

Introduction

Rational expressions are algebraic fractions containing polynomials in their numerators and denominators. Just as we can add, subtract, multiply, and divide numerical fractions, we can perform the same four operations with rational expressions.

For example, consider the numerical fractions:

1/3 + 1/6 = 1/2

The same principles apply when variables are involved:

1/x + 2/x = 3/x, where x ≠ 0.

However, operations with rational expressions require careful attention to factoring, common denominators, domain restrictions, and simplification.

These skills are important in algebra, calculus, physics, engineering, and economics. Rational expressions appear in problems involving rates, work, speed, electrical circuits, and other mathematical relationships.

In this lesson, we will learn how to add, subtract, multiply, and divide rational expressions, identify restrictions, and simplify final answers completely.

Understanding Rational Expressions

A rational expression is an expression that can be written as:

where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial.

Examples include:

• (x + 2)/(x − 3)

• 5/x

• (x² − 4)/(x + 2)

• (2x² + 3x − 1)/(x² − 9)

Before performing any operation, remember:

The denominator of a rational expression cannot equal zero.

For example:

is undefined when x = 5.

Therefore, x ≠ 5 is a restriction.

When combining rational expressions, we must preserve all restrictions from the original expressions. Division also requires that the expression being divided by is not zero.

The Four Operations: Overview

The least common denominator (LCD) is the smallest polynomial expression, up to a nonzero constant factor, that contains all the factors needed to form a common denominator.

Adding Rational Expressions

Adding Expressions with the Same Denominator

When two rational expressions have the same denominator, add their numerators and keep the denominator unchanged.

The general rule is:

where c ≠ 0.

Worked Example 1: Adding with a Common Denominator

Simplify:

Step 1: Identify restrictions.

x − 4 = 0

x = 4

Therefore, x ≠ 4.

Step 2: Combine the numerators.

Step 3: Collect like terms.

3x + 2 + x + 5 = 4x + 7

Final answer:

No common factors can be cancelled, so the expression is fully simplified.

Adding Expressions with Different Denominators

When denominators are different, we must first find a common denominator.

For numerical fractions:

1/3 + 1/4 = 4/12 + 3/12 = 7/12

The same method works for algebraic fractions.

Worked Example 2: Adding with Different Linear Denominators

Simplify:

Step 1: Identify restrictions.

x ≠ 0 and x ≠ −1.

Step 2: Find the LCD.

The denominators are x and x + 1.

LCD = x(x + 1)

Step 3: Rewrite each fraction.

Step 4: Add the numerators.

Step 5: Expand and simplify.

2x + 2 + 3x = 5x + 2

Final answer:

Worked Example 3: Adding with Factored Denominators

Simplify:

Step 1: Factor the denominators.

x² − 4 = (x − 2)(x + 2)

Step 2: Identify restrictions.

x ≠ 2 and x ≠ −2.

Step 3: Find the LCD.

LCD = (x − 2)(x + 2)

Step 4: Rewrite the second fraction.

Step 5: Combine numerators.

1 + 2(x − 2)

= 1 + 2x − 4

= 2x − 3

Final answer:

Subtracting Rational Expressions

Subtraction follows the same principles as addition.

When denominators are equal:

where c ≠ 0.

The important difference is that subtraction requires careful use of negative signs.

Worked Example 4: Subtracting with the Same Denominator

Simplify:

Step 1: Identify restrictions.

x ≠ −2.

Step 2: Combine numerators.

Step 3: Distribute the negative sign.

5x + 3 − 2x + 4

= 3x + 7

Final answer:

Worked Example 5: Subtracting with Different Denominators

Simplify:

Step 1: Identify restrictions.

x ≠ 1 and x ≠ −2.

Step 2: Find the LCD.

LCD = (x − 1)(x + 2)

Step 3: Rewrite.

Step 4: Combine numerators.

3(x + 2) − 2(x − 1)

Step 5: Expand.

3x + 6 − 2x + 2

= x + 8

Final answer:

Finding the Least Common Denominator

For more complicated expressions, finding the LCD is often the most important step.

Consider:

Factor the denominators:

x² − 9 = (x − 3)(x + 3)

x² + 3x = x(x + 3)

The LCD must contain every distinct factor, using the highest power needed.

Therefore:

LCD = x(x − 3)(x + 3)

The restrictions are:

x ≠ −3, x ≠ 0, and x ≠ 3.

Rewrite:

Combine:

Simplify:

Factor the numerator:

There are no common factors to cancel.

Final answer:

3(x − 2)/[x(x − 3)(x + 3)], where x ≠ −3, 0, 3.

Multiplying Rational Expressions

Multiplication is often simpler than addition or subtraction because a common denominator is not required.

The general rule is:

where b ≠ 0 and d ≠ 0.

The most efficient approach is usually to factor before multiplying, then cancel common factors.

Worked Example 6: Multiplying Simple Rational Expressions

Simplify:

Step 1: Identify restrictions.

The original denominators are x + 2 and 6.

Therefore, x ≠ −2.

Step 2: Multiply the fractions.

Step 3: Cancel the common factor (x + 2).

3x/6

Step 4: Simplify numerical coefficients.

x/2

Final answer: x/2, where x ≠ −2.

Worked Example 7: Multiplying with Quadratic Factors

Simplify:

Step 1: Identify restrictions.

x + 4 ≠ 0 and x − 3 ≠ 0.

Therefore:

x ≠ −4 and x ≠ 3.

Step 2: Factor the numerator.

x² − 9 = (x − 3)(x + 3)

Step 3: Rewrite.

Step 4: Cancel common factors.

Cancel (x − 3) and (x + 4).

The remaining expression is:

x + 3

Final answer: x + 3, where x ≠ −4 and x ≠ 3.

Worked Example 8: Multiplying Two Quadratic Rational Expressions

Simplify:

Step 1: Identify restrictions.

x² − 4 = (x − 2)(x + 2)

The other denominator is x + 3.

Therefore:

x ≠ 2, x ≠ −2, and x ≠ −3.

Step 2: Factor completely.

x² + 5x + 6 = (x + 2)(x + 3)

x² − 4 = (x − 2)(x + 2)

Rewrite:

Step 3: Cancel common factors.

Cancel (x + 2), (x + 3), and (x − 2).

All factors cancel, leaving:

1

Final answer: 1, where x ≠ −3, −2, and 2.

Even though the final expression is a constant, the original restrictions must remain.

Dividing Rational Expressions

Division of rational expressions follows the same rule as division of numerical fractions.

To divide by a fraction, multiply by its reciprocal.

For example:

The same rule applies to algebraic fractions.

where b, c, and d must all be nonzero.

There is an additional restriction when dividing rational expressions: the divisor itself cannot equal zero.

Worked Example 9: Dividing Simple Rational Expressions

Simplify:

Step 1: Identify restrictions.

The denominators require x ≠ 0.

The divisor (x + 2)/3 must not equal zero.

Therefore, x ≠ −2.

Step 2: Multiply by the reciprocal.

Step 3: Cancel (x + 2).

3/x

Final answer: 3/x, where x ≠ 0 and x ≠ −2.

Worked Example 10: Dividing with Factoring

Simplify:

Step 1: Identify restrictions.

The original denominators require:

x ≠ −3.

The divisor must also be nonzero:

(x − 4)/(x + 3) ≠ 0

Therefore:

x ≠ 4.

Step 2: Multiply by the reciprocal.

Step 3: Factor the difference of squares.

x² − 16 = (x − 4)(x + 4)

Rewrite:

Step 4: Cancel common factors.

The remaining expression is:

x + 4

Final answer: x + 4, where x ≠ −3 and x ≠ 4.

Worked Example 11: Dividing Quadratic Rational Expressions

Simplify:

Step 1: Identify restrictions.

The original denominators are:

x² − 9 = (x − 3)(x + 3)

and x − 3.

Therefore:

x ≠ −3 and x ≠ 3.

The divisor (x + 2)/(x − 3) must not equal zero.

Therefore, x ≠ −2.

Step 2: Rewrite using the reciprocal.

