3.5 - 1 Understand and analyze the graph of the reciprocal function Understand and analyze the graph of the function Learn how to find horizontal and vertical.

3.5 - 1

Understand and analyze the graph of the reciprocal function

Understand and analyze the graph of the function

Learn how to find horizontal and vertical asymptotes

Learn how to Graph Rational Functions

Learn how to model rational Function

1( )x

xf

2

1( )x

xf

Objectives

Students will -

3.5 - 2

Rational function

A function of the form

( )( ) ,

( )p x

xq x

f

where p(x) and q(x) are polynomials, with q(x) ≠ 0, is called a rational function.

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Rational Function

Some examples of rational functions are

2

2 2

1 1 3 3 6( ) , ( ) , and ( )

2 5 3 8 16x x x

x x xx x x x x

f f f

Since any values of x such that q(x) = 0 are excluded from the domain of a rational function, this type of function often has a discontinuous graph, that is, a graph that has one or more breaks in it.

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The Reciprocal Function

The simplest rational function with a variable denominator is the reciprocal function, defined by

1( ) .x

xf

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The Reciprocal Function

The domain of this function is the set of all real numbers except 0. The number 0 cannot be used as a value of x, but it is helpful to find values of for some values of (x) for some values of x very close to 0. We use the table feature of a graphing calculator to do this. The tables suggest that (x) gets larger and larger as x gets closer and closer to 0, which is written in symbols as ( ) as 0.x x f

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The Reciprocal Function(The symbol x 0 means that x approaches 0, without necessarily ever being equal to 0.) Since x cannot equal 0, the graph of will never intersect the vertical line x = 0.This line is called a vertical asymptote.On the other hand, as x gets larger and larger, the values of get closer and closer to 0, as shown in the tables. Letting x get larger and larger without bound (written x ) causes the graph to move closer and closer to the horizontal line. This line is called a horizontal asymptote.

1( )x

xf

1( )x

xf

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Domain: (– , 0) (0, ) Range: (– , 0) (0, )

RECIPROCAL FUNCTION

x y

– 2 – ½ – 1 – 1

– ½ – 2

0 undefined

½ 2

1 1

2 ½

decreases on the intervals (–,0) and (0, ).

1( )x

xf

xxf

1)(

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Domain: (– , 0) (0, ) Range: (– , 0) (0, )

RECIPROCAL FUNCTION

x y

– 2 – ½ – 1 – 1

– ½ – 2

0 undefined

½ 2

1 1

2 ½

It is discontinuous at x = 0.

xxf

1)(

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Domain: (– , 0) (0, ) Range: (– , 0) (0, )

RECIPROCAL FUNCTION

x y

– 2 – ½ – 1 – 1

– ½ – 2

0 undefined

½ 2

1 1

2 ½

The y-axis is a vertical asymptote, and the x-axis is a horizontal asymptote.

xxf

1)(

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Domain: (– , 0) (0, ) Range: (– , 0) (0, )

RECIPROCAL FUNCTION

x y

– 2 – ½ – 1 – 1

– ½ – 2

0 undefined

½ 2

1 1

2 ½

It is an odd function and its graph is symmetric with respect to the origin.

xxf

1)(

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Example 1 GRAPHING A RATIONAL FUNCTION

Solution

Graph Give the domain and range. 2.y

x

The expression can be written as or indicating that the graph may be obtained by stretching the graph of vertically by a factor of 2 and reflecting it across either the y-axis or x-axis. The x- and y-axes remain the horizontal and vertical asymptotes. The domain and range are both still (– , 0) (0, ).

2x

1

2x

12 ,

x

1y

x

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Solution

Graph Give the domain and range.

2( ) .

1x

x

f

The expression can be written as indicating that the graph may be obtained by shifting the graph ofto the left 1 unit and stretchingit vertically by a factor of 2.

21x 1

21x

1y

x

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Solution


2( ) .

1x

x

f

The horizontal shift affects the domain, which is now (– , – 1) (– 1, ) .The line x = – 1 is the vertical asymptote, and the line y = 0(the x-axis) remains the horizontal asymptote. The range is still (– , 0) (0, ).

