Discover How to Graph a Rational Function with Ease

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

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

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

Multiply $term$ by $divisor$

- \frac{t}{2} + 2 2 t^{2} + 4 t + 2 4 t^{2} + 3 t 4 t^{2} + 8 t + 4

Subtract down

- \frac{t}{2} + 2 2 t^{2} + 4 t + 2 - 5 t - 4

The given function can be rewritten as follows.

S (t) = - \frac{t}{2} + 2 + \frac{- 5 t - 4}{2 t ^{2} + 4 t + 2}

Therefore, the equation of the oblique asymptote is

O (t) = - \frac{t}{2} + 2 .

Discussion

Graphing Rational Functions

Rational functions may have more than one vertical asymptote, which can make graphing them a bit more difficult than graphing reciprocal functions. However, the asymptotes are useful when graphing a rational function because they can describe the end behavior. Consider graphing the following function.

f (x) = \frac{( x ^{2} + 5 x - 6 ) ( x - 3 )}{x ^{2} - 7 x + 1 2}

To graph the rational function, these four steps can be followed.

Draw the Asymptotes

First, the vertical asymptote(s) of the function will be found. If the numerator and the denominator of a rational function have no common factors, then the vertical asymptote is the

x -

value that makes the denominator zero. Start by factoring the denominator to see candidates for asymptotes.

f (x) = \frac{( x ^{2} + 5 x - 6 ) ( x - 3 )}{x ^{2} - 7 x + 1 2}

▼

Factor the denominator

WriteDiff

Write as a difference

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

FactorOut

Factor out $x$

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

FactorOut

Factor out $- 4$

f (x) = \frac{( x ^{2} + 5 x - 6 ) ( x - 3 )}{x ( x - 3 ) - 4 ( x - 3 )}

FactorOut

Factor out $(x - 3)$

f (x) = \frac{( x ^{2} + 5 x - 6 ) ( x - 3 )}{( x - 4 ) ( x - 3 )}

The rational function is undefined where

x - 4 = 0

and

x - 3 = 0 .

This means that

x = 4

and

x = 3

are not included in the domain. Notice that

x - 3

is a common factor between the numerator and the denominator. Next, check if there are any other common factors and cancel them.

f (x) = \frac{( x ^{2} + 5 x - 6 ) ( x - 3 )}{( x - 4 ) ( x - 3 )}

▼

Factor the numerator

WriteDiff

Write as a difference

f (x) = \frac{( x ^{2} + 6 x - x - 6 ) ( x - 3 )}{( x - 4 ) ( x - 3 )}

FactorOut

Factor out $x$

f (x) = \frac{[ x ( x + 6 ) - x - 6 ] ( x - 3 )}{( x - 4 ) ( x - 3 )}

FactorOut

Factor out $- 1$

f (x) = \frac{[ x ( x + 6 ) - 1 ( x + 6 ) ] ( x - 3 )}{( x - 4 ) ( x - 3 )}

FactorOut

Factor out $(x + 6)$

f (x) = \frac{( x - 1 ) ( x + 6 ) ( x - 3 )}{( x - 4 ) ( x - 3 )}

▼

Simplify

CrossCommonFac

Cross out common factors

f (x) = \frac{( x - 1 ) ( x + 6 ) ( x - 3 )}{( x - 4 ) ( x - 3 )}

CancelCommonFac

Cancel out common factors

f (x) = \frac{( x - 1 ) ( x + 6 )}{x - 4}, x \neq = 3

Since the factor

x - 3

was canceled, the graph has a hole at

x = 3 .

Also, the graph has a vertical asymptote at

x = 4

because

4

is not included in the domain.

Hole : Vertical Asymptote : x = 3 x = 4

The degrees of the polynomials in the numerator and denominator determine whether the function has a horizontal or oblique asymptote. In the following table,

m

and

n

are the degrees of the polynomials in the numerator and denominator, and

a

and

b

are their respective leading coefficients.

