2-5 rational functions - montville township public schools · find the domain of each function and...

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

1.f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 4. The real zeros of b(x) are 2 and 2.

Therefore, D = {x | x 2,2, x R}. Check for vertical asymptotes. Determine whether x = 2 is a point of infinite discontinuity. Find the limit as x approaches 2 from the left and the right.

Because x = 2 is a vertical asymptote of f .

Determine whether x = 2 is a point of infinite discontinuity. Find the limit as x approaches 2 from the left and the right.


Check for horizontal asymptotes. Use a table to examine the end behavior of f (x).

The table suggests Therefore,youknowthaty = 1 is a horizontal asymptote of f .

x 1.9 1.99 1.999 2 2.001 2.01 2.1 f (x) 4.1 49.1 499.1 undef 500.9 50.9 5.9

x 2.1 2.01 2.001 2 1.999 1.99 1.9 f (x) 5.9 50.9 500.9 undef 499.1 49.1 4.1

x 100 50 10 0 10 50 100 f (x) 1.0002 1.0008 1.02 0.5 1.02 1.0008 1.0002

2.h(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is 4. Therefore, D = {x | x 4, x R}. Check for vertical asymptotes.




The table suggests Therefore,theredoesnotappeartobeahorizontalasymptote.

x 4.1 4.01 4.001 4 3.999 3.99 3.9 f (x) 769.2 7248 72,048 undef 71,952 7152 673.2

x 100 50 10 0 10 50 100 f (x) 10417 2717 168 2 70.9 2314 9615

3.f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x 4). The real zeros of b(x) are 4 and3. Therefore, D = {x | x 4,3, x R}. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from the left and the right.






x 3.9 3.99 3.999 4 4.001 4.01 4.1 f (x) 570 6124 61667 undef 61762 6219 666

x 3.1 3.01 3.001 3 2.999 2.99 2.9 f (x) 21.7 175 1718 undef 1710 167 13.3

x 100 50 10 0 10 50 100 f (x) 9615 2314 71.8 0 166 2717 10417

4.g(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 5). The real zeros of b(x) are 3 and 5. Therefore, D = {x | x 3, 5, x R}. Check for vertical asymptotes.







x 5.1 5.01 5.001 5 4.999 4.99 4.9 f (x) 52.9 547 5498 undef 5502 552 57.4

x 3.1 3.01 3.001 3 2.999 2.99 2.9 f (x) 47.9 452 4503 undef 4497 447 42.4

x 100 50 10 0 10 50 100 f (x) 0.01 0.03 0.46 0.4 0.02 0.15 0.009

5.h(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 + 2x. The real zeros of b(x) are 0 and 2.

Therefore, D = {x | x 0,2, x R}. Check for vertical asymptotes.







x 0.1 0.01 0.001 0 0.001 0.01 0.1 f (x) 7.47 52.3 502 undef 498 47.8 2.95

x 2.1 2.01 2.001 2 1.999 1.99 1.9 f (x) 86.8 852 8501 undef 8498 848 82.3

x 1000 100 10 0 10 100 1000 f (x) 2.008 2.08 3.01 undef 1.34 1.92 1.992

6.f (x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x 4. The real zero of b(x) is 4. Therefore, D ={x | x 4,x R}. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from the left and the right.




x 3.9 3.99 3.999 4 4.001 4.01 4.1 f (x) 703 7183 71,983 undef 72,017 7217 737

x 100 50 10 0 10 50 100 f (x) 87.7 38.33 2.14 5 35 64.6 113.8

7.h(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x 2)(x + 4). The real zeros of b(x) are 2 and4. Therefore, D = {x | x 2,4, x R}. Check for vertical asymptotes. Determine whether x = 2 is a point of infinite discontinuity. Find the limit as x approaches 2 from the left and the right.






x 1.9 1.99 1.999 2 2.001 2.01 2.1 f (x) 7.5 825 83250 undef 83417 842 9.16

x 4.1 4.01 4.001 4 3.999 3.99 3.9 f (x) 42.5 4175 416,750 undef 416,583 4158 40.8

x 100 50 10 0 10 50 100 f (x) 0.0004 0.0004 0.019 0.016 0.008 0.0004 0.00009

8.g(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 1)(x 3). The real zeros of b(x) are 1 and 3. Therefore, D = {x | x 3,1, x R}. Check for vertical asymptotes. Determine whether x = 3 is a point of infinite discontinuity. Find the limit as x approaches 3 from the left and the right.






x 2.9 2.99 2.999 3 3.001 3.01 3.1 f (x) 13.8 126 1251 undef 1249 123 11.2

x 1.1 1.01 1.001 1 0.999 0.99 0.9 f (x) 11.2 123 1249 undef 1251 126 13.8

x 100 50 10 0 10 50 100 f (x) 0.999 0.998 0.957 2.67 0.935 0.998 0.999

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

9.f (x)=

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 4, 5, x R}. There are vertical asymptotes at x = 4 and x = 5. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercepts are 2 and 3, the zeros of the numerator. The y-intercept is becausef (0) = .

Graph the asymptotes and intercepts. Then find and plot points.