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The Function

The Function The rational function defined by

also has domain (– , 0) (0, ). We can use the table feature of a graphingcalculator to examine values of (x) for some x-values close to 0.

2

1( )x

xf

2

1( )x

xf

2

1( )x

xf

3.5 - 15

The Function

The tables suggest that (x) gets larger and larger as x gets closer and closer to 0. Notice that as x approaches 0 from either side, function values are all positive and there is symmetry with respect to the y-axis. Thus, (x) as x 0. The y-axis (x = 0) is the vertical asymptote.

2

1( )x

xf

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Domain: (– , 0) (0, ) Range: (0, )

RECIPROCAL FUNCTION

x y

3 2 ¼ 1 1 ½ 4

¼ 16

0 undefined

increases on the interval (–,0) and decreases on the interval (0, ).

2

1( )x

xf

19

2

1( )x

xf

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Domain: (– , 0) (0, ) Range: (0, )

RECIPROCAL FUNCTION

x y

3 2 ¼ 1 1 ½ 4

¼ 16

0 undefined

It is discontinuous at x = 0.

19

2

1( )x

xf

3.5 - 18

Domain: (– , 0) (0, ) Range: (0, )

RECIPROCAL FUNCTION

x y

3 2 ¼ 1 1 ½ 4

¼ 16

0 undefined

The y-axis is a vertical asymptote, and the x-axis is a horizontal asymptote.

19

2

1( )x

xf

3.5 - 19

Domain: (– , 0) (0, ) Range: (0, )

RECIPROCAL FUNCTION

x y

3 2 ¼ 1 1 ½ 4

¼ 16

0 undefined

It is an even function, and Its graph is symmetric with respect to the y-axis.

19

2

1( )x

xf

3.5 - 20


Solution The equation is equivalent to


2

11.

( 2)y

x

( 2) 1,y x f

where This indicates that the graph will be shifted 2 units to the left and 1 unit down.

2

1( ) .x

xf

3.5 - 21




2

11.

( 2)y

x

( 2) 1,y x f

The horizontal shift affects the domain, which is now (– , – 2) (– 2, ), while the vertical shift affects the range, now (– 1, ).

3.5 - 22




2

11.

( 2)y

x

( 2) 1,y x f

The vertical asymptote has equation x = – 2, and the horizontal asymptote has equation y = – 1.

3.5 - 23

Asymptotes

Let p(x) and q(x) define polynomials. For the rational function defined by written in lowest terms, and for real numbers a and b:

1. If (x) as x a, then the line is a vertical asymptote.2. If (x) b as x , then the line y = b is a horizontal asymptote.

( )( ) ,

( )p x

xq x

f

3.5 - 24

Determining Asymptotes

To find the asymptotes of a rational function defined by a rational expression in lowest terms, use the following procedures.1. Vertical Asymptotes Find any vertical asymptotes by setting the denominator equal to 0 and solving for x. If a is a zero of the denominator, then the line x = a is a vertical asymptote.

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2. Other AsymptotesDetermine any other asymptotes. Consider three possibilities:(a) If the numerator has lower degree than the denominator, then there is a horizontal asymptote y = 0 (the x-axis).

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2. Other AsymptotesDetermine any other asymptotes. Consider three possibilities:

(b) If the numerator and denominator have the same degree, and the function is of the form

where an, bn ≠ 0,

then the horizontal asymptote has equation.n

n

ay

b

0

0

( ) ,n

nn

n

a x ax

b x b

f

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2. Other AsymptotesDetermine any other asymptotes. Consider three possibilities:

(c) If the numerator is of degree exactly one more than the denominator, then there will be an oblique (slanted) asymptote. To find it, divide the numerator by the denominator and disregard the remainder. Set the rest of the quotient equal to y to obtain the equation of the asymptote.

3.5 - 28

Motion Problems

Note The graph of a rational function may have more than one vertical asymptote, or it may have none at all. The graph cannot intersect any verticalasymptote. There can be at most one other (nonvertical) asymptote, and the graph may intersect that asymptote as we shall see in Example 7.

3.5 - 29

Example 4 FINDING ASYMPTOTES OF GRAPHS OF RATIONAL FUNCTIONS

Solution To find the vertical asymptotes, set the denominator equal to 0 and solve.

a.

