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A reciprocal function can be considered as the quotient where the denominator is a polynomial with a degree of

1 .

In other words, there is a variable in the denominator. In this lesson, rational functions in which the numerator and the denominator have a degree of at least

1

will be graphed.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Challenge

Average Cost Function

LaShay's school, Jeffereson High, visits a local pencil factory. The average cost, in thousand dollars, of producing

x

thousand pencils is given by the following function.

A (x) = \frac{0 . 2 x ^{2} + 5 0 x + 5 0 0}{x}

a Graph the average cost function.

b Identify the domain and range of the function. Use a graphing calculator if necessary.

Discussion

Rational Functions

A rational function is a function that contains a rational expression. Any function that can be written as the quotient of two polynomial functions $p (x)$ and $q (x)$ is a rational function.

$f (x) = \frac{p ( x )}{q ( x )}, q (x) \neq = 0$

For any values of

x

where

q (x) = 0,

the rational function is undefined. These values are excluded from the domain of the function. One example of a rational function is the reciprocal function.

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

This function has two asymptotes, the

x -

axis and the

y -

axis. The applet shows the graph of the reciprocal function and some other rational functions.

Discussion

Discontinuity of Rational Functions

The graph of a rational function can be a smooth continuous curve, or it can have jumps, breaks, or holes. By looking at the graph of a function, functions can be categorized into two groups: continuous or discontinuous.

Concept

Discontinuous - Function

A function is said to be continuous if its graph can be drawn without lifting the pencil. Otherwise, the function is said to be discontinuous.

A discontinuous function can have holes, jumps, or asymptotes throughout its graph.

A point where a function is discontinuous is called a discontinuity. Function I has a point of discontinuity at

x = 1,

Function II has a point of discontinuity at

x = 2,

and Function III has a point of discontinuity at

x = 0 .

Concept

Point of Discontinuity

A point of discontinuity of a function is a point with the $x -$ coordinate that makes the function value undefined. A point of discontinuity can also be considered as an excluded value of the function rule.

Removable Discontinuity

If a function can be redefined so that its point of discontinuity is removed, it is called a removable discontinuity. A removable discontinuity may occur when there is a rational expression with common factors in the numerator and denominator. Consider the following rational function.

f (x) = \frac{x ^{2} - 4}{x - 2}

For this function,

x = 2

is a point of discontinuity because its denominator is

0

when

x = 2 .

Next, notice that the function can be simplified using a difference of squares.

f (x) = \frac{x ^{2} - 4}{x - 2}

▼

Simplify right-hand side

WritePow

Write as a power

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

FacDiffSquares

$a^{2} - b^{2} = (a + b) (a - b)$

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

SimpQuot

Simplify quotient

f (x) = x + 2, x \neq = 2

After simplifying, the graph behaves as a linear function except in the point of discontinuity, where its value is undefined.

Rational function with removable discontinuity

However, the function can be made continuous by redefining it at

x = 2 .

2

is substituted for

x

in the function

f (x) = x + 2,

it is found that

f (2) = 4 .

Therefore, the following function is continuous at

x = 2 .

g (x) = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{x ^{2} - 4}{x - 2}, 4, if x \neq = 2 if x = 2

This means that

x = 2

is a removable discontinuity. Note that a removable discontinuity is typically a hole in the graph.

Non-Removable Discontinuity

When a function cannot be redefined so that the point of discontinuity becomes a valid input, it is called a non-removable discontinuity. Consider, for example, $y = \frac{1}{x + 1} .$ Since $x = - 1$ is not in the domain of the function, there is a point of discontinuity at $x = - 1 .$

This is a non-removable discontinuity because there is no way to redefine the function so that it becomes continuous at

x = - 1 .

Example

Modeling Annual Income Per Capita

LaShay lives in Des Moines, where the total annual amount of personal income

I

in millions dollars, and the population

P

in millions, are modeled by the following functions.

