3.6 rational functions & their graphs...2014/11/03 · 3.6-rational functions & their...

3.6-Rational Functions & Their Graphs

What is a Rational Function? A rational function is a function that is the ratio of two polynomial functions. This definition is similar to a rational number which is a number that can be written as the ratio of two integers. Definition:

A rational function)(

)()(

xQ

xPxf = , is the ratio of two polynomial functions, P(x) and Q(x), where 0≠Q .

Example: The following are all rational functions. Notice that each numerator and each denominator is a polynomial function. The polynomial may be a monomial, a binomial, a trinomial, or a polynomial.

xxf

5)( =

13

2)(

2 +−−=

xx

xxg

72

42)(

25

−+−+=

x

xxxxh

Domain of a Rational Function: Because a rational function is a ratio we need to make sure that the function value of the denominator is never equal to zero. Remember that division by zero is mathematically undefined, that is, we simply cannot do it. Therefore we always need to restrict the output or range values of the denominator to non-zero numbers. This is most easily accomplished by setting the denominator equal to zero and solving the resulting equation. The rational function is therefore undefined at these values.

Example: Identify all the values for which the function 2

3)(

−=

xxf is undefined.

Solution: Set the function in the denominator equal to zero and solve.

2

02

==−

x

x

Therefore, the function is undefined at x = 2. This can be demonstrated by evaluating the function at this value.

0

3)2(

22

3)2(

2

3)(

=

−=

−=

f

f

xxf

Notice that the numerator is not involved in this problem at all.

Example: Identify all the values for which the function 25

3)(

2 −−=

x

xxg is undefined.


5

5

0)5)(5(

0252

−==

=+−=−

x

x

xx

x

Therefore, the function is undefined at x = 5, and x = -5. This can be demonstrated by evaluating the function at these values.

0

2)5(

255

35)5(

25

3)(

2

2

=

−−=

−−=

g

g

x

xxg

0

8)5(

25)5(

35)5(

25

3)(

2

2

−=−

−−−−=−

−−=

g

g

x

xxg

Example: Determine the domain of the function127

1)(

2 +−+=xx

xxQ .


0)4)(3(

01272

=−−=+−

xx

xx

3

03

==−

x

x and

4

04

==−

x

x

The domain is ),4()4,3()3,(: ∞∪∪−∞D .

Example: Determine the domain of the function372

5)(

2 ++=

xxxf .


0)12)(3(

0)3(1)3(2

0)3()62(

0362

0372

2

2

2

=++=+++=+++

=+++

=++

xx

xxx

xxx

xxx

xx

3

03

−==+

x

x And

21

12

012

−=−==+

x

x

x

The domain is ),21()2

1,3()3,(: ∞−∪−−∪−−∞D .

It should be quite obvious by now that factoring is an important part of these problems. In fact, factoring will be a major part of most problems involving rational expressions and equations. In the following problems, certain factors will cancel out.

Example: Determine the domain of the function 2

103)(

2

+−−=

x

xxxf

Solution: Notice that if we factor the numerator, a factor in the denominator cancels out leaving a simplified function with no denominator.

52

)2)(5(2

1032

−=+

+−=

+−−=

xx

xxx

xx

The domain is ),2()2,(: ∞−∪−−∞D . Even though the factors of (x+2) cancelled out, the domain must still be

restricted based on the original problem (before factoring and cancelling.)

Example: Determine the domain of the function )3)(3(

3)(

+−+=

xx

xxr

Solution: Notice that if we factor the denominator, a factor in the denominator cancels out.

3

1)(

)3)(3(

3)(

−=

+−+=

xxr

xx

xxr

The domain is ),3()3,3()3,(: ∞∪−∪−−∞D . Even though the factors of (x+3) cancelled out, the domain must

still be restricted based on the original problem (before factoring and cancelling.)

Vertical Asymptotes: You will recall that an asymptote is a line that a graph will come infinitely close to but never touch. A vertical asymptote is therefore a vertical line that the graph comes close to but never touches. Vertical asymptotes occur where there exists a restriction on the domain of a rational function. A vertical asymptote will always have an equation in the form ax = where a is any real number. In the following examples we will concentrate on finding the vertical asymptote rather than on graphing them.

Example: Find any vertical asymptotes of the function 2

3)(

−=

xxf

Solution: Because vertical asymptotes occur at restrictions to the domain, we must first determine the domain. By setting the denominator equal to zero, we can see that the domain excludes the value x = 2. Therefore, the equation of the vertical asymptote is 2=x .


3)(

2 −−=

x

xxg

Solution: Because vertical asymptotes occur at restrictions to the domain, we must first determine the domain. By setting the denominator equal to zero, we can see that the domain excludes the value7=x and 7−=x . Therefore, the equations of the vertical asymptotes are 7=x and 7−=x .