$y = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}$	Asymptote	Asymptote Type
$m < n$	$y = 0$	Horizontal
$m > n$	None	None
$m = n$	$y = \frac{a}{b}$	Horizontal
$m - n = 1$	$y =$ the quotient of the polynomials with no remainder	Oblique

Now consider the degrees of the numerator and denominator for the given function. The simplified form of the function can also be considered at this point because both have the same graph.

f (x) = \frac{( x - 1 ) ( x + 6 )}{x - 4}, x \neq = 3 ⇓ f (x) = \frac{x ^{2} + 5 x - 6}{x ^{1} - 4}, x \neq = 3

The degree of the numerator minus the degree of the denominator equals

1 .

Therefore, the function has an oblique asymptote. The equation of the oblique asymptote is found by dividing the numerator by the denominator without considering the remainder. To divide the polynomials, long division will be used.

x - 4 x^{2} + 5 x - 6

▼

Divide

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

x x - 4 x^{2} + 5 x - 6

Multiply $term$ by $divisor$

x x - 4 x^{2} + 5 x - 6 x^{2} - 4 x

Subtract down

x x - 4 9 x - 6

▼

Divide

$\frac{9 x}{x} = 9$

x + 9 x - 4 9 x - 6

Multiply $term$ by $divisor$

x + 9 x - 4 9 x - 6 9 x - 36

Subtract down

x + 9 x - 4 30

The equation of the oblique asymptote is

y = x + 9 .

Finally, draw the asymptotes on a coordinate plane.

Vertical asymptote x=4 and oblique asymptote y=x+9

The given function has one vertical asymptote. Therefore, its graph will consist of two parts, one to the left and the other to the right of the vertical asymptote. These parts may lie above or below the oblique asymptote.

Find the Intercepts

The

x -

intercepts are the points where the graph intersects the

x -

axis. The value of the

y -

coordinate is zero at these points. Substitute

0

for

f (x)

in the given function and solve for

x .

The simplified function can be used to find the zeros.

f (x) = \frac{( x - 1 ) ( x + 6 )}{x - 4}

Substitute

$f (x) = 0$

0 = \frac{( x - 1 ) ( x + 6 )}{x - 4}

▼

Solve for

x

MultEqn

$LHS \cdot (x - 4) = RHS \cdot (x - 4)$

0 = (x - 1) (x + 6)

ZeroProdProp

Use the Zero Product Property

x - 1 = 0 x + 6 = 0 (I) (II)

AddEqn

$(I):$ $LHS + 1 = RHS + 1$

x = 1 x + 6 = 0

SubEqn

$(II):$ $LHS - 6 = RHS - 6$

x = 1 x = - 6

The graph has two

x -

intercepts. The

y -

intercept is the point where the graph intersects the

y -

axis. At this point, the value of the

x -

coordinate is zero.

f (x) = \frac{( x - 1 ) ( x + 6 )}{x - 4}

Substitute

$x = 0$

f (x) = \frac{( 0 - 1 ) ( 0 + 6 )}{0 - 4}

▼

Evaluate right-hand side

IdPropAdd

Identity Property of Addition

f (x) = \frac{( - 1 ) ( 6 )}{- 4}

Multiply

f (x) = \frac{- 6}{- 4}

CalcQuot

Calculate quotient

f (x) = 1.5

There is a

y -

intercept at

(0, 1.5) .

Show the intercepts on the same coordinate plane.

The asymptotes and intercepts on the same coordinate plane

Make a Table of Values

Some additional points are needed to get a good sense of the shape of the graph. Make sure to only use values included in the domain of the function.