I (t) P (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{0 . 0 2 t + 1} = 0.36 t + 10800

In these functions,

t

represents the number of years that have passed since

1990 .

a Write a function

C (t)

modeling the annual income per capita in Des Moines.

b Identify the points of discontinuity of

C (t)

, if any.

Hint

a Divide the total amount of personal income by the population. Recall how to divide rational expressions.

b For which values of

t

is the function

C (t)

undefined?

Solution

a Take a look at what the given equations represent.

Total Annual Income	Population
$I (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{0 . 0 2 t + 1}$	$P (t) = 0.36 t + 10800$

In order to write a function for the annual income per person

C (t),

the total annual income

I (t)

must be divided by the population

P (t) .

This will give the annual income of a person.

C (t) = I (t) \div P (t)

SubstituteExpressions

Substitute expressions

C (t) = (\frac{5 0 0 0 t + 1 0 0 0 0 0}{0 . 0 2 t + 1}) \div (0.36 t + 10800)

DivFracSL

$\frac{a}{b} / c = \frac{a}{b \cdot c}$

C (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{( 0 . 0 2 t + 1 ) ( 0 . 3 6 t + 1 0 8 0 0 )}

The binomials in the denominator can be multiplied.

C (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{( 0 . 0 2 t + 1 ) ( 0 . 3 6 t + 1 0 8 0 0 )}

MultPar

Multiply parentheses

C (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{0 . 0 0 7 2 t ^{2} + 2 1 6 t + 0 . 3 6 t + 1 0 8 0 0}

AddTerms

Add terms

C (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{0 . 0 0 7 2 t ^{2} + 2 1 6 . 3 6 t + 1 0 8 0 0}

b Consider the function

C (t)

written in Part A.

C (t) = \frac{5 0 0 0 t + 1 0 0 0 0 0}{( 0 . 0 2 t + 1 ) ( 0 . 3 6 t + 1 0 8 0 0 )}

Recall that division by zero is not defined. Therefore, the rational function is undefined when either of the parentheses equals zero.

$C (t)$ is undefined
$0.02 t + 1 = 0$	$0.36 t + 10800 = 0$
$t = - 50$	$t = - 30000$

This means that

t = - 50

and

t = - 30000

are excluded values.

\underline{Domain} All real numbers except t = - 50 and t = - 30000

If a real number

a

is not in the domain of a function, then the function has a point of discontinuity at

x = a .

Since the domain is all real numbers except

t = - 50

and

t = - 30000,

there are two points of discontinuity of the function.

\underline{Points of Discontinuity} t = - 50 and t = - 30000

Example

Finding Point of Discontinuity

On the way to the pencil factory, LaShay noticed that road maintenance work is being carried out to fill a hole on the road.

A map that shows a lake, a forrest, some buildings, and a road with a hole in it

The equation below models the road to the factory.

R (x) = \frac{3 x ^{2} + 3 x}{x ^{3} + x ^{2} + x + 1}

a Find the

x -

coordinate of the point where the road maintenance is being completed.

b The municipality of Des Moines plans to build a new road, which is modeled by the following function.

N (x) = \frac{2 x ^{2} - x - 3}{1 0 x + 1 0}

Graph the function.

Answer

x = - 1

b Graph:

Hint

a Start by factoring the numerator and the denominator of the function. Find the point of discontinuity of the graph.

b Check if the numerator and the denominator have a common factor.

Solution

a The terms in the numerator of the function have a common factor of

3 x .

Also, the denominator can be factored by grouping.

R (x) = \frac{3 x ^{2} + 3 x}{x ^{3} + x ^{2} + x + 1}

FactorOut

Factor out $3 x$

R (x) = \frac{3 x ( x + 1 )}{x ^{3} + x ^{2} + x + 1}

FactorOut

Factor out $x^{2}$

R (x) = \frac{3 x ( x + 1 )}{x ^{2} ( x + 1 ) + x + 1}

FactorOut

Factor out $(x + 1)$

R (x) = \frac{3 x ( x + 1 )}{( x ^{2} + 1 ) ( x + 1 )}

Notice that there is no

x -

value that can make

x^{2} + 1

equal zero. However, the other factor in the denominator makes the function undefined when

x = - 1 .