Step 3: Factor completely.

Step 4: Cancel common factors.

Cancel (x + 2), (x + 3), and (x − 3).

The result is:

1

Final answer: 1, where x ≠ −3, −2, and 3.

Mixed Operations with Rational Expressions

Some problems involve more than one operation.

The correct order of operations is important:

1. Simplify expressions inside brackets.

2. Perform multiplication and division from left to right.

3. Perform addition and subtraction from left to right.

4. Factor and simplify the final result.

5. Preserve all original restrictions.

Worked Example 12: Combining Addition and Subtraction

Simplify:

Step 1: Identify restrictions.

x ≠ 2 and x ≠ 0.

Step 2: Combine fractions with the same denominator.

3/(x − 2) − 1/(x − 2) = 2/(x − 2)

The expression becomes:

2/(x − 2) + 4/x

Step 3: Find the LCD.

LCD = x(x − 2)

Step 4: Rewrite.

Step 5: Combine numerators.

2x + 4(x − 2)

= 2x + 4x − 8

= 6x − 8

Step 6: Factor the numerator.

6x − 8 = 2(3x − 4)

Final answer:

Worked Example 13: Combining Multiplication and Addition

Simplify:

Step 1: Identify restrictions.

x ≠ −1.

Step 2: Perform multiplication first.

The expression becomes:

x/2 + 3/2

Step 3: Add.

(x + 3)/2

Final answer: (x + 3)/2, where x ≠ −1.

Verifying Rational Expression Operations

A reliable way to check an answer is to substitute a permitted numerical value into both the original and simplified expressions.

Consider:

Choose x = 2.

Original:

2/2 + 3/3 = 1 + 1 = 2

Simplified:

(5(2) + 2)/[2(3)] = 12/6 = 2

Both expressions give the same value.

This supports the simplification.

However, numerical substitution alone is not a complete proof. Algebraic manipulation establishes equivalence across the permitted domain.

Real-World Applications

Application 1: Combining Work Rates

Suppose one machine can complete a job in 4 hours and another machine can complete the same job in 6 hours.

The first machine completes 1/4 of the job per hour.

The second completes 1/6 of the job per hour.

Their combined work rate is:

Find the LCD:

LCD = 12

Rewrite:

3/12 + 2/12 = 5/12

Together, the machines complete 5/12 of the job each hour.

The time needed to complete one job is:

1 ÷ (5/12) = 12/5 = 2.4 hours.

Therefore, the machines working together can complete the job in 2 hours and 24 minutes.

Application 2: Electrical Resistance

For two resistors connected in parallel:

Suppose:

R₁ = 6 Ω

R₂ = 3 Ω

Then:

1/R = 1/6 + 1/3

Find the LCD:

1/R = 1/6 + 2/6 = 3/6 = 1/2

Taking the reciprocal gives:

R = 2 Ω

The equivalent resistance is 2 Ω.

This demonstrates how adding rational expressions is used in electrical engineering.

Application 3: Average Speed

A vehicle travels a distance d at speed v₁ and returns over the same distance at speed v₂.

Total distance:

2d

Total time:

d/v₁ + d/v₂

Average speed:

Combine the denominator:

Therefore:

Cancel d:

For speeds of 60 km/h and 90 km/h:

Average speed = 2(60)(90)/(60 + 90)

= 10,800/150

= 72 km/h.

Notice that the average speed is not simply the arithmetic mean of the two speeds, because the vehicle spends different amounts of time travelling at each speed.

Common Mistakes

Mistake 1: Adding Denominators

Incorrect:

1/x + 1/(x + 2) = 2/(2x + 2)

Correct:

Use the LCD x(x + 2).

Mistake 2: Forgetting Negative Signs During Subtraction

Incorrect:

(3x + 2) − (x − 4) = 2x − 2

Correct:

3x + 2 − x + 4 = 2x + 6

Always distribute the negative sign across every term in the second numerator.

Mistake 3: Finding an LCD for Multiplication

A common denominator is not required when multiplying rational expressions.

Instead, multiply numerators and denominators, preferably after factoring and cancelling common factors.

Mistake 4: Forgetting to Use the Reciprocal

Incorrect:

(a/b) ÷ (c/d) = ac/bd

Correct:

(a/b) ÷ (c/d) = ad/bc

Mistake 5: Cancelling Terms Instead of Factors

Incorrect:

(x + 3)/(x + 5) = 3/5

Correct:

The expression cannot be simplified by cancellation because the x terms are part of sums.

Mistake 6: Forgetting Original Restrictions

Even when a denominator factor cancels, its zero must remain excluded.

For division, also exclude any value that makes the original divisor zero.

Mistake 7: Leaving an Answer Partially Simplified

For example:

(6x + 12)/(3x)

can be simplified by factoring the numerator:

6(x + 2)/(3x)

Then reducing 6/3 gives:

2(x + 2)/x, where x ≠ 0.

Always check whether additional common factors remain.

Key Terms

• Rational Expression: A fraction whose numerator and denominator are polynomials.

• Numerator: The expression above a fraction bar.

• Denominator: The expression below a fraction bar.

• Common Denominator: A denominator shared by two or more fractions.

• Least Common Denominator (LCD): The least polynomial common multiple of the denominators, up to a nonzero constant.

• Factor: An expression multiplied by another expression to form a product.

• Reciprocal: The multiplicative inverse of a nonzero number or expression.

• Equivalent Expressions: Expressions that produce the same output for every input in a specified common domain.

• Domain Restriction: A condition excluding inputs that make an expression undefined.

• Excluded Value: A value that makes an original denominator zero or, in a division problem, makes the divisor zero.

• Simplification: Rewriting an expression in an equivalent, less complicated form.

• Cancellation: Dividing numerator and denominator by a shared nonzero factor.

• Complex Fraction: A fraction containing one or more fractions in its numerator or denominator.

• Work Rate: The amount of work completed per unit of time.

Key Takeaways

1. Addition and subtraction of rational expressions require a common denominator.

2. When denominators are equal, combine the numerators and keep the denominator.

3. The LCD is found by factoring denominators and including every required factor at its highest power.

4. Multiplication involves multiplying numerators together and denominators together.

5. Division requires multiplying by the reciprocal of the divisor.

6. Factoring before multiplication or division often makes simplification easier.

7. Only common factors can be cancelled, not individual terms.

8. Restrictions from original denominators must always be preserved.

9. In division, the original divisor must also be nonzero.

10. Final answers should be factored or reduced wherever possible and checked for correctness.

Check Your Understanding

Question 1: Addition with a Common Denominator

Simplify:

Question 2: Addition with Different Denominators

Simplify:

Question 3: Subtraction

Simplify:

Question 4: Multiplication

Simplify:

Question 5: Division

Simplify:

Question 6: Adding with Factored Denominators

Simplify:

Question 7: Mixed Operations

Simplify:

Question 8: Real-World Application

One pump fills a tank in 5 hours. Another pump fills the same tank in 10 hours.

a) Write the work rate of each pump.

b) Calculate their combined work rate.

c) Determine how long the pumps take to fill the tank when working together.

Fully Worked Answers

Answer 1

Given:

(2x + 3)/(x − 4) + (x + 5)/(x − 4)

Combine numerators:

(2x + 3 + x + 5)/(x − 4)

= (3x + 8)/(x − 4)

Final answer: (3x + 8)/(x − 4), where x ≠ 4.

Answer 2

Given:

1/x + 2/(x + 3)

LCD = x(x + 3)

Rewrite:

(x + 3)/[x(x + 3)] + 2x/[x(x + 3)]

Combine:

(x + 3 + 2x)/[x(x + 3)]

= (3x + 3)/[x(x + 3)]

Factor:

Final answer: 3(x + 1)/[x(x + 3)], where x ≠ 0 and x ≠ −3.

Answer 3

Given:

5/(x + 1) − 2/(x − 2)

LCD = (x + 1)(x − 2)

Rewrite:

Expand:

5x − 10 − 2x − 2

= 3x − 12

Factor:

3(x − 4)

Final answer: 3(x − 4)/[(x + 1)(x − 2)], where x ≠ −1 and x ≠ 2.