$x$	$\frac{( x - 1 ) ( x + 6 )}{x - 4}$	$f (x) = \frac{( x - 1 ) ( x + 6 )}{x - 4}$
$- 11$	$\frac{( - 1 1 - 1 ) ( - 1 1 + 6 )}{- 1 1 - 4}$	$- 4$
$2$	$\frac{( 2 - 1 ) ( 2 + 6 )}{2 - 4}$	$- 4$
$6$	$\frac{( 6 - 1 ) ( 6 + 6 )}{6 - 4}$	$30$
$9$	$\frac{( 9 - 1 ) ( 9 + 6 )}{9 - 4}$	$24$
$19$	$\frac{( 1 9 - 1 ) ( 1 9 + 6 )}{1 9 - 4}$	$30$

It is almost done. Plot the points and imagine how the shape of the graph should look!

Asymptotes, intercepts and points from table of values on a coordinate plane

As can be seen, one part of the graph will lie to the left of the vertical asymptote and below the oblique asymptote. The other part of the graph will be to the right of the vertical asymptote and above the oblique asymptote.

Draw the Graph

The graph can now be drawn by connecting the points with a smooth curve. It must approach the asymptotes. Do not forget to plot the hole at $x = 3!$

Example

Graphing the Function for the Surface Area to Volume Ratio of a Box

The factory sells $6$ pencils in a rectangular box. The dimensions of the box are shown in the diagram.

6 pencils in a rectangular box height 2x+6, width x-2, depth 2

a Write a function of the form

E (x) = \frac{S ( x )}{V ( x )},

where

S (x)

and

V (x)

are the surface area and volume of the box, respectively.

b Graph the function

E (x) .

Answer

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

b Graph:

Hint

a The surface area of a prism is the sum of the areas of its surfaces. The volume of a prism is the product of its dimensions.

b Start by determining the vertical and horizontal asymptotes. Then, make a table of values.

Solution

a First, the surface area and volume of the given box will be found.

The surface area of a rectangular box is the total area of all the surfaces of the shape. Since the dimensions are given in terms of

x,

the surface area is a function of

x .

S (x) = 2 (2) (x - 2) + 2 (2 x + 6) (x - 2) + 2 (2) (2 x + 6)

▼

Simplify right-hand side

Multiply

S (x) = 4 (x - 2) + 2 (2 x + 6) (x - 2) + 4 (2 x + 6)

Distr

Distribute $4$

S (x) = 4 x - 8 + 2 (2 x + 6) (x - 2) + 8 x + 24

Distr

Distribute $2$

S (x) = 4 x - 8 + (4 x + 12) (x - 2) + 8 x + 24

MultPar

Multiply parentheses

S (x) = 4 x - 8 + 4 x^{2} - 8 x + 12 x - 24 + 8 x + 24

AddSubTerms

Add and subtract terms

S (x) = 4 x^{2} + 16 x - 8

Next, the volume of the pencil box will be calculated. Recall that the volume of a rectangular prism is the product of its dimensions.

V (x) = (2 x + 6) (x - 2) (2)

▼

Simplify right-hand side

Distr

Distribute $2$

V (x) = (2 x + 6) (2 x - 4)

MultPar

Multiply parentheses

V (x) = 4 x^{2} - 8 x + 12 x - 24

AddTerms

Add terms

V (x) = 4 x^{2} + 4 x - 24

Finally, the function

E (x),

the ratio of

S (x)

V (x),

can be written.

E (x) = \frac{S ( x )}{V ( x )}

SubstituteExpressions

Substitute expressions

E (x) = \frac{4 x ^{2} + 1 6 x - 8}{4 x ^{2} + 4 x - 2 4}

▼

Simplify right-hand side

FactorOut

Factor out $4$

E (x) = \frac{4 ( x ^{2} + 4 x - 2 )}{4 ( x ^{2} + x - 6 )}

ReduceFrac

$\frac{a}{b} = \frac{a / 4}{b / 4}$

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

b To graph the given rational function, its asymptotes and intercepts will be found. A table of values will then be used to draw the graph.

Asymptotes

Consider the function written in the previous part.

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

First, factor the denominator of this function to identify its domain.