10.g(x) =

SOLUTION:


The x-intercepts are and6,thezerosofthenumerator.They-intercept is 9 because g(0) =9.


11.f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 2, 2, x R}. There are vertical asymptotes at x = 2 and x = 2. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is 2 because f (0)=2. Graph the asymptotes and intercept. Then find and plot points.

12.f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 0,6,x R}. There are vertical asymptotes at x = 0 and x = 6. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 2, the zero of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptotes and intercept. Then find and plot points.

13.g(x) =

SOLUTION:

g(x) can be simplified to g(x) = .

The function is undefined at b(x) = 0, so D = {x | x5, 6, x R}. There are vertical asymptotes at x = 5 and x = 6, the real zeros of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 2, the zero of the simplified numerator. The y-intercept is because


14.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 2, 0, 5, x R}. There are vertical asymptotes at x = 2, x = 0 and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are 6 and 4, the zeros of the numerator. There is no y-intercept because h(x) is undefined for x = 0. Graph the asymptotes and intercepts. Then find and plot points.

15.h(x) =

SOLUTION:

h(x) can be written as h(x) = .

The function is undefined at b(x) = 0, so D = {x | x 3, 1, x R}. There are vertical asymptotes at x = 3 and x = 1. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote.

The x-intercepts are 5, 0, and 2, the zeros of the numerator. The y-intercept is 0 because h(0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

16.f (x) =

SOLUTION:

f(x) can be written as f (x) = .


The x-intercepts are 6, 0, and 4, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

17.f (x) =

SOLUTION:

The function is undefined at b(x) = 0. x2 + 4x + 5 yields no real zeros, so D = {x | x R}.

There are no vertical asymptotes. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 8, the zero of the numerator. The y-intercept is because

Graph the asymptote and intercepts. Then find and plot points.

18.

SOLUTION:

The function is undefined at b(x) = 0. x2 + 6 yields no real zeros, so D = {x | x R}.


Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is because

Graph the asymptote and intercept. Then find and plot points.

19.SALESThebusinessplanforanewcarwashprojectsthatprofitsinthousandsofdollarswillbemodeledbythe

function p (z) = , where z is the week of operation and z = 0 represents opening.

a. State the domain of the function. b. Determine any vertical and horizontal asymptotes and intercepts for p (z). c. Graph the function.

SOLUTION:

a. The function is undefined at b(x) = 0. 2z2 + 7z + 5 can be written as (2z + 5)(z+ 1). The zeros of this polynomial

are at 1 and . Since the car wash cannot be open for negative weeks, these zeros do not fall in the domain and

D = {z | z 0,z R}.

b. Though there are vertical asymptotes at x = 1 and , these values of x are not in the domain. Therefore,

there are no vertical asymptotes for this situation.

There is a horizontal asymptote at y = , the ratio of the leading coefficients of the numerator and denominator,

because the degrees of the polynomials are equal.

The x-intercept is 1, the zero of the numerator. The y-intercept is because .

c. Find and plot points to construct a graph. Since the domain is restricted to real numbers greater than or equal to 0, it is only necessary to show the first and fourth quadrants.

For each function, determine any asymptotes, holes, and intercepts. Then graph the function and state itsdomain.

20.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 0,x R}. There is a vertical asymptote at x = 0. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is , the zero of the numerator. There is no y-intercept because h(x) is undefined for x = 0.

Graph the asymptotes and intercept. Then find and plot points.

21.h(x) =

SOLUTION:The function is undefined at b(x) = 0.

Thus, .

There is a vertical asymptote at .

There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is becauseh

(0) = .


22.f (x) =

SOLUTION:



The x-intercepts are 5 and 3, the zeros of the numerator. The y-intercept is 5 because f (0) =5. Graph the asymptotes and intercepts. Then find and plot points.

23.g(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 4,x R}. There is a vertical asymptote at x = 4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.



24.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 3, x R}. There is a vertical asymptote at x = 3. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because h(0) = 0. Graph the asymptote and intercepts. Then find and plot points.

25.g(x) =

SOLUTION:

g(x) can be written as g(x) = .

The function is undefined at b(x) = 0, so D = {x | x 4,x R}. There is a vertical asymptote at x = 4. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote.

The x-intercepts are 2, 1, and 0, the zeros of the numerator. The y-intercept is 0 because g(0) = 0. Graph the asymptote and intercepts. Then find and plot points.

26.f (x) =

SOLUTION:

f(x) can be written as f (x) = or .

The function is undefined at b(x) = 0, so D = {x | x 3, 1, 2, x R}. There are vertical asymptotes at x = 1 and x = 2, the real zeros of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 7, the zero of the simplified numerator. The y-intercept is becauseh(0) = .

There is a hole at (3, 1) because the original function is undefined when x = 3. Graph the asymptotes and intercepts. Then find and plot points.

27.g(x) =

SOLUTION:

g(x) can be written as g(x) = or .

The function is undefined at b(x) = 0, so D = {x | x 2, 1, or 2, x R}. There is a vertical asymptote at x = 1, the real zero of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator. Since the simplified numerator has no real zeros, there are no x-intercepts. The y-intercept is 1 because g(0) =1.