Answer 4

Given:

(x² − 9)/(x + 2) × (x + 2)/(x − 3)

Factor:

Cancel common factors:

x + 3

Final answer: x + 3, where x ≠ −2 and x ≠ 3.

Answer 5

Given:

(x + 4)/x ÷ (x + 4)/2

Multiply by the reciprocal:

Cancel (x + 4):

2/x

The original denominator requires x ≠ 0.

The divisor must be nonzero, so x ≠ −4.

Final answer: 2/x, where x ≠ 0 and x ≠ −4.

Answer 6

Given:

1/(x² − 4) + 3/(x + 2)

Factor:

x² − 4 = (x − 2)(x + 2)

LCD = (x − 2)(x + 2)

Rewrite:

Combine:

[1 + 3(x − 2)]/[(x − 2)(x + 2)]

Expand:

1 + 3x − 6 = 3x − 5

Final answer: (3x − 5)/[(x − 2)(x + 2)], where x ≠ −2 and x ≠ 2.

Answer 7

Given:

2/x + 3/(x − 1) − 1/x

Combine like denominators:

2/x − 1/x = 1/x

Now:

1/x + 3/(x − 1)

LCD = x(x − 1)

Rewrite:

Combine:

(x − 1 + 3x)/[x(x − 1)]

= (4x − 1)/[x(x − 1)]

Final answer: (4x − 1)/[x(x − 1)], where x ≠ 0 and x ≠ 1.

Answer 8

a) Work rates

First pump:

1/5 tank per hour

Second pump:

1/10 tank per hour

b) Combined work rate

1/5 + 1/10

= 2/10 + 1/10

= 3/10 tank per hour

c) Time required

Time = 1 ÷ (3/10)

= 10/3 hours

= 3⅓ hours

= 3 hours and 20 minutes.

This completes Operations with Rational Expressions, covering addition, subtraction, multiplication, division, complete simplification, restrictions, and practical applications.

Introduction

Rational functions are used to describe mathematical relationships involving ratios, rates, inverse variation, and quantities that change in relation to one another.

Unlike many polynomial functions, rational functions can have graphs that are separated into different branches, approach certain lines without reaching them, or contain missing points called holes.

For example, consider:

f(x) = 1/x

As x approaches zero from the positive side, f(x) becomes increasingly large. As x approaches zero from the negative side, f(x) becomes increasingly negative. However, the function is never defined at x = 0.

This behaviour introduces one of the most important features of rational function graphs: asymptotes.

Understanding how to identify asymptotes, calculate intercepts, and analyze the behaviour of a function allows us to sketch rational graphs accurately without plotting hundreds of points.

In this lesson, we will investigate vertical, horizontal, and oblique asymptotes, calculate x- and y-intercepts, construct rational function graphs, and interpret their behaviour in mathematical and real-world situations.

Understanding the Shape of Rational Functions

A rational function has the general form:

f(x) = P(x)/Q(x)

where P(x) and Q(x) are polynomials and Q(x) is not the zero polynomial.

Consider the parent rational function:

f(x) = 1/x

Important observations:

• The graph has two separate branches.

• The function is undefined at x = 0.

• The graph approaches the y-axis without crossing it.

• The graph approaches the x-axis as x becomes very large in magnitude.

• The function is positive when x is positive and negative when x is negative.

These observations help us understand the behaviour of more complicated rational functions.

Vertical Asymptotes

A vertical asymptote is a vertical line x = a that a graph approaches as the function increases or decreases without bound near a particular input value.

For example:

f(x) = 1/(x − 3)

The denominator becomes zero when:

x − 3 = 0

x = 3

Therefore, the graph has a vertical asymptote at:

x = 3

As x approaches 3 from the right, f(x) approaches positive infinity.

As x approaches 3 from the left, f(x) approaches negative infinity.

Finding Vertical Asymptotes

1. Factor the numerator and denominator.

2. Identify and cancel any common factors, while recording the original restrictions.

3. Set the remaining denominator equal to zero.

4. Determine whether the function grows without bound near those values.

For a fully reduced rational function, real zeros of its denominator are vertical asymptotes.

Worked Example 1: Finding a Vertical Asymptote

Find the vertical asymptote of:

f(x) = (2x + 1)/(x − 4)

Step 1: Set the denominator equal to zero.

x − 4 = 0

Step 2: Solve.

x = 4

Step 3: Check the numerator.

At x = 4:

2(4) + 1 = 9

The numerator is not zero, so no factor cancels.

Answer: Vertical asymptote x = 4.

Worked Example 2: Multiple Vertical Asymptotes

Find the vertical asymptotes of:

f(x) = 3/(x² − 9)

Factor the denominator:

x² − 9 = (x − 3)(x + 3)

Set each factor equal to zero:

x − 3 = 0 → x = 3

x + 3 = 0 → x = −3

Neither factor cancels.

Answer: Vertical asymptotes x = −3 and x = 3.

Vertical Asymptotes Versus Holes

Not every zero of an original denominator produces a vertical asymptote.

Consider:

f(x) = (x² − 4)/(x − 2)

Factor:

f(x) = (x − 2)(x + 2)/(x − 2)

Simplify:

f(x) = x + 2, where x ≠ 2.

The graph is the line y = x + 2 with a missing point at (2, 4).

This is a hole, not a vertical asymptote.

Horizontal Asymptotes

A horizontal asymptote is a horizontal line y = L that a function approaches as x increases or decreases without bound.

Horizontal asymptotes describe the end behaviour of a function.

For rational functions, horizontal asymptotes can often be found by comparing the degrees of the numerator and denominator.

The degree of a polynomial is its highest power of x.

Rules for Horizontal Asymptotes

Let:

f(x) = P(x)/Q(x)

The leading coefficient is the coefficient of the highest-degree term.

Worked Example 3: Numerator Degree Smaller

Find the horizontal asymptote of:

f(x) = (3x + 2)/(x² + 1)

Numerator degree = 1.

Denominator degree = 2.

Since 1 is less than 2:

Horizontal asymptote: y = 0.

As x becomes very large in either direction, the denominator grows faster than the numerator, causing the fraction to approach zero.

Worked Example 4: Equal Degrees

Find the horizontal asymptote of:

f(x) = (4x² + 3x − 1)/(2x² + 5)

Both polynomials have degree 2.

Compare their leading coefficients:

4/2 = 2

Horizontal asymptote: y = 2.

Worked Example 5: Numerator Degree Greater

Find the horizontal asymptote of:

f(x) = (x² + 1)/(x − 2)

Numerator degree = 2.

Denominator degree = 1.

The numerator degree is greater.

Therefore, there is no horizontal asymptote.

However, because the degree difference is exactly 1, the function has an oblique asymptote.

Can a Graph Cross a Horizontal Asymptote?

Yes.

A horizontal asymptote describes what happens as x approaches positive or negative infinity. It does not necessarily prevent the graph from crossing the line.

For example:

f(x) = (x + 1)/(x² + 1)

has horizontal asymptote y = 0.

The function crosses y = 0 when x = −1.

Therefore, horizontal asymptotes are not barriers that a graph can never cross.

Oblique Asymptotes

An oblique asymptote, also called a slant asymptote, is a nonhorizontal straight line that a rational function approaches as x becomes very large in magnitude.

An oblique asymptote occurs when the degree of the numerator is exactly one greater than the degree of the denominator, after simplification.

To find it, divide the numerator by the denominator using polynomial division.

Worked Example 6: Finding an Oblique Asymptote

Find the oblique asymptote of:

f(x) = (x² + 3x + 1)/(x + 1)

Step 1: Divide x² + 3x + 1 by x + 1.

First term:

x² ÷ x = x

Multiply:

x(x + 1) = x² + x

Subtract:

(x² + 3x + 1) − (x² + x) = 2x + 1

Next term:

2x ÷ x = 2

Multiply:

2(x + 1) = 2x + 2

Subtract:

(2x + 1) − (2x + 2) = −1

Step 2: Rewrite the function.

f(x) = x + 2 − 1/(x + 1)

Step 3: Analyze end behaviour.