Therefore, this value is excluded from the domain.

\underline{Domain} All real numbers except x = - 1

Recall that the point of discontinuity occurs at

x = a

when

a

is not in the domain of a function. Consequently, there is a point of discontinuity at

x = - 1 .

\underline{Point of Discontinuity} x = - 1

In the context of the problem, this is the

x -

coordinate of the point where the road is being repaired.

Notice that it is also a removable discontinuity because $x + 1$ is the common factor in the numerator and denominator.

b Start by factoring the numerator and denominator to see if the function can be simplified.

N (x) = \frac{2 x ^{2} - x - 3}{1 0 x + 1 0}

▼

Factor

WriteDiff

Write as a difference

N (x) = \frac{2 x ^{2} + 2 x - 3 x - 3}{1 0 x + 1 0}

FactorOut

Factor out $2 x$

N (x) = \frac{2 x ( x + 1 ) - 3 x - 3}{1 0 x + 1 0}

FactorOut

Factor out $- 3$

N (x) = \frac{2 x ( x + 1 ) - 3 ( x + 1 )}{1 0 x + 1 0}

FactorOut

Factor out $(x + 1)$

N (x) = \frac{( 2 x - 3 ) ( x + 1 )}{1 0 x + 1 0}

FactorOut

Factor out $10$

N (x) = \frac{( 2 x - 3 ) ( x + 1 )}{1 0 ( x + 1 )}

The function is undefined when

x = - 1 .

This means that the domain of the function is all real numbers except

x = - 1 .

After simplification, the function will be a linear function except at

x = - 1 .

N (x) = \frac{( 2 x - 3 ) ( x + 1 )}{1 0 ( x + 1 )} = \frac{2 x - 3}{1 0}, x \neq = - 1

Note that

\frac{2 x - 3}{1 0} = \frac{2 x}{1 0} - \frac{3}{1 0} .

Therefore, the graph of the given function is the graph of the function

y = \frac{2 x}{1 0} - \frac{3}{1 0}

with a hole at

x = - 1 .

This graph also has a removable discontinuity at $x = - 1$ because $x + 1$ is the common factor in the numerator and denominator of the rational function.

Discussion

Asymptotes of Rational Functions

An asymptote of a graph is an imaginary line that the graph gets close to as $x$ goes to plus or minus infinity or a particular number. For example, the graph of the rational function $f (x) = \frac{1}{x}$ has two asymptotes — the $x -$ axis and the $y -$ axis.

Analyzing the diagram, the following can be observed.

As $x$ approaches infinity and as $x$ approaches negative infinity, the value of the function approaches $0 .$ Therefore, $y = 0,$ or the $x -$ axis, is a horizontal asymptote of the graph of $f .$
As $x$ approaches $0,$ the value of the function approaches either positive or negative infinity. Therefore, $x = 0,$ or the $y -$ axis, is a vertical asymptote of the graph of $f .$

In the coordinate plane below, the asymptotes for three different graphs are shown.

The graphs with asymptotes of three different functions

The applet demonstrates that asymptotes can be not only vertical and horizontal, but also oblique.

Consider a rational function where

p (x)

is a polynomial with a leading coefficient of

a

and a degree of

m

and where

q (x)

is a polynomial with a leading coefficient of

b

and a degree of

n .

f (x) = \frac{p ( x )}{q ( x )} = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}

To determine the asymptotes of the function, the denominator and the degrees of the polynomials in the numerator and denominator must be analyzed.

Identifying Vertical Asymptotes

To identify the vertical asymptotes, the function should be in its simplest form. That is, if $p (x)$ and $q (x)$ have common factors, the function should be simplified first. The vertical asymptotes will occur at the zeros of the denominator. Check out the following examples.