There are holes at (2, 1) and becausetheoriginalfunctionisundefinedwhenx = 2 and x = 2. Graph the

asymptotes and intercept. Then find and plot points.

28.f (x) =

SOLUTION:

f(x) can be written as f (x) =

The function is undefined at b(x) = 0, so D = {x | x 3, 1, x R}. There is a vertical asymptote at x = 3, the real zero of the simplified denominator. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 4, the zero of the simplified numerator. The y-intercept is becausef (0) = .

There is a hole at becausetheoriginalfunctionisundefinedwhenx = 1.


29.g(x) =

SOLUTION:


The function is undefined at b(x) = 0, so D = {x | x 4, 5, x R}. There is a vertical asymptote at x = 4, the real zero of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is , the zero of the simplified numerator.

The y-intercept is becauseg(0) = .

There is a hole at becausetheoriginalfunctionisundefinedwhenx = 5. Graph the asymptotes and

intercepts. Then find and plot points.

30.STATISTICSAnumberx is said to be the harmonic mean of y and z if istheaverageof and .

a. Write an equation for which the solution is the harmonic mean of 30 and 45. b. Find the harmonic mean of 30 and 45.

SOLUTION:

a. Write an equation that represents astheaverageof and .

=

Substitute y = 30 and z = 45.

b. Solve for x for the equation found in part a.

The harmonic mean of 30 and 45 is 36.

31.OPTICSThelensequationis = + , where f is the focal length, di is the distance from the lens to the

image, and do is the distance from the lens to the object. Suppose the object is 32 centimeters from the lens and the

focal length is 8 centimeters.

a. Write a rational equation to model the situation. b. Find the distance from the lens to the image.

SOLUTION:a. Substitute f = 8 and do = 32 into the equation.

b. Solve for di for the equation found in part a.

The distance from the lens to the image is 10 centimeters.

Solve each equation.

32.y + =5

SOLUTION:

y = 3 or y = 2

33. z = 4

SOLUTION:

34. + =1

SOLUTION:

x = 1 or x = 8

35.

SOLUTION:

Because the original equation is not defined when y = 2, you can eliminate this extraneous solution. So,

hasnosolution.

36. + =

SOLUTION:

37.

SOLUTION:

38.

SOLUTION:

x = 4 or x = 5

39.

SOLUTION:

40. + =3

SOLUTION:

x = 7 or x = 1

41.

SOLUTION:

42.WATERThecostperdaytoremovex percent of the salt from seawater at a desalination plant is c(x) = ,

where 0 x < 100. a. Graph the function using a graphing calculator. b. Graph the line y = 8000 and find the intersection with the graph of c(x) to determine what percent of salt can be removed for $8000 per day. c. According to the model, is it feasible for the plant to remove 100% of the salt? Explain your reasoning.

SOLUTION:a.

b. Use the CALC menu to find the intersection of c(x) and the line y = 8000.

For $8000 per day, about 88.9% of salt can be removed. c. No; sample answer: The function is not defined when x = 100. This suggests that it is not financially feasible to remove 100% of the salt at the plant.

Write a rational function for each set of characteristics.43.x-intercepts at x = 0 and x = 4, vertical asymptotes at x = 1 and x = 6, and a horizontal asymptote at y = 0

SOLUTION:Sample answer: For the function to have x-intercepts at x = 0 and x = 4, the zeros of the numerator need to be 0 and

4.Thus, factors of the numerator are x and (x 4). For there to be vertical asymptotes at x = 1 and x = 6, the zeros of the denominator need to be 1 and 6. Thus, factors of the denominator are (x 1) and (x 6). For there to be a horizontal asymptote at y = 0, the degree of the denominator needs to be greater than the degree of the numerator.

Raising the factor (x 6) to the second power will result in an asymptote at y = 0.

f(x) =

44.x-intercepts at x = 2 and x = 3, vertical asymptote at x = 4, and point discontinuity at (5, 0)

SOLUTION:

Sample answer: For the function to have x-intercepts at x = 2 and x = 3, the zeros of the numerator need to be 2 and 3.Thus, factors of the numerator are (x 2) and (x + 3). For there to be a vertical asymptote at x = 4, a zero of the denominator needs to be 4. Thus, a factor of the denominator is (x 4). For there to be point discontinuity at x= 5, (x + 5) must be a factor of both the numerator and denominator. Additionally, for the point discontinuity to be at (5, 0), the numerator must have a zero at x = 5. Thus, the (x + 5) in the numerator must have a power of 2 so that x = 5 remains a zero of the numerator after the function is simplified.

f(x) =

45.TRAVELWhen distance and time are held constant, the average rates, in miles per hour, during a round trip can

be modeled by , where r1 represents the average rate during the first leg of the trip and r2 represents the

average rate during the return trip. a. Find the vertical and horizontal asymptotes of the function, if any. Verify your answer graphically. b. Copy and complete the table shown.

c. Is a domain of r1 > 30 reasonable for this situation? Explain your reasoning.