As x approaches positive or negative infinity:

−1/(x + 1) approaches zero.

Therefore:

Oblique asymptote: y = x + 2.

The vertical asymptote is x = −1.

Worked Example 7: Another Oblique Asymptote

Find the oblique asymptote of:

f(x) = (2x² − x + 5)/(x − 2)

Polynomial division gives:

2x² − x + 5 = (x − 2)(2x + 3) + 11

Therefore:

f(x) = 2x + 3 + 11/(x − 2)

As x becomes very large in magnitude, the remainder term approaches zero.

Oblique asymptote: y = 2x + 3.

What About Larger Degree Differences?

If the numerator degree exceeds the denominator degree by 2 or more, polynomial division may produce a quadratic or higher-degree polynomial asymptote rather than a straight oblique asymptote.

For example:

f(x) = (x³ + 1)/x

Simplify:

f(x) = x² + 1/x

As |x| becomes very large, 1/x approaches zero.

Therefore, the graph approaches the parabola:

y = x²

This is a polynomial asymptote, not an oblique line.

Finding Intercepts

Intercepts help us locate where a rational function crosses the coordinate axes.

Finding x-Intercepts

An x-intercept occurs where:

f(x) = 0

For a rational function, a fraction equals zero when its numerator equals zero and its denominator is nonzero.

Therefore:

1. Set the numerator equal to zero.

2. Solve for x.

3. Check that each solution is in the original domain.

Worked Example 8: Finding x-Intercepts

Find the x-intercepts of:

f(x) = (x² − 9)/(x + 1)

Set the numerator equal to zero:

x² − 9 = 0

Factor:

(x − 3)(x + 3) = 0

Solve:

x = 3 or x = −3

The denominator is zero only at x = −1, so both solutions are valid.

x-intercepts: (−3, 0) and (3, 0).

Finding the y-Intercept

A y-intercept occurs when x = 0.

Substitute x = 0 into the function, provided zero belongs to its domain.

Worked Example 9: Finding the y-Intercept

For:

f(x) = (x² − 9)/(x + 1)

Substitute x = 0:

f(0) = (0 − 9)/(0 + 1)

f(0) = −9

y-intercept: (0, −9).

When an Intercept Does Not Exist

Consider:

f(x) = 1/x

There is no x-intercept because the numerator never equals zero.

There is no y-intercept because x = 0 is excluded from the domain.

A rational function may have multiple intercepts, one intercept, or none.

A Complete Method for Sketching Rational Functions

A systematic approach makes graphing much easier.

1. Factor the function. Identify any common factors.

2. Determine the domain. Record all values that make the original denominator zero.

3. Identify holes. Locate cancelled factors and calculate their missing coordinates.

4. Find vertical asymptotes. Solve the remaining denominator equal to zero.

5. Find horizontal or oblique asymptotes. Compare polynomial degrees and use division if necessary.

6. Calculate intercepts. Find valid x-intercepts and the y-intercept.

7. Test values in different intervals. Determine where branches lie and how they behave.

8. Sketch the graph. Draw smooth curves consistent with the asymptotes, intercepts, holes, and end behaviour.

Worked Example 10: Sketching a Rational Function with Two Asymptotes

Sketch:

f(x) = (x + 1)/(x − 2)

Step 1: Determine the domain.

x − 2 = 0

x = 2

Domain: x ≠ 2.

Step 2: Find the vertical asymptote.

x = 2.

Step 3: Find the horizontal asymptote.

Both polynomials have degree 1.

The ratio of leading coefficients is 1/1.

Horizontal asymptote: y = 1.

Step 4: Find the x-intercept.

x + 1 = 0

x = −1

x-intercept: (−1, 0).

Step 5: Find the y-intercept.

f(0) = 1/(−2) = −1/2

y-intercept: (0, −0.5).

Step 6: Rewrite the function.

f(x) = 1 + 3/(x − 2)

This reveals the graph's behaviour.

When x approaches 2 from the left, the function approaches negative infinity.

When x approaches 2 from the right, the function approaches positive infinity.

As x approaches either positive or negative infinity, the function approaches y = 1.

Worked Example 11: Sketching a Function with Two Vertical Asymptotes

Sketch:

f(x) = 1/(x² − 4)

Step 1: Factor the denominator.

x² − 4 = (x − 2)(x + 2)

Step 2: Find vertical asymptotes.

x = −2 and x = 2.

Step 3: Find the horizontal asymptote.

Numerator degree = 0.

Denominator degree = 2.

Horizontal asymptote: y = 0.

Step 4: Find intercepts.

The numerator is 1, so there are no x-intercepts.

For the y-intercept:

f(0) = 1/(−4) = −1/4

y-intercept: (0, −0.25).

Step 5: Determine signs.

Step 6: Interpret the graph.

The graph has three branches.

The middle branch lies below the x-axis and approaches negative infinity near both vertical asymptotes.

The outer branches lie above the x-axis, approach positive infinity near their respective vertical asymptotes, and approach zero as x moves far away.

Worked Example 12: Sketching a Function with an Oblique Asymptote

Sketch:

f(x) = (x² + 1)/(x − 1)

Step 1: Find the vertical asymptote.

x − 1 = 0

x = 1.

Step 2: Find the oblique asymptote.

Divide x² + 1 by x − 1:

f(x) = x + 1 + 2/(x − 1)

Therefore:

Oblique asymptote: y = x + 1.

Step 3: Find the x-intercepts.

x² + 1 = 0

There are no real solutions.

Therefore, there are no x-intercepts.

Step 4: Find the y-intercept.

f(0) = 1/(−1) = −1

y-intercept: (0, −1).

Step 5: Interpret the behaviour.

As x approaches 1 from the left, f(x) approaches negative infinity.

As x approaches 1 from the right, f(x) approaches positive infinity.

For large positive x, the graph approaches y = x + 1 from above.

For large negative x, it approaches the line from below.

The graph has two branches separated by x = 1.

Interpreting Asymptotic Behaviour

Asymptotes help us describe what happens near critical input values and at the extremes of the graph.

Behaviour Near Vertical Asymptotes

Consider:

f(x) = 1/(x − 2)

As x approaches 2 from the left, f(x) decreases without bound.

As x approaches 2 from the right, f(x) increases without bound.

Using limit notation:

The minus and plus signs on the approach indicate the left and right sides.

Behaviour Near Horizontal Asymptotes

Consider:

f(x) = (2x + 3)/(x + 1)

Rewrite:

f(x) = 2 + 1/(x + 1)

As x approaches positive infinity, 1/(x + 1) approaches zero.

As x approaches negative infinity, the same term also approaches zero.

Therefore:

The horizontal asymptote is y = 2.

Behaviour Near Oblique Asymptotes

Consider:

f(x) = x + 3 + 5/(x − 2)

As x becomes very large in magnitude:

5/(x − 2) approaches zero.

Therefore, the difference between f(x) and the line y = x + 3 approaches zero.

This is the defining behaviour of an oblique asymptote.

Real-World Applications of Rational Function Graphs

Rational functions are useful for modelling situations where one quantity depends on the ratio of two changing quantities.

Application 1: Average Production Cost

Suppose a factory has a fixed cost of $1,200 and a variable production cost of $15 per item.

The total cost of producing x items is:

C(x) = 1200 + 15x

The average cost per item is:

A(x) = (1200 + 15x)/x

Simplify:

A(x) = 1200/x + 15

The mathematical function has a vertical asymptote at x = 0 and a horizontal asymptote at y = 15.

In the real-world situation, x must be a positive whole number.

Interpretation: As more items are produced, the fixed cost is spread across more units. The average cost decreases toward $15 but remains above $15 for every finite positive production quantity.

Application 2: Travel Time and Speed

The time needed to travel a fixed distance is:

Time = Distance / Speed

For a journey of 180 km:

t(v) = 180/v

where v is speed in km/h.