Identifying Horizontal Asymptotes

The degrees of polynomials in the numerator and denominator of a rational function will determine if the graph has a horizontal asymptote.

	$f (x) = \frac{p ( x )}{q ( x )} = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}$
Horizontal Asymptote	If $m > n,$ there is no horizontal asymptote.
	If $m < n,$ the horizontal asymptote is $y = 0 .$
	If $m = n,$ the horizontal asymptote is $y = \frac{a}{b} .$

The graph of a rational function may have one or more vertical asymptotes and at most one horizontal asymptote.

Functions y=x/(x^2-9), y=3x^2/(x^2-x), and y=-x^4/(5x^2-5) are graphed with the vertical and horizontal asymptotes

Notice that the graph of a rational function will never cross a vertical asymptote, whereas the graph may or may not cross a horizontal asymptote.

Example

Identifying Asymptotes

LaShay takes a travel-sickness pill about

45

minutes before the school bus leaves for the field trip. The amount

A

in milligrams of the medication in her bloodstream is modeled by the following rational function.

A (t) = \frac{6 t ^{2} - 2 4 t}{t ^{3} - t ^{2} - 1 0 t - 8}

Here,

t

represents the time in hours after one pill is taken. Identify the asymptotes of the function.

Answer

Vertical Asymptotes: $t = - 2$ and $t = - 1$
Horizontal Asymptote: $y = 0$

Hint

Factor the numerator. What does the Factor Theorem state?

Solution

To identify the asymptotes, the numerator and denominator of the function

A (t)

need to be factored. Notice that the terms in the numerator have a common factor of

6 t .

A (t) = \frac{6 t ^{2} - 2 4 t}{t ^{3} - t ^{2} - 1 0 t - 8}

FactorOut

Factor out $6 t$

A (t) = \frac{6 t ( t - 4 )}{t ^{3} - t ^{2} - 1 0 t - 8}

To factor the polynomial in the denominator, use the Factor Theorem.

Factor Theorem
The expression $x - a$ is a factor of a polynomial if and only if the value $a$ is a zero of the related polynomial function.

Now, the aim is to find all real zeros of the polynomial. Before doing so, identify the degree of the polynomial.

p (t) = t^{3} - t^{2} - 10 t - 8

It is a polynomial with a degree of

3 .

Subsequently, by the Fundamental Theorem of Algebra, it is known know that

p (t)

has exactly three roots. Additionally, by the Rational Root Theorem, integer roots must be factors of the constant term. Since the constant term of

p (t)

- 8,

the possible integer roots are

\pm 1,

\pm 2,

\pm 4,

and

\pm 8 .

$t$	$t^{3} - t^{2} - 10 t - 8$	$p (t) = t^{3} - t^{2} - 10 t - 8$
$1$	$1^{3} - (1)^{2} - 10 (1) - 8$	$- 18$
$- 1$	$- 1^{3} - (- 1)^{2} - 10 (- 1) - 8$	$0$
$2$	$2^{3} - (2)^{2} - 10 (2) - 8$	$- 24$
$- 2$	$- 2^{3} - (- 2)^{2} - 10 (- 2) - 8$	$0$
$4$	$4^{3} - 6 (4)^{2} - 7 (4) + 60$	$0$

Three roots for

p (t)

were found, so there is no need for further investigation. In conclusion,

- 1,

- 2,

and

4

are all the real roots of

t^{3} - t^{2} - 10 t - 8 .

Next, apply the Factor Theorem to write its factored form.

t^{3} - t^{2} - 10 t - 8 = (t + 1) (t + 2) (t - 4)

Now that the factored form of the denominator has been found, the given function can be simplified.

