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.