103)(

2

+−−=

x

xxxf

Solution: In a previous example we saw that if we factor the numerator, a factor in the denominator cancels out leaving a simplified function with no denominator.

52

)2)(5(2

1032

−=+

+−=

+−−=

xx

xxx

xx

The domain is ),2()2,(: ∞−∪−−∞D . Even though the factor of 2+x cancelled out, the domain must still be

restricted based on the original problem.

However, because the factor 2+x cancelled out, there will be no vertical asymptote at this point. Rather, there

will be a hole in the graph at the point where 2−=x . This is called a removable discontinuity.

Example: Find any vertical asymptotes of the function )3)(3(

3)(

+−+=

xx

xxr

Solution: In a previous example we saw that if we factor the denominator, a factor in the denominator cancels out leaving only one factor in the denominator.

3

1)(

)3)(3(

3)(

−=

+−+=

xxr

xx

xxr

The domain is ),3()3,3()3,(: ∞∪−∪−−∞D . Even though the factors of 3+x cancelled out, the domain must

still be restricted based on the original problem.

There will only be one vertical asymptote at 3=x . Once again, because the factor 3+x cancelled out, there will

be no vertical asymptote at this point. Rather, there will be a hole in the graph at the point where 3−=x .

Horizontal Asymptotes:

A horizontal asymptote is a horizontal line that the graph comes infinitely close to but never touches. Because it is a horizontal line, the equation of a horizontal asymptote will always be of the form by = where b is any real

number. To find any horizontal asymptotes of a rational function we need to use the following guidelines.

Given the rational function)(

)()(

xQ

xPxf = ,then if,

1. The degree of P(x) < the degree of Q(x), the horizontal asymptote is the equation 0=y .

2. The degree of P(x) = the degree of Q(x), the horizontal asymptote is the ratio of the leading coefficients.

3. The degree of P(x) > the degree of Q(x), the horizontal asymptote does not exist.

Example: Find any horizontal asymptotes of the function53

2)(

2 −+=

x

xxf .

Solution: Because the degree of the numerator P(x) is less than the degree of the denominator Q(x), the horizontal asymptote has the equation 0=y

Example: Find any horizontal asymptotes of the function 72

14)(

−+=

x

xxf .

Solution: Because the degree of the numerator is equal to the degree of the denominator, the horizontal

asymptote is the ratio of the leading coefficients 2

4=y or 2=y

Example: Find any horizontal asymptotes of the function83

4)(

2

++=

x

xxf .

Solution: Because the degree of the numerator P(x) is greater than the degree of the denominator Q(x), there are no horizontal asymptotes.

Oblique Asymptotes: When the degree of the numerator P(x) is greater than the degree of the denominator Q(x) we already know that there are no horizontal asymptotes. However, there exists a special case where the degree of the numerator is exactly one degree higher than the denominator. In this case there will be a slant or oblique asymptote. The equation of the oblique asymptote is written in slope

intercept form bmxy += . The equation is the quotient

found by dividing the denominator into the numerator using long division.

Example: Find any oblique asymptotes of the function 15

31710)(

2

+++=

x

xxxf

Solution: Because the degree of the numerator is exactly one degree greater than the degree of the denominator, there will be an oblique asymptote. Divide the denominator into the numerator using long division. The resulting quotient will provide the equation of the oblique asymptote.

32

0......................

315............

315............

210

31710152

2

+

++

++++

x

x

x

xx

xxx

The equation of the oblique asymptote is 32 += xy

Example: Find any oblique asymptotes of the function 3

4194)(

2

23

−−−=

x

xxxf

Solution: Because the degree of the numerator is exactly one degree greater than the degree of the denominator, there will be an oblique asymptote. Divide the denominator into the numerator using long division. The resulting quotient will provide the equation of the oblique asymptote.

194

5312.....................

57.........19........

41219.......

12..........4

401943

2

2

3

232

−

+−−−+−

−−+−−x

x

x

xx

xx

xxxx

The equation of the oblique asymptote is 194 −= xy .

Graphing Rational Functions: The following strategy can be used to sketch the graph rational functions.

A. If the graph is a transformation ofx

xf1

)( = , then use transformations to sketch the graph of the

rational function. B. If the graph is not a transformation of this function, then:

a. Find any vertical asymptotes. b. Find any horizontal or oblique asymptotes. c. Find the x-intercepts d. Find the y-intercepts e. Plot at least one point between each x-intercept and vertical asymptote.

Example: Sketch the graph of the function 42

2)( +

−=

xxf

Solution: This is a transformation ofx

xf1

)( = .