SOLUTION:a. There is a vertical asymptote at the real zero of the denominator r1 = 30. There is a horizontal asymptote at r2 =

orr2 = 30, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the

polynomials are equal.

b. Evaluate the function for each value of r1 to determine r2.

c. No; sample answer: As r1 approaches infinity, r2 approaches 30. This suggests that the average speeds reached

during the first leg of the trip have no bounds. Being able to reach an infinite speed is not reasonable. The same

holds true about r2 as r1 approaches 30 from the right.

Use your knowledge of asymptotes and the provided points to express the function represented by each graph.

46.

SOLUTION:

Sample answer: Since there are vertical asymptotes at x = 6 and x = 1, (x + 6) and (x 1) are factors of the denominator. Since x = 4 is a x-intercept, (x 4) is a factor of the numerator. Because there is a horizontal asymptote at y = 1, the degrees of the polynomials in the numerator and denominator are equal and the ratio of their leading coefficients is 1. There also appears to be an additional x-intercept that is not labeled. Let x = a be the

second x-intercept. (x a) is then a factor of the numerator. A function that meets these characteristics is

f(x) = .

Substitute the point (5, 6) into the equation and solve for a.

Substitute a = 1 into the original equation.

A function that can represent the graph is f (x) = .

47.

SOLUTION:



f(x) = .




Use the intersection feature of a graphing calculator to solve each equation.

48. = 8

SOLUTION:

Graph y = andy = 8 using a graphing calculator. Use the CALC menu to find the intersection of the

graphs.

The intersections occur at x 1.59 and x10.05.Thus,thesolutionsto =8areabout 1.59 and

about 10.05.

49. = 1

SOLUTION:


graphs.


about 2.65.

50. = 2

SOLUTION:


graphs.


about 0.90.

51. = 6

SOLUTION:

Graph y = andy = 6 using a graphing calculator. Use the CALC menu to find the intersection of

the graphs.

The intersections occur at x 3.87, x1.20,andx3.70.Thus,thesolutionsto =6areabout

3.87, about 1.20, and about 3.70.

52.CHEMISTRYWhena60%aceticacidsolutionisaddedto10litersofa20%aceticacidsolutionina100-liter tank, the concentration of the total solution changes.

a. Show that the concentration of the solution is f (a) = , where a is the volume of the 60% solution.

b. Find the relevant domain of f (a) and the vertical or horizontal asymptotes, if any. c. Explain the significance of any domain restrictions or asymptotes. d. Disregarding domain restrictions, are there any additional asymptotes of the function? Explain.

SOLUTION:a. Sample answer: The concentration of the total solution is the sum of the amount of acetic acid in the original 10

liters and the amount in the a liters of the 60% solution, divided by the total amount of solution or .

Multiplying both the numerator and the denominator by 5 gives or .

b. Since a cannot be negative and the maximum capacity of the tank is 100 liters, the relevant domain of a is 0 a 90. There is a horizontal asymptote at y = 0.6, the ratio of the leading coefficients of the numerator and denominator,because the degrees of the polynomials are equal. c. Sample answer: Because the tank already has 10 liters of solution in it and it will only hold a total of 100 liters, the amount of solution added must be less than or equal to 90 liters. It is also impossible to add negative amounts of solution, so the amount added must be greater than or equal to 0. As you add more of the 60% solution, the concentration of the total solution will get closer to 60%, but because the solution already in the tank has a lower concentration, the concentration of the total solution can never reach 60%. Therefore, there is a horizontal asymptote at y = 0.6.

d. Yes; sample answer: The function is not defined at a = 10, but because the value is not in the relevant domain, the asymptote does not pertain to the function. If there were no domain restrictions, there would be a vertical

asymptote at a = 10.

53.MULTIPLEREPRESENTATIONS In this problem, you will investigate asymptotes of rational functions. a. TABULARCopyandcompletethetablebelow.Determinethehorizontalasymptoteofeachfunctionalgebraically.

b. GRAPHICAL Graph each function and its horizontal asymptote from part a. c. TABULAR Copy and complete the table below. Use the Rational Zero Function to help you find the real zeros of the numerator of each function.

d. VERBAL Make a conjecture about the behavior of the graph of a rational function when the degree of the denominator is greater than the degree of the numerator and the numerator has at least one real zero.

SOLUTION:a. Since the degree of the denominator is greater than the degree of the numerator, f (x), h(x), and g(x) will have horizontal asymptotes at y = 0.

b. Use a graphing calculator to graph f (x), h(x), and g(x).

c. c. To find the real zeros of the numerator of f (x), use the Rational Zero Theorem and factoring. Because the leading coefficient is 1, the possible rational zeros are the integer

factors of the constant term 4. Therefore, the possible rational zeros of f are 1, 2, and 4. Factoring x2 5x + 4 results in (x 4)(x 1). Thus, the real zeros of the numerator of f (x) are 4 and 1. To find the real zeros of the numerator of h(x), use the Rational Zero Theorem and synthetic division. Because the

leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 12. Therefore, the possible rational zeros of f are 1, 2, 3, 4, 6, and 12. By using synthetic division, it can be determined that x = 3 is a rational zero.