For each rational function , find all asymptotes.

1( )

(2 1)( 3)x

xx x

f

(2 1)( 3) 0x x

2 1 0 or 3 0x x Zero-property

1 or 3

2x x Solve each

equation.

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The equations of the vertical asymptotes are x = ½ and x = – 3.To find the equation of the horizontal asymptote, divide each term by the greatest power of x in the expression. First, multiply the factors in the denominator.

2

1 1( )

(2 1)( 3) 2 5 3x x

xx x x x

f

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2 2

2

2

2 2 2

2

1

( )2

1 1

25 35 3

x x

x x

x

xx

xx

x

x

x

x

f

Now divide each term in the numerator and denominator by x2 since 2 is the greatest power of x.

Stop here. Leave the

expression in complex

form.

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As x gets larger and larger, the quotients

all approach 0, and the value of (x) approaches

2 2

1 1 5 3, , , and

x x x x

0 00 0

00.

2 2

The line y = 0 (that is, the x-axis) is therefore the horizontal asymptote.

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Solution Set the denominator x – 3 = 0 equal to 0 to find that the vertical asymptote has equation x = 3. To find the horizontal asymptote, divide each term in the rational expression by x since the greatest power of x in the expression is 1.

b.


2 1( )

3x

xx

f

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Solution

b.


2 1( )

3x

xx

f

2 122 1

( )33

1

13

xx

xxxx x x

x xx

f

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As x gets larger and larger, both approach 0, and (x) approaches

1 3 and

x x

2 22,

100 1

so the line y = 2 is the horizontal asymptote.

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Solution Setting the denominator x – 2 equal to 0 shows that the vertical asymptote has equation x = 2. If we divide by the greatest power of x as before ( inthis case), we see that there is no horizontal asymptote because

c.


2 1( )

2x

xx

f

3.5 - 37


2

2 2 2 2

2 2 2

1 111

( )2 1 22

xx x x xx

xxx x x x

f

does not approach any real number as x , since is undefined. This happens whenever the degree of the numerator is greater than the degree of the denominator. In such cases, divide the denominator into the numerator to write the expression in another form. We use synthetic division.

10

3.5 - 38


We use synthetic division.

The result allows us to write the function as

2 1 0 12 4

1 2 5

5( ) 2 .

2x x

x

f

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For very large values of x, is close to 0, and the graph approaches the line y = x + 2. This line is an oblique asymptote (slanted, neither vertical nor horizontal) for the graph of the function.

52x

3.5 - 40

Steps for Graphing Functions

A comprehensive graph of a rational function exhibits these features:1. all x- and y-intercepts;2. all asymptotes: vertical, horizontal, and/or oblique;3. the point at which the graph intersects its nonvertical asymptote (if there is any such point);4. the behavior of the function on each domain interval determined by the verticalasymptotes and x-intercepts.

3.5 - 41

Graphing a Rational Function

Let define a function where p(x) and q(x) are polynomials and the rational expression is written in lowest terms. To sketch its graph, follow these steps.Step 1 Find any vertical asymptotes.Step 2 Find any horizontal or oblique asymptotes.Step 3 Find the y-intercept by evaluating (0).

( )( )

( )p x

xq x

f

3.5 - 42


Step 4 Find the x-intercepts, if any, by solving (x) = 0 . (These will be the zeros of the numerator, p(x).)

Step 5 Determine whether the graph will intersect its nonvertical asymptote y = b or y = mx + b by solving (x) = b or(x) = mx + b.

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Step 6 Plot selected points, as necessary. Choose an x-value in each domain interval determined by the vertical asymptotes and x-intercepts.

Step 7 Complete the sketch.

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Example 5 GRAPHING A RATIONAL FUNCTION WITH THE x-AXIS AS HORIZONTAL ASYMPTOTE

Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 1 Since 2x2 + 5x – 3 = (2x – 1)(x + 3), from Example 4(a), the vertical asymptotes have equations x = ½ and x = – 3.

Step 2 Again, as shown in Example 4(a), the horizontal asymptote is the x-axis.