The graph will shift to the right 2 unit and shift upward 4 units. The 2 in the numerator will simply stretch the graph although this will have a negligible effect on the appearance of the graph.


2)(

−=

x

xxf

Solution: Because this is not a transformation, use procedure B to sketch the graph.

a. Vertical Asymptotes: 1=x is restricted from the domain.

b. Horizontal Asymptotes: degree of P(x) = degree of Q(x), therefore 2=y

c. Find x-intercepts: Let y = 0 and solve for x.

01

20

=−

=

xx

x

The x-intercept is (0, 0)

d. Find y-intercept: Let x = 0 and solve for y.

01

2

=−

=

yx

xy

The y-intercept is (0, 0)

Since we already knew that the point (0,0) was an intercept, this step was really unnecessary.

e. To plot a point between the x-intercept and the vertical asymptote, let2

1=x and solve

for y.

( )

2

1

2

21

21

−=−

=

y

y

Plot the point

−2,2

1


3)(

2

2

−=

x

xxf

Solution: Because this is not a transformation, use procedure B to sketch the graph.

a. Vertical Asymptotes: 2=x and 2−=x are restricted from the domain.

b. Horizontal Asymptotes: degree of P(x) = degree of Q(x), therefore 3=y


04

30

2

2

=−

=

xx

x


d. The y-intercept is (0, 0).

e. To plot a point between each x-intercept and the vertical asymptotes, let 1±=x and solve for y.

1

4)1(

)1(32

2

−=−

=

y

y

1

4)1(

)1(32

2

−=−−

−=

y

y

Plot the points ( )1,1− and ( )1,1−−


)(2 −+

=xx

xxf

Solution: Once again, use procedure B to sketch the graph.

a. Vertical Asymptotes: 3=x and 4−=x are restricted from the domain.

b. Horizontal Asymptotes: degree of P(x) < degree of Q(x), therefore 0=y


012

02

=−+

=

xxx

x


d. The y-intercept is (0, 0).

e. o plot a point between each x-intercept and the vertical asymptotes, let 2±=x and solve for y.

3

11222

212

2

2

−=

−+=

−+=

y

y

xx

xy

5

1

12)2()2(

212

2

2

=

−−+−−=

−+=

y

y

xx

xy

Plot the points ( )31,1− and ( )5

1,1−


1)(

2 −−=

x

xxf

Solution: This function simplifies to 1

1)(

+=

xxf

a. Vertical Asymptotes: 1=x and 1−=x are restricted from the domain. However, because 1=x is a removable discontinuity there will not be a vertical asymptote at this point. Instead, there will be a hole in the graph.

b. Horizontal Asymptotes: degree of P(x) < degree of Q(x), therefore 0=y


101

10

=+

=x

There is no x-intercept.

d. Find y-intercepts: Let x = 0 and solve for y.

110

102

=−

−=

y

y

The y-intercept is (0, 1).

e. Because there are no x-intercepts, we may omit this step.


54)(

2

−−−=

x

xxxf

Solution:

a. Vertical Asymptotes: 3=x is restricted from the domain. b. Horizontal Asymptotes: P(x) > Q(x), therefore none

Oblique Asymptotes: 1−= xy

1

8..........

3.....

5.....

3

5432

2

−

−+−−−

−−−−

x

x

x

xx

xxx


1,5

0)1)(5(

054

3

540

2

2

−===+−

=−−−

−−=

xx

xx

xx

x

xx

The x-intercepts are (-1, 0), and (5, 0)

d. Find y-intercepts: Let x = 0 and solve for y.

3

530

5)0(402

=

−−−=

y

y

The y-intercept is (0, 5/3)

e. To plot a point between each x-intercept and the vertical asymptotes, let 2−=x and 4=x . Solve for y.

Plot the points ( )57,2 −− and ( )5,4−

3.6-Applications

Example: Jordan paid $100.00 for a lifetime membership to the zoo, so that he could gain admittance to the zoo for only $1.00 per visit. Write Jordan’s average cost per visit C as a function of the number of visits when he has visited x times. What is his average cost per visit when he has visited the zoo 100 times? Graph the function for x > 0. What happens to the day? Solution: Since x equals the number of visits we can create the following function.

When Jordan has visited the zoo 100 times his average cost will be.

The average cost will be $2.00 per visit.

According to the graph, as the number of visits init will never reach zero.

The line y = 1 is a horizontal asymptote.

Applications

Jordan paid $100.00 for a lifetime membership to the zoo, so that he could gain admittance to the zoo for only $1.00 per visit. Write Jordan’s average cost per visit C as a function of the number of visits

imes. What is his average cost per visit when he has visited the zoo 100 times? Graph the function for x > 0. What happens to the average cost per visit if he starts when he is young as visits every

of visits we can create the following function.

x

xxC

100)(

+=

When Jordan has visited the zoo 100 times his average cost will be.