The mathematical graph has asymptotes at v = 0 and t = 0.

In the physical model, only positive speeds are meaningful.

As speed increases, travel time decreases toward zero, but it cannot equal zero for a finite positive speed.

Application 3: Concentration and Dilution

Suppose 50 grams of a dissolved substance are contained in a solution with a volume of V litres.

The concentration is:

C(V) = 50/V

The function has:

• A vertical asymptote at V = 0.

• A horizontal asymptote at C = 0.

As the volume increases, concentration decreases.

This relationship helps explain dilution in chemistry, although the model assumes the amount of dissolved substance remains constant.

Application 4: Electrical Resistance

For a constant voltage of 24 V, Ohm's law gives:

I(R) = 24/R

where I is current and R is resistance.

For positive resistance, increasing resistance decreases the current.

The graph approaches the horizontal asymptote I = 0 as resistance becomes very large.

This helps describe how electrical current responds to changes in resistance.

Common Mistakes

Mistake 1: Treating Every Denominator Zero as a Vertical Asymptote

Incorrect: The function (x² − 9)/(x − 3) has a vertical asymptote at x = 3.

Correct: Factoring and cancelling gives x + 3, where x ≠ 3. The graph has a hole at (3, 6), not a vertical asymptote.

Mistake 2: Comparing the Wrong Polynomial Degrees

For f(x) = (3x² + 1)/(2x² − 5), both degrees are 2.

The horizontal asymptote is y = 3/2, not y = 0.

Mistake 3: Assuming a Graph Cannot Cross a Horizontal Asymptote

A graph can cross a horizontal asymptote. The asymptote describes end behaviour rather than a forbidden boundary.

Mistake 4: Finding x-Intercepts Without Checking the Domain

A zero of the numerator is not a valid x-intercept if the original denominator is also zero at that value.

Always check for cancelled factors and excluded values.

Mistake 5: Confusing Vertical and Horizontal Asymptotes

Vertical asymptotes are equations of the form x = a.

Horizontal asymptotes are equations of the form y = b.

Mistake 6: Drawing Branches on the Wrong Side of an Asymptote

Knowing where an asymptote is located does not tell us whether the graph approaches positive or negative infinity.

Use sign analysis or test values on either side of the asymptote.

Mistake 7: Assuming Every Rational Function Has a Horizontal Asymptote

A rational function may have a horizontal, oblique, or higher-degree polynomial asymptote, depending on the degrees of its polynomials.

Key Terms

• Rational Function: A function expressed as the quotient of two polynomials.

• Domain: The set of permitted input values.

• Excluded Value: An input that makes the original denominator zero.

• Vertical Asymptote: A vertical line approached by a graph as its function values increase or decrease without bound.

• Horizontal Asymptote: A horizontal line approached by a graph as x tends toward positive or negative infinity.

• Oblique Asymptote: A nonhorizontal straight line approached by a graph as x becomes very large in magnitude.

• Polynomial Asymptote: A polynomial curve or line approached by a function at infinity.

• Degree: The highest exponent of a variable in a polynomial.

• Leading Coefficient: The coefficient of the highest-degree term.

• x-Intercept: A point where the graph crosses or touches the x-axis.

• y-Intercept: A point where the graph crosses the y-axis.

• Hole: A missing point caused by a removable discontinuity.

• End Behaviour: The behaviour of a function as x approaches positive or negative infinity.

• One-Sided Limit: The value or behaviour approached by a function from one side of an input.

• Polynomial Division: A method for dividing one polynomial by another.

Key Takeaways

1. Rational function graphs can have multiple branches, holes, and asymptotes.

2. Vertical asymptotes are found from the zeros of the denominator after cancelling common factors.

3. A cancelled denominator factor may create a hole instead of a vertical asymptote.

4. Horizontal asymptotes are determined by comparing the degrees of the numerator and denominator.

5. When the numerator degree is smaller, the horizontal asymptote is y = 0.

6. When the degrees are equal, the horizontal asymptote is the ratio of leading coefficients.

7. When the numerator degree is exactly one greater, polynomial division gives an oblique asymptote.

8. x-intercepts occur at valid zeros of the numerator; y-intercepts are found by substituting x = 0.

9. Accurate graph sketches require asymptotes, intercepts, holes, sign analysis, and end behaviour.

10. Asymptotes help interpret long-term trends and extreme behaviour in real-world models.

Check Your Understanding

Question 1: Vertical Asymptotes

Find the vertical asymptotes of:

f(x) = 5/(x² − 16)

Question 2: Horizontal Asymptotes

Find the horizontal asymptote of:

f(x) = (6x² + 3x − 1)/(2x² + 7)

Question 3: Oblique Asymptotes

Find the oblique asymptote of:

f(x) = (x² + 4x + 2)/(x + 1)

Question 4: Intercepts

Find the x- and y-intercepts of:

f(x) = (x − 4)/(x + 2)

Question 5: Holes and Asymptotes

Analyze:

f(x) = (x² − 9)/(x − 3)

Determine whether the graph has a hole or vertical asymptote and identify its coordinates or equation.

Question 6: Complete Graph Analysis

For:

f(x) = (2x + 4)/(x − 1)

Find:

a) The domain.

b) The vertical asymptote.

c) The horizontal asymptote.

d) The x-intercept.

e) The y-intercept.

f) The behaviour near the vertical asymptote.

Question 7: Multiple Vertical Asymptotes

For:

f(x) = x/(x² − 4)

Determine the vertical and horizontal asymptotes, all intercepts, and the sign of the function on each interval separated by its vertical asymptotes.

Question 8: Real-World Interpretation

A business models its average cost per item using:

A(x) = 500/x + 20

where x is the number of items produced.

a) Identify the mathematical vertical asymptote.

b) Identify the horizontal asymptote.

c) Calculate the average cost for 50 items.

d) Explain the practical meaning of the horizontal asymptote.

Fully Worked Answers

Answer 1: Vertical Asymptotes

Given:

f(x) = 5/(x² − 16)

Set the denominator equal to zero:

x² − 16 = 0

Factor:

(x − 4)(x + 4) = 0

Solve:

x = 4 or x = −4

Neither factor cancels.

Answer: x = −4 and x = 4.

Answer 2: Horizontal Asymptote

Given:

f(x) = (6x² + 3x − 1)/(2x² + 7)

Both polynomials have degree 2.

The ratio of leading coefficients is:

6/2 = 3

Answer: y = 3.

Answer 3: Oblique Asymptote

Given:

f(x) = (x² + 4x + 2)/(x + 1)

Divide x² + 4x + 2 by x + 1.

First:

x² ÷ x = x

Subtract x² + x from the numerator:

3x + 2

Next:

3x ÷ x = 3

Subtract 3x + 3:

Remainder = −1

Therefore:

f(x) = x + 3 − 1/(x + 1)

As |x| becomes large, the remainder term approaches zero.

Answer: y = x + 3.

Answer 4: Intercepts

Given:

f(x) = (x − 4)/(x + 2)

x-intercept:

Set numerator equal to zero:

x − 4 = 0

x = 4

Therefore, (4, 0).

y-intercept:

Substitute x = 0:

f(0) = −4/2 = −2

Therefore, (0, −2).

Answer: x-intercept (4, 0); y-intercept (0, −2).

Answer 5: Hole or Asymptote

Given:

f(x) = (x² − 9)/(x − 3)

Factor:

f(x) = (x − 3)(x + 3)/(x − 3)

Simplify:

f(x) = x + 3, where x ≠ 3.

The graph follows y = x + 3 but is missing the point where x = 3.

At x = 3, the corresponding line value is:

y = 3 + 3 = 6

Answer: Hole at (3, 6). There is no vertical asymptote.

Answer 6: Complete Graph Analysis

Given:

f(x) = (2x + 4)/(x − 1)

a) Domain

x − 1 ≠ 0

x ≠ 1.

b) Vertical asymptote

x = 1.

c) Horizontal asymptote

Both polynomials have degree 1.