A (t) = \frac{6 t ( t - 4 )}{t ^{3} - t ^{2} - 1 0 t - 8}

Substitute

$t^{3} - t^{2} - 10 t - 8 = (t + 1) (t + 2) (t - 4)$

A (t) = \frac{6 t ( t - 4 )}{( t + 1 ) ( t + 2 ) ( t - 4 )}

CrossCommonFac

Cross out common factors

A (t) = \frac{6 t ( t - 4 )}{( t + 1 ) ( t + 2 ) ( t - 4 )}

CancelCommonFac

Cancel out common factors

A (t) = \frac{6 t}{( t + 1 ) ( t + 2 )}

Since the numerator and the denominator share the factor

t - 4,

there is a hole at

t = 4 .

Also, the denominator becomes zero when

t = - 1

and

t = - 2 .

These values make the function undefined. Therefore, the function has two vertical asymptotes.

\underline{Vertical Asymptotes} t = - 1 and t = - 2

To find the horizontal asymptote, compare the degrees of the polynomials in the numerator and denominator.

$y = \frac{a x ^{m}}{b x ^{n}}$	Asymptote
$m < n$	$y = 0$
$m > n$	none
$m = n$	$y = \frac{a}{b}$

Look at the degrees of the numerator and denominator of the function.

A (t) = \frac{6 t ^{2} - 2 4 t}{t ^{3} - t ^{2} - 1 0 t - 8}

The degree of the denominator is greater than the degree of the numerator,

m < n .

Therefore, the horizontal asymptote is

y = 0 .

\underline{Horizontal Asymptotes} y = 0

Discussion

Oblique Asymptote

An asymptote can be neither vertical nor horizontal. It can be a slanted line. In such cases, it is called an oblique asymptote or slant asymptote.

An oblique asymptote of a rational function occurs when the degree of the numerator is

1

more than that of the denominator. Therefore, a function with an oblique asymptote can never have a horizontal asymptote. To find the equation of the oblique asymptote of a rational function

f (x) = \frac{p ( x )}{q ( x )},

p (x)

is divided by

q (x)

using long division to get a quotient

a x + b

with a remainder

r (x) .

f (x) = \frac{p ( x )}{q ( x )} ⇓ f (x) = a x + b + r (x)

The equation of the oblique asymptote is of the form

y = a x + b,

where

a

is a non-zero real number.

Example

To illustrate how the technique is used, the asymptotes of the following function will be found.

f (x) = \frac{x ^{3} + 2 x}{4 x ^{2} - 1}

The difference between the degrees of the numerator and denominator is

1 .

This means that its graph does not have a horizontal asymptote but an oblique asymptote. To find the oblique asymptote, polynomial long division will be used.

4 x^{2} - 1 x^{3} + 2 x

▼

Divide

$\frac{x ^{3}}{4 x ^{2}} = \frac{x}{4}$

\frac{x}{4} 4 x^{2} - 1 x^{3} + 2 x

Multiply $term$ by $divisor$

\frac{x}{4} 4 x^{2} - 1 x^{3} + 2 x x^{3} - \frac{x}{4}

Subtract down

\frac{x}{4} 4 x^{2} - 1 \frac{9 x}{4}

The quotient is

\frac{x}{4}

with a remainder of

\frac{9 x}{4} .

f (x) = \frac{x ^{3} + 2 x}{4 x ^{2} - 1} ⇓ f (x) = \frac{x}{4} + \frac{\frac{9 x}{4}}{4 x ^{2} - 1}

Therefore,

y = \frac{x}{4}

is the oblique asymptote of the rational function.

Example

Modeling Interest in The Trip

The students spent about $3$ hours at the factory. At the end of this period, LaShay's teacher made a survey about the trip and drew a graph relating the students' interest in the trip and the time elapsed at the factory. The greater the $S (t)$ value, the greater the interest of the students.

The graph of the function S(t) in the first quadrant

The graph above is the part of the graph of the following function in the first quadrant.

S (t) = \frac{t}{t + 1} - \frac{t ^{3}}{2 ( t + 1 ) ^{2}}

Identify the asymptotes of the function.