The remaing quadratic factor (x2 + 4) yields no real zeros. Thus, the real zero of the numerator of h(x) is 3.

To find the real zeros of the numerator of g(x), use the Rational Zero Theorem and factoring. Because the leading

coefficient is 1, the possible rational zeros are the integer factors of the constant term 1. Therefore, the possible

rational zeros of f are 1. Factoring x4 1 results in (x2 1)(x2 + 1) or (x + 1)(x 1)(x2 + 1). The quadratic factor

(x2 + 1) yields no real zeros. Thus, the real zeros of the numerator of g(x) are 1 and 1.

d. Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least on real zero, the graph of the function will have y = 0 as an asymptote and will intersect the asymptote atthe real zeros of the numerator.

54.REASONING Given f (x) = , will f (x) sometimes, always, or never have a horizontal asymptote at

y = 1 if a, b, c, d, e, and f are constants with a0andd0?Explain.

SOLUTION:Sometimes; sample answer: When a = d, the function will have a horizontal asymptote at y = 1. When ad, the function will not have a horizontal asymptote at y = 1.

55.PREWRITE Design a lesson plan to teach the graphing rational functions topics in this lesson. Make a plan that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion.

SOLUTION:See studentswork.

56.CHALLENGE Write a rational function that has vertical asymptotes at x = 2 and x = 3 and an oblique asymptotey = 3x.

SOLUTION:

Sample answer: Since there are vertical asymptotes at x = 2 and x = 3, (x + 2) and (x 3) are factors of the denominator. Since there is an oblique asymptote y = 3x, the degree of the numerator is exactly 1 greater than the degree of the denominator. Additionally, when the numerator is divided by the denominator, the quotient polynomial q

(x) is q(x) = 3x. So, the function can be written as f (x) = , where a(x) is the numerator of the

function. We know that =3x + , where r(x) is the remainder. We can write the

denominator of f (x) as x2 x 6 and can use the equation to solve for a(x).

The degree of the remainder has to be less than the degree of the denominator, so it will either be 1 or 0. Thus, the

sum of 18x + r(x) cannot be determined, but the first two terms of a(x) must be 3x3 3x

2. Substitute this

expression for a(x) and use long division to verify that the quotient is 3x.

The quotient is 3x. There is a remainder but it has no bearing on quotient, which is the oblique asymptote. Thus, a

function that has vertical asymptotes at x = 2 and x = 3 and an oblique asymptote y = 3x is f (x) = .

57.Writing in Math Use words, graphs, tables, and equations to show how to graph a rational function.


58.CHALLENGE Solve for k so that the rational equation has exactly one extraneous solution and one real solution.

= +

SOLUTION:

Let k = 3 so that the denominator factors to (x 1)(x 3).

The solutions are x = 3 and . Because the original equation is not defined when x = 3, this is the extraneous

solution. Thus, when k = 3, the rational equation has exactly one extraneous solution and one real solution.

59.Writing in Math Explain why all of the test intervals must be used in order to get an accurate graph of a rational function.

SOLUTION:Sample answer: The test intervals are used to determine the location of points on the graph. Because many rational functions are not continuous, one interval may include y-values that are vastly different than the next interval. Therefore, at least one, and preferably more than one, point is needed for every interval in order to sketch a reasonably accurate graph of the function.

List all the possible rational zeros of each function. Then determine which, if any, are zeros.

60.f (x) = x3 + 2x2 5x 6

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 6. Therefore, the possible rational zeros of g are 1, 2, 3, and 6. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 1) is a factor of f (x), we can use the quotient to write a factored form of f (x) as f (x) = (x + 1)(x2 + x

6). Factoring the quadratic expression yields f (x) = (x + 1)(x + 3)(x 2). Thus, the rational zeros of f are 1, 3, and 2.

61.f (x) = x3 2x2 + x + 18

SOLUTION:Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros of g are 1, 2, 3, 6, 9, and 18. By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x + 2) is a factor of f (x), we can use the quotient to write a factored form of f (x) as f (x) = (x + 2)(x2 4x

+ 9). The quadratic expression (x2 4x + 9) yields no rational zeros. Thus, the rational zero of f is 2.

62.f (x) = x4 5x3 + 9x2 7x + 2

SOLUTION:Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 2. Therefore, the possible rational zeros of g are 1 and 2. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x 1) and (x 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x 1)(x 2)(x2 2x + 1). Factoring the quadratic expression yields f (x) = (x 1)(x 2)(x 1)(x 1) or f (x) =

(x 1)3(x 2). Thus, the rational zeros of f are 1 (multiplicity: 3) and 2.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

63.f (x) = x4 2x3 13x2 + 14x + 24; x 3, x 2

SOLUTION:

Use synthetic division to test each factor, (x 3) and (x 2).

Because the remainder when f (x) is divided by (x 3) is 72, (x 3) is not a factor. Test the second factor, (x 2).

Because the remainder when f (x) is divided by (x 2) is 0, (x 2) is a factor. Because (x 2) is a factor of f (x), we can use the quotient of f (x)(x 2) to write a factored form of f (x) as f (x)

= (x 2)(x3 13x 12).