2)100(100

100100)100(

=

+=

C

C

The average cost will be $2.00 per visit.

According to the graph, as the number of visits increases the cost per visit will come close to $1.00 although

The line y = 1 is a horizontal asymptote.

Jordan paid $100.00 for a lifetime membership to the zoo, so that he could gain admittance to the zoo for only $1.00 per visit. Write Jordan’s average cost per visit C as a function of the number of visits

imes. What is his average cost per visit when he has visited the zoo 100 times? Graph verage cost per visit if he starts when he is young as visits every

creases the cost per visit will come close to $1.00 although

Example: The cost of renting a car for one day is $19.00 plus $0.30 per mile. Write the average cost per mile C as a function of the number of miles driven in one day x. What is the maximum number of miles the car can be driven so as not to exceed an average cost What happens to C as the number of miles gets very large?

Solution: Since x equals the number of miles we can create the following function.

To find the maximum number of miles the carper mile, let C(x) =0.35 and solve for x.

At 380 miles, the average cost will be $0.35 per mile. will approximate $0.30 per mile.

The line y = 0.30 is a horizontal asymptote.

The cost of renting a car for one day is $19.00 plus $0.30 per mile. Write the average cost per mile C as a function of the number of miles driven in one day x. What is the maximum number of miles the car can be driven so as not to exceed an average cost of $0.35 per mile. Graph the function for x > 0. What happens to C as the number of miles gets very large?

Since x equals the number of miles we can create the following function.

x

xxC

193.0)(

+=

To find the maximum number of miles the car can be driven so as not to exceed an average cost of $0.35 per mile, let C(x) =0.35 and solve for x.

380

1905.0

1930.035.0

193.035.0

==

+=

+=

x

x

xxx

x

At 380 miles, the average cost will be $0.35 per mile. As the number of miles increases, the average cost

The line y = 0.30 is a horizontal asymptote.

The cost of renting a car for one day is $19.00 plus $0.30 per mile. Write the average cost per mile C as a function of the number of miles driven in one day x. What is the maximum number of miles

of $0.35 per mile. Graph the function for x > 0.

Since x equals the number of miles we can create the following function.

can be driven so as not to exceed an average cost of $0.35

s the number of miles increases, the average cost

Example:: The cost o f renting a twin engine Cessna 402 is $200.00 per day plus $150.00 per hour of flight time.

1. Write the average cost per hour F as a function of the number of hours flown in one day h. 2. What is the maximum number of hours the Cessna may be flown per day so as not to exceed an

average cost of $200.00 per hour. 3. Graph the function for h > 0. 4. What is the lowest average hourly cost? At how many hours does this average cost occur?

Solution:

1. The average cost per hour

2. To find the maximum number of hours the Cessna may be flown so as not to exceed an average cost of $200.00 per hour, create an inequality and solve.

So the Cessna must be flown at least 4 hours per day.

3. Graph the function for h >

4. The lowest possible average cost in a 24 hour day will be approximately $158.00 per hour, which occurs at exactly 24 hours. This value is approaching the horizontal asymptote of 150.

The cost o f renting a twin engine Cessna 402 is $200.00 per day plus $150.00 per hour of

Write the average cost per hour F as a function of the number of hours flown in one day h. What is the maximum number of hours the Cessna may be flown per day so as not to exceed an average cost of $200.00 per hour.

the lowest average hourly cost? At how many hours does this average cost occur?

The average cost per hour F as a function of the number of hours flown in one day

h

hhF

200150)(

+=

To find the maximum number of hours the Cessna may be flown so as not to exceed an average cost of $200.00 per hour, create an inequality and solve.

4

20050

200200150

200200150

≥≥

≤+

≤+

h

h

hhh

h

So the Cessna must be flown at least 4 hours per day.

Graph the function for h > 0.

The lowest possible average cost in a 24 hour day will be approximately $158.00 per hour, which occurs at exactly 24 hours. This value is approaching the horizontal asymptote

The cost o f renting a twin engine Cessna 402 is $200.00 per day plus $150.00 per hour of

Write the average cost per hour F as a function of the number of hours flown in one day h. What is the maximum number of hours the Cessna may be flown per day so as not to exceed an

the lowest average hourly cost? At how many hours does this average cost occur?

as a function of the number of hours flown in one day h is:

To find the maximum number of hours the Cessna may be flown so as not to exceed an average

3.6 rational functions & their graphs...2014/11/03 · 3.6-rational functions & their...

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