Answer

Vertical Asymptotes: $t = - 1$
Oblique Asymptote: $O (t) = - \frac{t}{2} + 2$

Hint

Start by subtracting the rational expressions on the right-hand side of the function.

Solution

First, the rational expressions on the right-hand side will be subtracted. The least common denominator of the expressions is

2 (t + 1)^{2} .

Therefore, the first expression needs to be multiplied by

\frac{2 ( t + 1 )}{2 ( t + 1 )}

so that both have the same denominator.

S (t) = \frac{t}{t + 1} - \frac{t ^{3}}{2 ( t + 1 ) ^{2}}

▼

Simplify right-hand side

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2 ( t + 1 )}{b \cdot 2 ( t + 1 )}$

S (t) = \frac{t \cdot 2 ( t + 1 )}{( t + 1 ) \cdot 2 ( t + 1 )} - \frac{t ^{3}}{2 ( t + 1 ) ^{2}}

Multiply

S (t) = \frac{2 t ( t + 1 )}{2 ( t + 1 ) ^{2}} - \frac{t ^{3}}{2 ( t + 1 ) ^{2}}

SubFrac

Subtract fractions

S (t) = \frac{2 t ( t + 1 ) - t ^{3}}{2 ( t + 1 ) ^{2}}

Distr

Distribute $2 t$

S (t) = \frac{2 t ^{2} + 2 t - t ^{3}}{2 ( t + 1 ) ^{2}}

CommutativePropAdd

Commutative Property of Addition

S (t) = \frac{- t ^{3} + 2 t ^{2} + 2 t}{2 ( t + 1 ) ^{2}}

▼

Rewrite the denominator

ExpandPosPerfectSquare

$(a + b)^{2} = a^{2} + 2 a b + b^{2}$

S (t) = \frac{- t ^{3} + 2 t ^{2} + 2 t}{2 ( t ^{2} + 2 t + 1 )}

Distr

Distribute $2$

S (t) = \frac{- t ^{3} + 2 t ^{2} + 2 t}{2 t ^{2} + 4 t + 2}

The numerator and the denominator do not have any common factors. Therefore, the function does not have a point of discontinuity, or hole, in its graph. However, it has a vertical asymptote at

t = - 1

because the function is undefined at that value.

\underline{Vertical Asymptote} t = - 1

Now, to determine if the given rational function has a horizontal or oblique asymptote, the degrees of the polynomials in the numerator and the denominator will be compared.

S (t) = \frac{- t ^{3} + 2 t ^{2} + 2 t}{2 ( t + 1 ) ^{2}} ⇕ S (t) = \frac{- t ^{3} + 2 t ^{2} + 2 t}{2 t ^{2} + 4 t + 2}

The degree of the numerator is greater than the degree of the denominator, so the graph has no horizontal asymptotes. However, since the degree of the numerator is

1

more than that of the denominator, the graph has an oblique asymptote. Its equation is found by dividing the numerator by the denominator. Use long division to do this.

2 t^{2} + 4 t + 2 - t^{3} + 2 t^{2} + 2 t

▼

Divide

$\frac{- t ^{3}}{2 t ^{2}} = - \frac{t}{2}$

- \frac{t}{2} 2 t^{2} + 4 t + 2 - t^{3} + 2 t^{2} + 2 t

Multiply $term$ by $divisor$

- \frac{t}{2} 2 t^{2} + 4 t + 2 - t^{3} + 2 t^{2} + 2 t - t^{3} - 2 t^{2} - t

Subtract down

- \frac{t}{2} 2 t^{2} + 4 t + 2 4 t^{2} + 3 t

▼

Divide

$\frac{4 t ^{2}}{2 t ^{2}} = 2$

-

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Catch-Up and Review

Removable Discontinuity

Non-Removable Discontinuity

Hint

Solution

Answer

Hint

Solution

Identifying Vertical Asymptotes

Identifying Horizontal Asymptotes

Answer

Hint

Solution

Example

Answer

Hint

Solution