Leading coefficient ratio:

2/1 = 2

Therefore, y = 2.

d) x-intercept

2x + 4 = 0

2x = −4

x = −2

x-intercept: (−2, 0).

e) y-intercept

f(0) = 4/(−1) = −4

y-intercept: (0, −4).

f) Behaviour near the vertical asymptote

Rewrite:

f(x) = 2 + 6/(x − 1)

As x approaches 1 from the left, f(x) approaches negative infinity.

As x approaches 1 from the right, f(x) approaches positive infinity.

Answer 7: Multiple Asymptotes

Given:

f(x) = x/(x² − 4)

Factor the denominator:

x² − 4 = (x − 2)(x + 2)

Vertical asymptotes:

x = −2 and x = 2.

Horizontal asymptote:

Numerator degree = 1.

Denominator degree = 2.

Therefore, y = 0.

Intercepts:

Set the numerator equal to zero:

x = 0

The denominator at x = 0 is −4, so the point is valid.

Both the x-intercept and y-intercept are (0, 0).

Sign analysis:

These signs help determine the positions of the graph's branches.

Answer 8: Real-World Interpretation

Given:

A(x) = 500/x + 20

a) Vertical asymptote

x = 0.

b) Horizontal asymptote

As x becomes very large, 500/x approaches zero.

Therefore, y = 20.

c) Average cost for 50 items

A(50) = 500/50 + 20

= 10 + 20

= $30 per item.

d) Interpretation

As production increases, the fixed cost is distributed across more items.

The average cost approaches $20 per item, representing the variable production cost per unit in this model.

The average cost remains above $20 for every finite positive production quantity.

This completes Graphing Rational Functions, including vertical, horizontal, and oblique asymptotes, intercepts, graph sketching, and interpretation of asymptotic behaviour.

Introduction

Rational equations are equations that contain rational expressions—fractions whose numerators and denominators are polynomials.

They appear in many areas of mathematics and in real-world situations involving speed, time, work rates, concentration, and electrical circuits.

For example, suppose two machines can complete a job in different amounts of time. How long would they take if they worked together? Or suppose a vehicle travels a fixed distance at two different speeds. How much faster must it travel to save a certain amount of time?

Both problems can be solved using rational equations.

However, rational equations have an important challenge: some algebraic solutions may not be valid in the original equation.

These invalid results are called extraneous solutions. They can arise when we multiply both sides of an equation by an expression containing a variable.

To solve rational equations correctly, we must identify restrictions, eliminate denominators, solve the resulting equation, and verify every possible solution.

In this lesson, we will develop these skills through worked examples, investigate why extraneous solutions occur, and apply rational equations to practical problems.

What Is a Rational Equation?

A rational equation is an equation containing one or more rational expressions.

Examples include:

• 3/x = 6

• 2/(x + 1) = 5/(x − 2)

• 1/x + 1/(x + 3) = 1/2

• (x + 2)/(x − 1) = (x − 3)/(x + 4)

Rational equations differ from rational expressions.

A rational expression is an algebraic fraction, such as:

(x + 2)/(x − 3)

A rational equation contains an equals sign:

(x + 2)/(x − 3) = 4

Solving the equation means finding all values of x that make the original equation true.

The Most Important Rule: Denominators Cannot Equal Zero

Before solving any rational equation, we must identify the values that make its denominators zero.

Consider:

The denominator is x − 3.

Set it equal to zero:

x − 3 = 0

x = 3

Therefore, x = 3 is excluded.

Even if our algebraic work later produces x = 3, it cannot be accepted as a solution.

This restriction must be preserved throughout the entire solving process.

Finding Restrictions with Multiple Denominators

Consider:

The denominators are x + 1 and x − 4.

Set each equal to zero:

x + 1 = 0 → x = −1

x − 4 = 0 → x = 4

Therefore:

Restrictions: x ≠ −1 and x ≠ 4.

These values cannot be solutions to the original equation.

A Reliable Method for Solving Rational Equations

Use the following six-step process.

1. Identify restrictions. Find all values that make any original denominator zero.

2. Factor denominators. Factor completely to identify the least common denominator (LCD).

3. Multiply by the LCD. Multiply every term on both sides of the equation by the LCD.

4. Solve the resulting equation. Use algebraic techniques such as expanding, collecting terms, factoring, or the quadratic formula.

5. Check possible solutions. Reject any excluded values and substitute valid candidates into the original equation.

6. Interpret the result. In applications, determine whether each solution makes sense in context.

The LCD method is particularly useful because it converts a rational equation into a polynomial equation.

Worked Example 1: Solving a Simple Rational Equation

Solve:

Step 1: Identify restrictions.

x ≠ 0.

Step 2: Multiply both sides by x.

12 = 3x

Step 3: Solve.

x = 4

Step 4: Verify in the original equation.

12/4 = 3

3 = 3

The equation is satisfied.

Final answer: x = 4.

Worked Example 2: Solving with a Linear Denominator

Solve:

Step 1: Identify restrictions.

x − 1 = 0

x = 1

Therefore, x ≠ 1.

Step 2: Multiply both sides by x − 1.

2x + 3 = 5(x − 1)

Step 3: Expand.

2x + 3 = 5x − 5

Step 4: Collect like terms.

8 = 3x

x = 8/3

Step 5: Verify.

Original left side:

[2(8/3) + 3]/(8/3 − 1)

= (16/3 + 9/3)/(5/3)

= (25/3)/(5/3)

= 5

The right side is also 5.

Final answer: x = 8/3.

Worked Example 3: Solving with Two Rational Expressions

Solve:

Step 1: Identify restrictions.

x ≠ −2 and x ≠ 1.

Step 2: Multiply by the LCD.

LCD = (x + 2)(x − 1)

Multiplying both sides:

3(x − 1) = 5(x + 2)

Step 3: Expand.

3x − 3 = 5x + 10

Step 4: Solve.

−13 = 2x

x = −13/2

Step 5: Verify.

Left side:

3/(−13/2 + 2) = 3/(−9/2) = −2/3

Right side:

5/(−13/2 − 1) = 5/(−15/2) = −2/3

Both sides agree.

Final answer: x = −13/2.

Worked Example 4: Solving with Three Rational Terms

Solve:

Step 1: Identify restrictions.

x ≠ 0 and x ≠ −1.

Step 2: Find the LCD.

LCD = x(x + 1)

Step 3: Multiply every term by the LCD.

(x + 1) + 2x = x(x + 1)

Step 4: Expand and simplify.

3x + 1 = x² + x

Rearrange:

x² − 2x − 1 = 0

Step 5: Use the quadratic formula.

For ax² + bx + c = 0:

Here:

a = 1, b = −2, c = −1

Therefore:

Step 6: Check restrictions.

Neither value is 0 or −1.

To verify algebraically, notice that both roots satisfy:

x² − 2x − 1 = 0

Thus:

x² + x = 3x + 1

Since x(x + 1) is nonzero for both roots, dividing by it recovers the original equation.

Final answers:

x = 1 + √2 or x = 1 − √2.

Extraneous Solutions

An extraneous solution is a value obtained during algebraic solving that does not satisfy the original equation.

Extraneous solutions commonly arise when we multiply an equation by an expression that may equal zero.

For example, multiplying both sides by x − 3 is reversible only when x ≠ 3.

If x = 3 is allowed into the transformed equation, it may appear to be a solution even though the original rational equation is undefined.

Worked Example 5: Identifying an Extraneous Solution

Solve:

Step 1: Identify restrictions.

x − 2 = 0

x = 2

Therefore, x ≠ 2.

Step 2: Multiply both sides by x − 2.

x + 1 = 3

Step 3: Solve.

x = 2

Step 4: Check the restriction.

The solution x = 2 is excluded.

Substituting x = 2 into the original equation gives:

3/0 = 3/0

Both sides are undefined.

Therefore, x = 2 is an extraneous solution.

Final answer: No solution.

This example shows why checking restrictions is essential.

Worked Example 6: A Quadratic Equation with an Extraneous Root

Solve:

Step 1: Identify restrictions.

x ≠ 2.

Step 2: Multiply by x − 2.

x² − 4 = 0

Step 3: Factor.