64.f (x) = 2x4 5x3 11x2 4x; x 4, 2x 1

SOLUTION:

Use synthetic division to test each factor, (x 4) and (2x 1).

Because the remainder when f (x) is divided by (x 4) is 0, (x 4) is a factor.

For (2x 1), use the depressed polynomial 2x3 + 3x

2 + x and rewrite the division expression so that the divisor is of

the form x c.

Because Setupthesyntheticdivisionasfollows.Thenfollowthesyntheticdivisionprocedure.

Because the remainder when the depressed polynomial is divided by (2x 1) is , (2x 1) is not a factor.

Because (x 4) is a factor of f (x), we can use the quotient of f (x)(x 4) to write a factored form of f (x) as f (x)

= (x 4)(2x3 + 3x2 + x). Factoring the cubic expression 2x3 + 3x2 + x results in x(2x2 + 3x + 1) or x(2x + 1)(x + 1).

Thus, f (x) = (x 4)(2x3 + 3x

2 + x) or f (x) = x(2x + 1)(x + 1)(x 4).

65.f (x) = 6x4 + 59x3 + 138x2 45x 50; 3x 2, x 5

SOLUTION:

Use synthetic division to test each factor, (3x 2) and (x 5). For (3x 2), rewrite the division expression so that the divisor is of the form x c.


Because the remainder when f (x) is divided by (3x 2) is 0, (3x 2) is a factor. Test the second factor, (x 5),

with the depressed polynomial 2x3 + 21x

2 + 60x + 25.

Because the remainder when the depressed polynomial is divided by (x 5) is 1100, (x 5) is not a factor of f (x). Because (3x 2) is a factor of f (x), we can use the quotient of f (x)(3x 2) to write a factored form of f (x) as f

(x) = (3x 2)(2x3 + 21x2 + 60x + 25).

Synthetic division can be used to factor 2x3 + 21x

2 + 60x + 25. By using synthetic division, it can be determined that

x = 5 is a rational zero.

The remaining quadratic expression (2x2 + 11x + 5) can be written as (x + 5)(2x + 1). Thus, f (x) = (3x 2) (2x3 +

21x2 + 60x + 25) or f (x) = (3x 2)(2x + 1)(x + 5)

2.

66.f (x) = 4x4 3x3 12x2 + 17x 6; 4x 3; x 1

SOLUTION:




with the depressed polynomial x3 3x + 2.

Because the remainder when the depressed polynomial is divided by (x 1) is 0, (x 1) is a factor of f (x). Because (4x 3) and (x 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f

(x) = (4x 3)(x 1)(x2 + x 2). Factoring the quadratic expression yields f (x) = (4x 3)(x 1)(x 1)(x + 2) or (4x

3)(x 1)2(x + 2).

67.f (x) = 4x5 + 15x4 + 12x3 4x2; x + 2, 4x + 1

SOLUTION:Use synthetic division to test each factor, (x + 2) and (4x + 1).

Because the remainder when f (x) is divided by (x + 2) is 0, (x + 2) is a factor of f (x).

For (4x + 1), use the depressed polynomial 4x4 + 7x

3 2x2 and rewrite the division expression so that the divisor is

of the form x c.


Because the remainder when the depressed polynomial is divided by (4x + 1) is , (4x + 1) is not a factor.

Because (x + 2) is a factor of f (x), we can use the quotient of f (x)(x + 2) to write a factored form of f (x) as f (x)

= (x + 2)(4x4 + 7x

3 2x2). The factor (4x4 + 7x3 2x2) can be written as x2(x + 2)(4x 1). Thus, f (x) = (x + 2)

(4x4 + 7x

3 2x

2) or f (x) = x

2(x + 2)

2(4x 1).

68.f (x) = 4x5 8x4 5x3 + 10x2 + x 2; x + 1, x 1

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x 1).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x 1), with

the depressed polynomial 4x4 12x

3 + 7x

2 + 3x 2.

Because the remainder when the depressed polynomial is divided by (x 1) is 0, (x 1) is a factor of f (x). Because (x + 1) and (x 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x 1)(4x3 8x2 x + 2). The factor (4x3 8x2 x + 2) can be written as (x 2)(2x + 1)(2x 1). Thus, f

(x) = (x + 1)(x 1) (4x3 8x

2 x + 2) or f (x) = (x 2)(2x 1)(2x+ 1)(x + 1)(x 1).

Graph each function.

69.f (x) = (x + 7)2

SOLUTION:

The graph of f (x) = (x + 7)2 is the graph of y = x

2 translated 7 units to the left.

70.f (x) = (x 4)3

SOLUTION:

The graph of f (x) = (x 4)3 is the graph of y = x

3 translated 4 units to the right.

71.f (x) = x4 5

SOLUTION:

This is an even-degree function, so its graph is similar to the graph of y = x2. The graph of f (x) = x

4 5 is the graph

of y = x4 translated 5 units down.