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

WriteDiff

Write as a difference

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + 3 x - 2 x - 6}

▼

Factor out

x & - 2

FactorOut

Factor out $x$

E (x) = \frac{x ^{2} + 4 x - 2}{x ( x + 3 ) - 2 x - 6}

FactorOut

Factor out $- 2$

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

FactorOut

Factor out $(x + 3)$

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

The rational function is undefined where

x - 2 = 0

and

x + 3 = 0

because division by zero is not defined. This means that

x = 2

and

x = - 3

are not included in the domain.

\underline{Domain} All real numbers except x = - 3 and x = 2

Asymptotes can be vertical, horizontal, or slanted lines. Once again, consider the function.

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

It seems that the denominator and the numerator do not have a common factor. To find the factors of the numerator, its zeros can be found using the Quadratic Formula.

E (x) = \frac{( x + 2 + 6 ) ( x + 2 - 6 )}{( x - 2 ) ( x + 3 )}

As shown, they do not have common factors. Therefore, there are no holes in the graph. Recall that if the real number

a

is not included in the domain, there is a vertical asymptote at

x = a .

In this case, there are two vertical asymptotes, one at

x = - 3

and the other at

x = 2 .

\underline{Vertical Asymptotes} x = - 3 and x = 2

To find whether the graph has a horizontal or oblique asymptote, review how its type is determined using the degrees of the polynomials in the numerator and the denominator.

$y = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}$	Asymptote
$m < n$	$y = 0$
$m > n$	none
$m = n$	$y = \frac{a}{b}$
$m - n = 1$	$y =$ the quotient of the polynomials with no remainder

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

E (x) = \frac{1 x ^{2} + 4 x - 2}{1 x ^{2} + x - 6}

The degree of the denominator is equal to the degree of the numerator. Therefore, the line

y = \frac{1}{1} = 1

is the horizontal asymptote of the function.

Intercepts

To find the

x -

intercepts,

0

is substituted for

E (x)

into the function. It will be better to use the form in which the numerator and the denominator are factored.

E (x) = \frac{( x + 2 + 6 ) ( x + 2 - 6 )}{( x - 2 ) ( x + 3 )}

Substitute

$E (x) = 0$

0 = \frac{( x + 2 + 6 ) ( x + 2 - 6 )}{( x - 2 ) ( x + 3 )}

▼

Solve for

x

MultEqn

$LHS \cdot (x - 2) (x + 3) = RHS \cdot (x - 2) (x + 3)$

0 = (x + 2 + 6) (x + 2 - 6)

ZeroProdProp

Use the Zero Product Property

x + 2 + 6 = 0 x + 2 - 6 = 0 (I) (II)

SubEqn

$(I):$ $LHS - 2 - 6 = RHS - 2 - 6$

x = - 2 - 6 x + 2 - 6 = 0

SubEqn

$(II):$ $LHS - 2 + 6 = RHS - 2 + 6$

x = - 2 - 6 x = - 2 + 6

UseCalc

Use a calculator

x = - 4.449489 \dots x = 0.449489 \dots

RoundDec

Round to $1$ decimal place(s)

x \approx - 4.4 x \approx 0.4

To find the

y -

intercept, substitute

0

for

x

in the function and solve for

E (0) .

This time the form of the function found in Part A can be used.

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

Substitute

$x = 0$

E (0) = \frac{0 ^{2} + 4 ( 0 ) - 2}{0 ^{2} + 0 - 6}

▼

Solve for

E (0)

CalcPow

Calculate power

E (0) = \frac{0 + 4 \cdot 0 - 2}{0 + 0 - 6}

ZeroPropMult

Zero Property of Multiplication

E (0) = \frac{0 + 0 - 2}{0 + 0 - 6}

IdPropAdd

Identity Property of Addition

y = \frac{- 2}{- 6}

ReduceFrac

$\frac{a}{b} = \frac{a / ( - 2 )}{b / ( - 2 )}$

y = \frac{1}{3}

There is a

y -

intercept at

(0, \frac{1}{3}) .