(x − 2)(x + 2) = 0

Step 4: Solve.

x = 2 or x = −2.

Step 5: Check both solutions.

For x = 2:

The original denominator becomes zero.

Therefore, x = 2 is extraneous.

For x = −2:

[(−2)² − 4]/(−2 − 2)

= (4 − 4)/(−4)

= 0

This satisfies the original equation.

Final answer: x = −2 only.

Visualizing an Extraneous Solution

The rational expression from Example 6 simplifies to:

(x² − 4)/(x − 2) = x + 2, where x ≠ 2.

Its graph is a straight line with a hole at (2, 4).

Multiplying by x − 1 gives:

x + 2 = 4

x = 2

Check in the original equation:

Left side = (2 + 2)/(2 − 1) = 4

Right side = 4/(2 − 1) = 4

The solution is valid.

Three Verification Checks

A complete verification should answer:

1. Does the value satisfy all original domain restrictions?

2. Does substituting the value make both sides of the original equation equal?

3. If the equation represents a real-world situation, is the value physically meaningful?

A solution must pass all applicable checks.

Solving More Complex Rational Equations

Worked Example 7: Equations with Factored Denominators

Solve:

Step 1: Identify restrictions.

x² − 9 = (x − 3)(x + 3)

Therefore:

x ≠ 3 and x ≠ −3.

Step 2: Find the LCD.

LCD = (x − 3)(x + 3)

Step 3: Multiply every term by the LCD.

2(x + 3) + (x − 3) = x

Step 4: Expand.

2x + 6 + x − 3 = x

3x + 3 = x

Step 5: Solve.

2x = −3

x = −3/2

Step 6: Verify.

Substitute x = −3/2.

Left side:

2/(−9/2) + 1/(3/2)

= −4/9 + 2/3

= −4/9 + 6/9

= 2/9

Right side:

(−3/2)/(9/4 − 9)

= (−3/2)/(−27/4)

= 2/9

Both sides agree.

Final answer: x = −3/2.

Worked Example 8: A Rational Equation with No Solution

Solve:

Restriction:

x ≠ 4.

Multiply both sides by x − 4:

x = 4

But x = 4 is excluded.

Final answer: No solution.

Worked Example 9: An Equation with Two Valid Solutions

Solve:

Step 1: Restrictions.

x ≠ 0 and x ≠ 1.

Step 2: Multiply by x(x − 1).

x(x + 3) = 6(x − 1)

Step 3: Expand.

x² + 3x = 6x − 6

x² − 3x + 6 = 0

Step 4: Calculate the discriminant.

b² − 4ac = (−3)² − 4(1)(6)

= 9 − 24

= −15

The discriminant is negative.

Therefore, the equation has no real solutions.

Final answer: No real solution.

Over the complex numbers, the solutions are (3 ± i√15)/2. For this lesson, we are solving over the real numbers.

Rational Equations in Real-World Applications

Many rational equations arise from relationships involving quantities divided by other quantities.

The most common applications involve:

• Work rates and shared tasks.

• Distance, speed, and time.

• Flow rates and filling containers.

• Average costs and production.

• Electrical circuits.

The important skill is translating a real-world situation into a mathematical equation before solving.

Application 1: Two Workers Completing a Job

One worker can complete a job in 6 hours. A second worker takes 4 hours.

How long will they take working together?

Step 1: Identify individual rates.

First worker's rate = 1/6 job per hour.

Second worker's rate = 1/4 job per hour.

Step 2: Let t represent the time working together.

Their combined rate is 1/t.

Therefore:

Step 3: Find a common denominator.

1/6 + 1/4 = 2/12 + 3/12

= 5/12

Therefore:

5/12 = 1/t

Step 4: Cross-multiply.

5t = 12

t = 12/5

t = 2.4 hours

Step 5: Interpret the result.

0.4 hours = 24 minutes.

Answer: 2 hours and 24 minutes.

This is reasonable because two workers together should finish faster than either worker working alone.

Application 2: A Journey at Different Speeds

A driver travels 180 km.

If the driver increases the speed by 15 km/h, the journey takes one hour less.

Find the original speed.

Step 1: Define the variable.

Let x = original speed in km/h.

New speed = x + 15.

The physical restriction is x greater than zero.

Step 2: Express the journey times.

Original time = 180/x

New time = 180/(x + 15)

The difference is one hour.

Therefore:

Step 3: Multiply by x(x + 15).

180(x + 15) − 180x = x(x + 15)

Step 4: Expand.

180x + 2700 − 180x = x² + 15x

2700 = x² + 15x

x² + 15x − 2700 = 0

Step 5: Factor.

(x + 60)(x − 45) = 0

Therefore:

x = −60 or x = 45.

Step 6: Interpret the solutions.

A negative speed is not meaningful in this context.

Therefore, reject x = −60.

Original speed = 45 km/h.

Step 7: Verify.

Original journey time:

180/45 = 4 hours.

New speed:

45 + 15 = 60 km/h.

New journey time:

180/60 = 3 hours.

Difference = 1 hour.

Final answer: The original speed was 45 km/h.

Application 3: Filling a Water Tank

One pipe fills a tank in 5 hours.

A second pipe takes 3 hours longer than the first.

How long will the two pipes take to fill the tank together?

Step 1: Determine individual times.

First pipe = 5 hours.

Second pipe = 8 hours.

Step 2: Let t be the combined filling time.

Step 3: Add the rates.

1/5 + 1/8 = 8/40 + 5/40

= 13/40

Therefore:

13/40 = 1/t

Step 4: Solve.

13t = 40

t = 40/13

t ≈ 3.08 hours.

Step 5: Interpret.

0.08 hours is approximately 5 minutes.

Answer: Approximately 3 hours and 5 minutes.

The combined time is less than either individual filling time, so the result is reasonable.

Application 4: Average Production Cost

A company has a fixed production cost of $600 and a variable cost of $10 per item.

How many items must be produced for the average cost to equal $16 per item?

Step 1: Write the total cost.

C(x) = 600 + 10x

Step 2: Write the average cost.

A(x) = (600 + 10x)/x

Step 3: Form the rational equation.

Step 4: Identify restrictions.

x ≠ 0.

In context, x must be a positive whole number.

Step 5: Multiply by x.

600 + 10x = 16x

600 = 6x

x = 100

Step 6: Verify.

Total cost:

600 + 10(100) = $1,600

Average cost:

1600/100 = $16

Final answer: The company must produce 100 items.

Interpreting Solutions in Context

An algebraically valid solution may still be unsuitable for a real-world situation.

For example, suppose an equation produces:

x = −5 or x = 12

If x represents time, a negative value may not make sense for the situation.

If x represents the number of manufactured items, the answer may need to be a nonnegative integer.

When interpreting rational equations, consider the following.

A mathematical solution is not automatically a meaningful practical answer.

Common Mistakes

Mistake 1: Forgetting Restrictions

Incorrect:

3/(x − 2) = 3/(x − 2) has all real numbers as solutions.

Correct:

The equality holds for every real x except x = 2.

The solution set is all real numbers excluding 2.

Mistake 2: Multiplying Only Some Terms by the LCD

Consider:

1/x + 2/(x + 1) = 3

Multiplying by x(x + 1) requires multiplying all three terms.

Correct:

(x + 1) + 2x = 3x(x + 1)

Mistake 3: Assuming Every Algebraic Root Is Valid

Multiplying by a denominator can produce an excluded root.

Always substitute candidate solutions into the original equation.

Mistake 4: Cancelling Terms Instead of Factors

Incorrect:

(x + 3)/(x + 5) = 3/5

Correct:

The x terms cannot be cancelled because they are not common factors.

Mistake 5: Incorrect Cross-Multiplication

For:

a/b = c/d

the correct relationship is:

ad = bc

provided b and d are nonzero.

Mistake 6: Checking Only the Simplified Equation

A candidate might satisfy a transformed equation while making the original expression undefined.

Always check the original equation.

