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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.
Solution: Set the function in the denominator equal to zero and solve.
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 .
Solution: Set the function in the denominator equal to zero and solve.
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 .
Solution: Set the function in the denominator equal to zero and solve.
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 .
Example: Find any vertical asymptotes of the function 49
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 .
Example: Find any vertical asymptotes of the function 2
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.
Example: Sketch the graph of the function 1
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
Example: Sketch the graph of the function 4
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
c. Find x-intercepts: Let y = 0 and solve for x.
04
30
2
2
=−
=
xx
x
The x-intercept is (0, 0)
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−−
Example: Sketch the graph of the function 12
)(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
c. Find x-intercepts: Let y = 0 and solve for x.
012
02
=−+
=
xxx
x
The x-intercept is (0, 0)
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−
Example: Sketch the graph of the function 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
c. Find x-intercepts: Let y = 0 and solve for x.
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.
Example: Sketch the graph of the function 3
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
c. Find x-intercepts: Let y = 0 and solve for x.
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