Finally, all the intercepts can be plotted on the coordinate plane.

$Asymptotes and intercepts (-2-\sqrt{6},0), (-2+\sqrt{6},0), and (0,1/3) are graphed on the coordinate plane$

Table of Values

Now, make a table of values to graph the given function.

$x$	$\frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}$	$E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}$
$- 6$	$\frac{( - 6 ) ^{2} + 4 ( - 6 ) - 2}{( - 6 ) ^{2} + ( - 6 ) - 6}$	$\approx 0.417$
$- 3.5$	$\frac{( - 3 . 5 ) ^{2} + 4 ( - 3 . 5 ) - 2}{( - 3 . 5 ) ^{2} + ( - 3 . 5 ) - 6}$	$\approx - 1.364$
$- 2$	$\frac{( - 2 ) ^{2} + 4 ( - 2 ) - 2}{( - 2 ) ^{2} + ( - 2 ) - 6}$	$1.5$
$- 1$	$\frac{( - 1 ) ^{2} + 4 ( - 1 ) - 2}{( - 1 ) ^{2} + ( - 1 ) - 6}$	$\approx 0.833$
$1$	$\frac{( 1 ) ^{2} + 4 ( 1 ) - 2}{( 1 ) ^{2} + ( 1 ) - 6}$	$- 0.75$
$3$	$\frac{( 3 ) ^{2} + 4 ( 3 ) - 2}{( 3 ) ^{2} + ( 3 ) - 6}$	$\approx 3.167$
$4$	$\frac{( 4 ) ^{2} + 4 ( 4 ) - 2}{( 4 ) ^{2} + ( 4 ) - 6}$	$\approx 2.143$
$5$	$\frac{( 5 ) ^{2} + 4 ( 5 ) - 2}{( 5 ) ^{2} + ( 5 ) - 6}$	$\approx 1.792$

Finally, the graph of the function can be drawn by plotting the found points and connecting them with a smooth curve. Recall that a rational function can cross the horizontal asymptote but cannot cross the vertical asymptotes. In this case, the horizontal asymptote is crossed.

Extra

Factoring The Numerator of

E (x)

Recall the Factor Theorem.

$p (x)$ has a factor $(x - a)$ if and only if $p (a) = 0 .$

In other words, the polynomial function has zeros at the

a -

values. Since the numerator of

E (x)

is a quadratic expression its zeros can be found by using the Quadratic Formula.

E (x) = \frac{x ^{2} + 4 x - 2}{x ^{2} + x - 6}

The values of

a,

b,

and

c

of the numerator are

1,

4,

and

- 2,

respectively. Substitute these values into the formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- 4 \pm 4 ^{2} - 4 ( 1 ) ( - 2 )}{2 ( 1 )}

▼

Evaluate right-hand side

IdPropMult

Identity Property of Multiplication

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

CalcPow

Calculate power

x = \frac{- 4 \pm 1 6 - 4 ( - 2 )}{2}

MultNegNegOnePar

$- a (- b) = a \cdot b$

x = \frac{- 4 \pm 1 6 + 8}{2}

AddTerms

Add terms

x = \frac{- 4 \pm 2 4}{2}

SplitIntoFactors

Split into factors

x = \frac{- 4 \pm 4 \cdot 6}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{- 4 \pm 4 \cdot 6}{2}

CalcRoot

Calculate root

x = \frac{- 4 \pm 2 \cdot 6}{2}

FactorOut

Factor out $2$

x = \frac{2 ( - 2 \pm 6 )}{2}

SimpQuot

Simplify quotient

x = - 2 \pm 6

The zeros of the numerator are

x = - 2 - 6

and

x = - 2 + 6 .

By the Factor Theorem, the following is the factored form of the numerator.

x^{2} + 4 x - 2 ⇕ (x + 2 + 6) (x + 2 - 6)

Example

Modeling the Amount of Water Pumped Into the Factory

Factory managers plan to pump water into the factory from a pond near the factory. The managers modeled the amount of water pumped in thousands of barrels per year

x

years after pumping starts to be as follows.