72.RETAIL Sara is shopping at a store that offers $10 cash back for every $50 spent. Let andh(x) =

10x, where x is the amount of money Sara spends. a. If Sara spends money at the store, is the cash back bonus represented by f [h(x)] or h[f (x)]? Explain your reasoning. b. Determine the cash back bonus if Sara spends $312.68 at the store.

SOLUTION:

a. representstheamountoftimes$50wasspent.h(x) = 10x represents the cash back bonus. The

cash back bonus is found after calculating the amount of times $50 was spent. So, the cash back bonus is represented by h[f (x)]. b.

The cash back bonus is $60.

73.INTERIOR DESIGN Adrienne Herr is an interior designer. She has been asked to locate an oriental rug for a new corporate office. The rug should cover half of the total floor area with a uniform width surrounding the rug.

a. If the dimensions of the room are 12 feet by 16 feet, write an equation to model the area of the rug in terms of x. b. Graph the related function. c. What are the dimensions of the rug?

SOLUTION:a. The area of the rug is given by A = wl. The width of the room is 12 feet. Since the rug will be x feet from the edge of the room, the width of the rug is (12 2x). The length of the room is 16 feet, so the length of the rug is (16 2x). The area of the rug is A = (12 2x)(16 2x). Since the rug is to cover half the area of the total floor, we know

that the area of the rug should be or96squarefeet.Thus,anequationtomodeltheareaoftheruginterms

of x is (12 2x)(16 2x) = 96. b. Write (12 2x)(16 2x) = 96 as a function of f (x).

Evaluate the equation for several x-values.

Use these points and several others to construct a graph.

c. Solve the expression found in part a.

x = 12 and x = 2. If x = 12, the rug would have dimensions with negative lengths. Thus, x = 2. The width of the rug is 12 2(2) or 8 feet and the length is 16 2(2) or 12 feet.

x 0 2 4 6 8 10 12 f (x) 96 0 64 96 96 64 0

Simplify.

74.i10 + i2

SOLUTION:

75.(2 + 3i) + (6 + i)

SOLUTION:

76.(2.3 + 4.1i) (1.2 6.3i)

SOLUTION:

77.SAT/ACT A company sells ground coffee in two sizes of cylindrical containers. The smaller container holds 10 ounces of coffee. If the larger container has twice the radius of the smaller container and 1.5 times the height, how

many ounces of coffee does the larger container hold? (The volume of a cylinder is given by the formula V = r2h.)

A 30 B 45 C 60 D 75 E 90

SOLUTION:

The volume of a cylinder is given by the formula V = r2h. Thus, the volume of the small container is 10 = r

2h.

The larger container has a radius of 2r and a height of 1.5h. Substituting these values in the formula for the volume

of a cylinder results in V = (2r)2(1.5h) or 6r2h. Substitute 10 = r2h into the formula for the volume of the larger cylinder.

The volume of the large container is 60 ounces. The correct answer is C.

78.What are the solutions of ?

F x = 1, x = 2 G x = 2, x = 1

H x = 1 + , x = 1

J x = , x =

SOLUTION:Solve for x.

Use the quadratic equation.

eSolutions Manual - Powered by Cognero Page 1

2-5 Rational Functions

Find the domain of each function and the equations of the vertical or horizontal asymptotes, if any.

1.f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 4. The real zeros of b(x) are 2 and 2.

Therefore, D = {x | x 2,2, x R}. Check for vertical asymptotes. Determine whether x = 2 is a point of infinite discontinuity. Find the limit as x approaches 2 from the left and the right.






x 1.9 1.99 1.999 2 2.001 2.01 2.1 f (x) 4.1 49.1 499.1 undef 500.9 50.9 5.9

x 2.1 2.01 2.001 2 1.999 1.99 1.9 f (x) 5.9 50.9 500.9 undef 499.1 49.1 4.1

x 100 50 10 0 10 50 100 f (x) 1.0002 1.0008 1.02 0.5 1.02 1.0008 1.0002

2.h(x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x + 4. The real zero of b(x) is 4. Therefore, D = {x | x 4, x R}. Check for vertical asymptotes.





x 4.1 4.01 4.001 4 3.999 3.99 3.9 f (x) 769.2 7248 72,048 undef 71,952 7152 673.2

x 100 50 10 0 10 50 100 f (x) 10417 2717 168 2 70.9 2314 9615

3.f (x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x 4). The real zeros of b(x) are 4 and3. Therefore, D = {x | x 4,3, x R}. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from the left and the right.






x 3.9 3.99 3.999 4 4.001 4.01 4.1 f (x) 570 6124 61667 undef 61762 6219 666

x 3.1 3.01 3.001 3 2.999 2.99 2.9 f (x) 21.7 175 1718 undef 1710 167 13.3

x 100 50 10 0 10 50 100 f (x) 9615 2314 71.8 0 166 2717 10417

4.g(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 3)(x + 5). The real zeros of b(x) are 3 and 5. Therefore, D = {x | x 3, 5, x R}. Check for vertical asymptotes.







x 5.1 5.01 5.001 5 4.999 4.99 4.9 f (x) 52.9 547 5498 undef 5502 552 57.4

x 3.1 3.01 3.001 3 2.999 2.99 2.9 f (x) 47.9 452 4503 undef 4497 447 42.4

x 100 50 10 0 10 50 100 f (x) 0.01 0.03 0.46 0.4 0.02 0.15 0.009

5.h(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = x2 + 2x. The real zeros of b(x) are 0 and 2.