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Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 3 The y-intercept is – ⅓, since

2( ) .2( ) 5

1(

00

0 01

3 3)

f

The y-intercept is the ratio of the constant terms.

3.5 - 46


Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 4 The x-intercept is found by solving (x) = 0.

2

12 5 3

0x

x x

If a rational expression is equal to 0, then its numerator must equal 0.

1x The x-intercept is – 1.

1 0x

3.5 - 47


Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 5 To determine whether the graph intersects its horizontal asymptote, solve

y-value of horizontal asymptote

Since the horizontal asymptote is the x-axis, the solution of this equation was found in Step 4. The graph intersects its horizontal asymptote at (– 1, 0).

)(xf

3.5 - 48


Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 6 Plot a point in each of the intervals determined by the x-intercepts and vertical asymptotes, to get an idea of how the graph behaves in each interval.

3.5 - 49


Interval Test Point

Value of (x)

Sign of (x) Graph Above or Below

x-Axis

(– , – 3) – 4 Negative Below

(– 3, – 1) – 2 Positive Above

(– 1, ½ ) 0 Negative Below

(½ , ) 2 Positive Above

15

15

13

13

3.5 - 50


Solution

Graph 2

1( ) .

2 5 3x

xx x

f

Step 7 Complete the sketch.

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Example 6 GRAPHING A RATIONAL FUNCTION THAT DOES NOT INTERSECT ITS HORIZONTAL ASYMPTOTE

Solution

Graph2 1

( ) .3

xx

x

f

Step 1 and 2 As determined in Example 4(b), the equation of the vertical asymptote is x = 3. The horizontal asymptote has equation y = 2.

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Solution

Graph2 1

( ) .3

xx

x

f

Step 3 (0) = – ⅓, so the y-intercept is – ⅓.

3.5 - 53


Solution

Graph2 1

( ) .3

xx

x

f

Step 4 Solve (x) = 0 to find any x-intercepts.2

30

1xx

If a rational expression is equal to 0, then its numerator must equal 0.2 1 0x

12

x x-intercept

3.5 - 54


Solution

Graph2 1

( ) .3

xx

x

f

Step 5 The graph does not intersect its horizontal asymptote since (x) = 2 has no solution.

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Solution

Graph2 1

( ) .3

xx

x

f

Step 6 and 7 The points (– 4, 1), (1, – 3/2), and (6, 13/3) are on the graph and can be used to complete the sketch.

3.5 - 56

Example 7 GRAPHING A RATIONAL FUNCTION THAT INTERSECTS ITS HORIZONTAL ASYMPTOTE

Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 1 To find the vertical asymptote(s), solve x2 + 8x + 16 = 0.

2 8 16 0x x Set the denominator equal to 0.

2( 4) 0x Factor.

4x Zero-factor property.

3.5 - 57


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Since the numerator is not 0 when x = – 4, the vertical asymptote has the equation x = – 4.

4x Zero-factor property.

3.5 - 58


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 2 We divide all terms by x2 to get the equation of the horizontal asymptote.

Leading coefficient of numerator

Leading coefficient of denominator

31

y

or 3.y

3.5 - 59


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 3 The y-intercept is (0) = – 3/8.

3.5 - 60


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 4 To find the x-intercept(s), if any, we solve (x) = 0.

2

2

3 3 68 16

0x x

x x

23 3 6 0x x Set the numerator equal

to 0.

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Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 4 23 3 6 0x x Set the numerator equal to 0.

2 2 0x x Divide by 3.

( 2)( 1) 0x x Factor.

2 or 1x x Zero-factor property

The x-intercepts are – 1 and 2.

3.5 - 62


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 5 We set (x) = 3 and solve to locate the point where the graph intersects the horizontal asymptote.

Multiply by x2 + 8x + 16.

2

2

3 3 68 16

3x x

x x

2 23 3 6 3 24 48x x x x

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Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 5

Multiply by x2 + 8x + 16.2 23 3 6 3 24 48x x x x

3 6 24 48x x Subtract 3x2.

27 54x Subtract 24x; add 6.

2x Divide by – 27.

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Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 52x Divide by – 27.

The graph intersects its horizontal asymptote at (– 2, 3).

3.5 - 65


Solution

Graph2

2

3 3 6( ) .