W (x) = \frac{- x ^{3} + 1 4 0 x ^{2} + 1 5 0 0 x + 5 2 5 0}{1 5 x ^{2} + 1 5 0 x}

a Use a graphing calculator to find the

x -

intercepts rounded to the nearest integer. Interpret the

x -

intercepts.

b Determine if the graph has horizontal or oblique asymptotes. If the graph has an oblique asymptote, find its equation.

Answer

x - intercept :

(150, 0)

Interpretation: In about $150$ years, there will be no water left in the pond.

b The graph has an oblique asymptote because the difference between the degrees of the numerator and the denominator is

1 .

Equation: $y = - \frac{x}{1 5} + 10$

Hint

a Use the zero option in the calculator. It may be necessary to re-size the viewing window in order to see the entire graph.

b Compare the degrees of the numerator and the denominator.

Solution

a In order to find any

x -

intercepts, or zeros, of the function, look at the graph of the function using a graphing calculator. To draw the function on the calculator, push the

Y =

button and type the equation in the first row.

Now, push $GRAPH$ to draw the function.

It appears that only one part of the graph is shown, but it is not possible to know that until the size of the viewing window is changed.

The graph does not seem to have any zeros. However, note that as $x$ increases in the first quadrant, the graph gets closer to the $x -$ axis and could cross it. To check this, zoom in on this part by changing the window settings. Push $WINDOW$ and change the settings as shown below. Then push $GRAPH$ once more to draw the equation with these new settings.

Now it can be seen that the graph intersects the $x -$ axis and there is a zero. To find it, use the zero option in the calculator. It can be found by pressing $2ND$ and then $CALC .$

After selecting the zero option, choose left and right boundaries for the zero. Finally, the calculator asks for a guess where the zero might be. After that, it will calculate the exact point.

The function intersects the $x -$ axis at about $(150, 0) .$ Since $x$ represents the numbers of years that have passed since pumping starts, the $x -$ intercept means that in about $150$ years, there will be no water left in the pond.

b Recall the table used to determine what type of asymptote a rational function can have.

$y = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}$	Asymptote	Asymptote Type
$m < n$	$y = 0$	Horizontal
$m > n$	None	None
$m = n$	$y = \frac{a}{b}$	Horizontal
$m - n = 1$	$y =$ the quotient of the polynomials with no remainder	Oblique

Now, compare the degrees of the numerator and denominator of the given function.

W (x) = \frac{- x ^{3} + 1 4 0 x ^{2} + 1 5 0 0 x + 5 2 5 0}{1 5 x ^{2} + 1 5 0 x}

The degree of the numerator minus the degree of the denominator equals

1 .

Therefore, the function has an oblique asymptote. The equation of this asymptote is found by dividing the numerator by the denominator and disregarding any remainder. To divide the polynomials, use long division.

15 x^{2} + 150 x - x^{3} + 140 x^{2} + 1500 x + 5250

▼

Divide

$\frac{- x ^{3}}{1 5 x ^{2}} = - \frac{x}{1 5}$

- \frac{x}{1 5} 15 x^{2} + 150 x - x^{3} + 140 x^{2} + 1500 x + 5250

Multiply $term$ by $divisor$

- \frac{x}{1 5} 15 x^{2} + 150 x - x^{3} + 140 x^{2} + 1500 x + 5250 - x^{3} - 10 x^{2}

Subtract down

- \frac{x}{1 5} 15 x^{2} + 150 x 150 x^{2} + 1500 x + 5250

▼

Divide

$\frac{1 5 0 x ^{2}}{1 5 x ^{2}} = 10$

- \frac{x}{1 5} + 10 15 x^{2} + 150 x 150 x^{2} + 1500 x + 5250

Multiply $term$ by $divisor$

- \frac{x}{1 0} + 10 15 x^{2} + 150 x 150 x^{2} + 1500 x + 5250 150 x^{2} + 1500 x

Subtract down

- \frac{x}{1 5} + 10 15 x^{2} + 150 x 5250

The given function can be rewritten as follows.