Mistake 7: Ignoring Real-World Meaning

If an equation models the number of students, a solution of −4 or 2.5 students may be mathematically possible in the transformed equation but unsuitable for the situation.

Mistake 8: Confusing No Real Solution with an Extraneous Solution

An equation may have no real solutions because its quadratic discriminant is negative, even when no excluded root has been produced.

For instance, in Worked Example 9, the quadratic x² − 3x + 6 = 0 has no real roots. Its heading should therefore be read as an equation with no real solutions, rather than an equation with two valid solutions.

Key Terms

• Rational Equation: An equation containing one or more rational expressions.

• Rational Expression: A fraction whose numerator and denominator are polynomials.

• Domain Restriction: A condition excluding values that make an expression undefined.

• Excluded Value: A value that makes an original denominator zero.

• Least Common Denominator (LCD): The least common polynomial multiple of all denominators, up to a nonzero constant.

• Extraneous Solution: A candidate solution that does not satisfy the original equation.

• Valid Solution: A value that satisfies the original equation and all domain restrictions.

• Cross-Multiplication: A method for solving an equality of two fractions by multiplying opposite numerator-denominator pairs.

• Quadratic Equation: An equation that can be written as ax² + bx + c = 0, where a ≠ 0.

• Verification: Checking a candidate solution in the original equation.

• Work Rate: The fraction or amount of a task completed per unit of time.

• Mathematical Model: An equation or mathematical relationship used to represent a real-world situation.

• Contextual Restriction: A limitation on possible solutions imposed by the meaning of a problem.

• Solution Set: The set of all values that satisfy an equation.

Key Takeaways

1. Rational equations contain fractions involving polynomials.

2. Every original denominator must be nonzero.

3. Restrictions should be identified before solving.

4. Multiplying every term by the LCD eliminates denominators and usually produces a polynomial equation.

5. Extraneous solutions can arise when multiplying by expressions that may equal zero.

6. Every candidate solution must be checked against the original restrictions.

7. Substitution into the original equation verifies whether a candidate is valid.

8. Some rational equations have multiple solutions, one solution, or no solution.

9. Real-world applications often involve rates, time, speed, cost, or concentration.

10. Practical solutions must be interpreted using the conditions of the situation.

Check Your Understanding

Question 1: Simple Rational Equation

Solve and verify:

Question 2: Linear Rational Equation

Solve:

Question 3: Two Rational Expressions

Solve:

Question 4: Multiple Denominators

Solve:

Question 5: Extraneous Solutions

Solve and identify any extraneous solutions:

Question 6: No Solution

Solve:

Question 7: Work Rate Application

One machine completes a job in 8 hours. Another completes the same job in 12 hours.

How long will they take working together?

Question 8: Travel Time Application

A car travels 120 km.

Increasing its speed by 20 km/h reduces the journey time by one hour.

Find the original speed.

Question 9: Production Cost Application

A factory has a fixed cost of $900 and a variable cost of $15 per item.

How many items must be produced for the average cost to be $24 per item?

Question 10: Challenge Problem

Solve:

Identify all restrictions and verify the valid solutions.

Fully Worked Answers

Answer 1

Given:

18/x = 6

Restriction: x ≠ 0.

Multiply by x:

18 = 6x

x = 3

Verify:

18/3 = 6

Answer: x = 3.

Answer 2

Given:

(3x + 2)/(x − 1) = 5

Restriction: x ≠ 1.

Multiply by x − 1:

3x + 2 = 5(x − 1)

3x + 2 = 5x − 5

7 = 2x

x = 7/2

Verify:

[3(7/2) + 2]/(7/2 − 1)

= (25/2)/(5/2)

= 5

Answer: x = 7/2.

Answer 3

Given:

4/(x + 1) = 6/(x − 2)

Restrictions:

x ≠ −1 and x ≠ 2.

Cross-multiply:

4(x − 2) = 6(x + 1)

4x − 8 = 6x + 6

−14 = 2x

x = −7

Verify:

Left side = 4/(−6) = −2/3

Right side = 6/(−9) = −2/3

Answer: x = −7.

Answer 4

Given:

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

Restrictions:

x ≠ 0 and x ≠ −2.

LCD = 4x(x + 2)

Multiply every term:

4(x + 2) + 4x = 3x(x + 2)

Expand:

4x + 8 + 4x = 3x² + 6x

Rearrange:

3x² − 2x − 8 = 0

Factor:

(3x + 4)(x − 2) = 0

Therefore:

x = −4/3 or x = 2

Neither is excluded.

Verify x = 2:

1/2 + 1/4 = 3/4

Verify x = −4/3:

1/(−4/3) + 1/(2/3)

= −3/4 + 3/2

= 3/4

Answers: x = −4/3 and x = 2.

Answer 5

Given:

(x² − 16)/(x − 4) = 0

Restriction:

x ≠ 4.

Multiply by x − 4:

x² − 16 = 0

Factor:

(x − 4)(x + 4) = 0

Candidates:

x = 4 or x = −4

x = 4 is extraneous because it makes the denominator zero.

Check x = −4:

(16 − 16)/(−8) = 0

Answer: x = −4. Extraneous solution: x = 4.

Answer 6

Given:

(x + 2)/(x − 3) = 5/(x − 3)

Restriction:

x ≠ 3.

Multiply by x − 3:

x + 2 = 5

x = 3

But x = 3 is excluded.

Answer: No solution.

Answer 7

First machine's rate:

1/8 job per hour.

Second machine's rate:

1/12 job per hour.

Let t represent the combined time.

1/8 + 1/12 = 1/t

Find the LCD of 8 and 12:

LCD = 24

3/24 + 2/24 = 1/t

5/24 = 1/t

5t = 24

t = 24/5 = 4.8 hours

0.8 hours = 48 minutes.

Answer: 4 hours and 48 minutes.

Answer 8

Let x be the original speed in km/h.

Original time = 120/x

New time = 120/(x + 20)

The difference is one hour:

Restrictions:

x ≠ 0 and x ≠ −20.

Multiply by x(x + 20):

120(x + 20) − 120x = x(x + 20)

Expand:

120x + 2400 − 120x = x² + 20x

x² + 20x − 2400 = 0

Factor:

(x + 60)(x − 40) = 0

Therefore:

x = −60 or x = 40

A negative speed is not meaningful here.

Verify x = 40:

Original time = 120/40 = 3 hours.

New time = 120/60 = 2 hours.

Difference = 1 hour.

Answer: Original speed = 40 km/h.

Answer 9

Total cost:

C(x) = 900 + 15x

Average cost:

A(x) = (900 + 15x)/x

Set average cost equal to 24:

(900 + 15x)/x = 24

Restriction:

x ≠ 0.

Multiply by x:

900 + 15x = 24x

900 = 9x

x = 100

Verify:

Total cost = 900 + 15(100) = $2,400

Average cost = 2400/100 = $24

Answer: 100 items.

Answer 10

Given:

x/(x − 2) + 2/(x + 2) = 8/(x² − 4)

Step 1: Restrictions.

x² − 4 = (x − 2)(x + 2)

Therefore:

x ≠ 2 and x ≠ −2.

Step 2: Multiply by the LCD.

LCD = (x − 2)(x + 2)

x(x + 2) + 2(x − 2) = 8

Step 3: Expand.

x² + 2x + 2x − 4 = 8

x² + 4x − 12 = 0

Step 4: Factor.

(x + 6)(x − 2) = 0

Candidates:

x = −6 or x = 2.

Step 5: Check restrictions.

x = 2 is excluded.

For x = −6:

Left side:

−6/(−8) + 2/(−4)

= 3/4 − 1/2

= 1/4

Right side:

8/(36 − 4) = 8/32 = 1/4

The original equation is satisfied.

Final answer: x = −6.

The candidate x = 2 is extraneous.

This completes Solving Rational Equations, covering all five learning targets through algebraic methods, extraneous solutions, verification, real-world modelling, and contextual interpretation.

The lesson is ready for adaptation into yEdu, with copyable explanations and equations, a supporting graph, worked examples, common mistakes, key terms, and fully worked practice answers.