8 16x x

xx x

f

Step 6 and 7 Some of the other points that lie on the graph are

These are used to complete the graph.

1 210,9 , 8,13 , and 5, .

8 3

3.5 - 66

Behavior of Graphs of Rational Functions Near Vertical Asymptotes

Suppose that (x) is defined by a rational expression in lowest terms. If n is the largest positive integer such that (x – a)n is a factor of the denominator of (x), the graph will behave in the manner illustrated.

3.5 - 67

Behavior of Graphs

Previously we observed that the behavior of the graph of a polynomial function near its zeros is dependent on the multiplicity of the zero. The same statement can be made for rational functions. Suppose that (x) is defined by a rational expression in lowest terms. If n is the greatest positive integer such that (x – c)n is a factor of the numerator of (x), the graph will behave in the manner illustrated.

3.5 - 68

Behavior of Graphs

3.5 - 69

Example 8 GRAPHING A RATIONAL

FUNCTION WITH AN OBLIQUE ASYMPTOTE

Solution In Example 4, the vertical asymptote has equation x = 2, and the graph has an oblique asymptote with equation y = x + 2. Refer to the previous discussion to determine the behavior near the vertical asymptote x = 2.

Graph 2 1

( ) .2

xx

x

f

3.5 - 70



Solution The y-intercept is – ½ , and the graph has no x-intercepts since the numerator, x2 + 1, has no real zeros. The graph does not intersect its oblique asymptote because

Graph 2 1

( ) .2

xx

x

f

3.5 - 71



Solution

Graph 2 1

( ) .2

xx

x

f

2 12

2x

xx

has no solution. Using the y-intercept, asymptotes, the points andand the general behavior of the graph near its asymptotes leads to this graph.

174,

2

21, ,

3

3.5 - 72



Solution

Graph 2 1

( ) .2

xx

x

f

3.5 - 73

As mentioned earlier, a rational function must be defined by an expression in lowest terms before we can use the methods discussed in this section to determine the graph. A rational function that is not in lowest terms usually has a “hole,” or point of discontinuity, in its graph.

3.5 - 74

Example 9 GRAPHING A RATIONAL FUNCTION DEFINED BY AN EXPRESSION THAT IS NOT IN LOWEST TERMS

Solution The domain of this function cannotinclude 2. The expression should be written in lowest terms.

Graph 2 4

( ) .2

xx

x

f

2 42

xx

2( )2

2)(xx

x

Factor.

( 2 2) ,x x x

3.5 - 75

Example 9 GRAPHING A RATIONAL FUNCTION DEFINED BY AN EXPRESSION THAT IS NOT IN LOWEST TERMS

Solution The graph of this function will be the same as the graph of y = x + 2 (a straight line), with the exception of the point with x-value 2. A “hole” appears in the graph at (2, 4).

Graph 2 4

( ) .2

xx

x

f

3.5 - 76

Example 10 MODELING TRAFFIC INTENSITY WITH A RATIONAL FUNCTION

Vehicles arrive randomly at a parking ramp at an average rate of 2.6 vehicles per minute. The parking attendant can admit 3.2 vehicles per minute. However, since arrivals are random, lines form at various times.

3.5 - 77


(a) The traffic intensity x is defined as the ratio of the average arrival rate to the average admittance rate. Determine x for this parking ramp.

The average arrival rate is 2.6 vehicles and the average admittance rate is3.2 vehicles, so 2.6

3.2.8125x

Solution

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(b) The average number of vehicles waiting in line to enter the ramp is given by

2.8125.8125

.( ) 1.76 vehicles

2(1 )8125

f

2

( ) ,2(1 )

xx

x

f

where 0 x < 1 is the traffic intensity. Graph (x) and compute (.8125) for this parking ramp.

Solution

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(c) What happens to the number of vehicles waiting as the traffic intensity approaches 1?


SolutionFrom the graph we see that as x approaches 1, y = (x) gets very large; that is, the average number of waiting vehicles gets very large. This is what we would expect.

3.5 - 1 Understand and analyze the graph of the reciprocal function Understand and analyze the graph of the function Learn how to find horizontal and vertical.

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