W (x) = - \frac{x}{1 5} + 10 + \frac{5 2 5 0}{1 5 x ^{2} + 1 5 0 x}

Therefore, the equation of the oblique asymptote is

y = - \frac{x}{1 5} + 10 .

Closure

Graphing the Average Cost Function

Rational functions can have more than one vertical asymptote but at most one horizontal or oblique asymptote. Also, if the numerator and denominator have a common factor

(x - a),

then there is a hole at

x = a .

With these tips in mind, the graph of the function that LaShay wrote at the beginning of the lesson can now be graphed.

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

The function gives the average cost

A (x)

in thousands dollars for producing

x

thousand pencils.

a Graph the average cost function.

b Identify the domain and range of the function.

Answer

a Graph:

b Domain:

(- \infty, 0) \cup (0, \infty)

Range:

(- \infty, - 15] \cup [25, \infty)

Hint

a Start by identifying asymptotes. Then, make a table of values using values that are around the vertical asymptote.

b Examine the graph and see if the graph is symmetric about the point of intersection of the asymptotes. Determine all the

y -

values for which the function is graphed.

Solution

a The steps to draw a rational function can be listed as follows.

Draw the asymptotes.
Find any intercepts.
Make a table of values.
Draw the graph.

Drawing the Asymptotes

The vertical asymptotes of a rational function are the

x -

values where the denominator of the function is zero.

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

The denominator is

0

when

x = 0 .

This means that this value is not included in the domain.

\underline{Domain} All real numbers except x = 0

Since the numerator and denominator do not share a common factor, the graph does not contain any holes. Additionally, the graph has a vertical asymptote at the

x -

value excluded from the domain.

\underline{Vertical Asymptote} x = 0

Next, recall the table for determining the other asymptotes.

$y = \frac{a x ^{m} + \dots}{b x ^{n} + \dots}$	Asymptote	Asymptote Type
$m < n$	$y = 0$	Horizontal
$m > n$	None	None
$m = n$	$y = \frac{a}{b}$	Horizontal
$m - n = 1$	$y =$ the quotient of the polynomials with no remainder	Oblique

Examine the degrees of the numerator and denominator.

A (x) = \frac{0 . 2 x ^{2} + 5 x + 5 0 0}{x ^{1}}

The degree of the numerator minus the degree of the denominator equals

1 .

Therefore, the function has an oblique asymptote. To find its equation, divide the numerator by the denominator using long division.

x 0.2 x^{2} + 5 x + 500

▼

Divide

$\frac{0 . 2 x ^{2}}{x} = 0.2 x$

0.2 x x 0.2 x^{2} + 5 x + 500

Multiply $term$ by $divisor$

0.2 x x 0.2 x^{2} + 5 x + 500 0.2 x^{2}

Subtract down

0.2 x x 5 x + 500

▼

Divide

$\frac{5 x}{x} = 5$

0.2 x + 5 x 5 x + 500

Multiply $term$ by $divisor$

0.2 x + 5 x 5 x + 500 5 x

Subtract down

0.2 x + 5 x 500

The equation of the oblique asymptote is

y = 0.2 x + 5 .

\underline{Oblique Asymptote} y = 0.2 x + 5

The asymptotes can be drawn on a coordinate plane.

Finding the Intercepts

The

x -

intercept and

y -

intercept are the points where a graph crosses the

x -

and

y -

axes, respectively. Subsequently, the

x -

intercepts occurs when

A (x) = 0 .

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

Substitute

$A (x) = 0$

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

MultEqn

$LHS \cdot x = RHS \cdot x$

0 = 0.2 x^{2} + 5 x + 500

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a = 0.2, b = 5, c = 500$

x = \frac{- 5 \pm 5 ^{2} - 4 ( 0 . 2 ) ( 5 0 0 )}{2 ( 0 . 2 )}

CalcPowProd

Calculate power and product

x = \frac{- 5 \pm 2 5 - 4 0 0}{0 . 4}

SubTerm

Subtract term

x = \frac{- 5 \pm - 3 7 5}{0 . 4}

Notice that there is a negative number under the radical symbol. Therefore, the function has no real solutions. In other words, its graph does not cross the

x -

axis. Furthermore, the graph has no

y -

intercept as the domain does not contain

0 .

Making a Table of Values

A table of values will be used to get a rough idea of the shape of the graph.

$x$	$\frac{0 . 2 x ^{2} + 5 x + 5 0 0}{x}$	$A (x) = \frac{0 . 2 x ^{2} + 5 x + 5 0 0}{x}$
$- 75$	$\frac{0 . 2 ( - 7 5 ) ^{2} + 5 ( - 7 5 ) + 5 0 0}{- 7 5}$	$\approx - 16.7$
$- 25$	$\frac{0 . 2 ( - 2 5 ) ^{2} + 5 ( - 2 5 ) + 5 0 0}{- 2 5}$	$- 20$
$- 10$	$\frac{0 . 2 ( - 1 0 ) ^{2} + 5 ( - 1 0 ) + 5 0 0}{- 1 0}$	$- 47$
$10$	$\frac{0 . 2 ( 1 0 ) ^{2} + 5 ( 1 0 ) + 5 0 0}{1 0}$	$57$
$25$	$\frac{0 . 2 ( 2 5 ) ^{2} + 5 ( 2 5 ) + 5 0 0}{2 5}$	$30$
$75$	$\frac{0 . 2 ( 7 5 ) ^{2} + 5 ( 7 5 ) + 5 0 0}{7 5}$	$\approx 26.7$

Plot the points and imagine how the shape of the graph should look!

Finally, draw the graph by connecting the points. It must approach, but not cross, the asymptotes.

b The domain of the function was identified when graphing the function.

Domain : (- \infty, 0) \cup (0, \infty)

To identify the range of the function, take a look at its graph.

As shown, the range does not contain all real numbers. It appears that the minimum point of the part of the graph in the first quadrant is $(50, 25) .$ Notice also that the graph is symmetric about the point of intersection of the asymptotes, $(0, 5) .$ Using this fact, the maximum point of the other part can be identified.

It is the point

(- 50, - 15) .

Therefore, the range is all the

y -

values less than or equal to

- 15

and greater than or equal to

25 .

Range : (\infty, - 15] \cup [25, \infty)

Checking Our Answer

Use a Graphing Calculator

A graphing calculator can be used to check the answers. The graph of the function will be drawn first. Press the $Y =$ button and type the equation in the first row.

By pushing $GRAPH,$ the calculator will draw the equation. For this function to be visible on the screen, re-size the standard window by pushing the $WINDOW$ button. Change the settings to a more appropriate size and then push $GRAPH .$

Next, the local maximum and local minimum will be found. To do so, push $2nd,$ then $TRACE,$ and choose the maximum option.

When using the maximum feature, choose the left and right bounds. The calculator will then provide a best guess as to where the maximum might be.

The maximum point of the graph in the third quadrant is $(- 50, - 15) .$ To find the minimum, repeat the same process but choose the minimum option.

The minimum point of the part in the first quadrant is $(50, 25) .$ As a result, the range is $(\infty, - 15] \cup [25, \infty),$ and the domain is all real numbers except $0 .$

	14 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Graphing Rational Functions

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

Answer

Hint

Solution

Asymptotes

Intercepts

Table of Values

Extra

Answer

Hint

Solution

Answer

Hint

Solution

Drawing the Asymptotes

Finding the Intercepts

Making a Table of Values

Checking Our Answer

Graphing Rational Functions

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