Therefore, D = {x | x 0,2, x R}. Check for vertical asymptotes.







x 0.1 0.01 0.001 0 0.001 0.01 0.1 f (x) 7.47 52.3 502 undef 498 47.8 2.95

x 2.1 2.01 2.001 2 1.999 1.99 1.9 f (x) 86.8 852 8501 undef 8498 848 82.3

x 1000 100 10 0 10 100 1000 f (x) 2.008 2.08 3.01 undef 1.34 1.92 1.992

6.f (x) =

SOLUTION:

The function is undefined at the real zero of the denominator b(x) = x 4. The real zero of b(x) is 4. Therefore, D ={x | x 4,x R}. Check for vertical asymptotes. Determine whether x = 4 is a point of infinite discontinuity. Find the limit as x approaches 4 from the left and the right.




x 3.9 3.99 3.999 4 4.001 4.01 4.1 f (x) 703 7183 71,983 undef 72,017 7217 737

x 100 50 10 0 10 50 100 f (x) 87.7 38.33 2.14 5 35 64.6 113.8

7.h(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x 2)(x + 4). The real zeros of b(x) are 2 and4. Therefore, D = {x | x 2,4, x R}. Check for vertical asymptotes. Determine whether x = 2 is a point of infinite discontinuity. Find the limit as x approaches 2 from the left and the right.






x 1.9 1.99 1.999 2 2.001 2.01 2.1 f (x) 7.5 825 83250 undef 83417 842 9.16

x 4.1 4.01 4.001 4 3.999 3.99 3.9 f (x) 42.5 4175 416,750 undef 416,583 4158 40.8

x 100 50 10 0 10 50 100 f (x) 0.0004 0.0004 0.019 0.016 0.008 0.0004 0.00009

8.g(x) =

SOLUTION:

The function is undefined at the real zeros of the denominator b(x) = (x + 1)(x 3). The real zeros of b(x) are 1 and 3. Therefore, D = {x | x 3,1, x R}. Check for vertical asymptotes. Determine whether x = 3 is a point of infinite discontinuity. Find the limit as x approaches 3 from the left and the right.






x 2.9 2.99 2.999 3 3.001 3.01 3.1 f (x) 13.8 126 1251 undef 1249 123 11.2

x 1.1 1.01 1.001 1 0.999 0.99 0.9 f (x) 11.2 123 1249 undef 1251 126 13.8

x 100 50 10 0 10 50 100 f (x) 0.999 0.998 0.957 2.67 0.935 0.998 0.999

For each function, determine any asymptotes and intercepts. Then graph the function and state its domain.

9.f (x)=

SOLUTION:


The x-intercepts are 2 and 3, the zeros of the numerator. The y-intercept is becausef (0) = .


10.g(x) =

SOLUTION:


The x-intercepts are and6,thezerosofthenumerator.They-intercept is 9 because g(0) =9.


11.f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 2, 2, x R}. There are vertical asymptotes at x = 2 and x = 2. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is 2 because f (0)=2. Graph the asymptotes and intercept. Then find and plot points.

12.f (x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 0,6,x R}. There are vertical asymptotes at x = 0 and x = 6. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 2, the zero of the numerator. There is no y-intercept because f (x) is undefined for x = 0. Graph the asymptotes and intercept. Then find and plot points.

13.g(x) =

SOLUTION:

g(x) can be simplified to g(x) = .

The function is undefined at b(x) = 0, so D = {x | x5, 6, x R}. There are vertical asymptotes at x = 5 and x = 6, the real zeros of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 2, the zero of the simplified numerator. The y-intercept is because


14.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 2, 0, 5, x R}. There are vertical asymptotes at x = 2, x = 0 and x = 5. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercepts are 6 and 4, the zeros of the numerator. There is no y-intercept because h(x) is undefined for x = 0. Graph the asymptotes and intercepts. Then find and plot points.

15.h(x) =

SOLUTION:

h(x) can be written as h(x) = .


The x-intercepts are 5, 0, and 2, the zeros of the numerator. The y-intercept is 0 because h(0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

16.f (x) =

SOLUTION:



The x-intercepts are 6, 0, and 4, the zeros of the numerator. The y-intercept is 0 because f (0) = 0. Graph the asymptotes and intercepts. Then find and plot points.

17.f (x) =

SOLUTION:

The function is undefined at b(x) = 0. x2 + 4x + 5 yields no real zeros, so D = {x | x R}.


The x-intercept is 8, the zero of the numerator. The y-intercept is because

Graph the asymptote and intercepts. Then find and plot points.

18.

SOLUTION:

The function is undefined at b(x) = 0. x2 + 6 yields no real zeros, so D = {x | x R}.


Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is because

Graph the asymptote and intercept. Then find and plot points.

19.SALESThebusinessplanforanewcarwashprojectsthatprofitsinthousandsofdollarswillbemodeledbythe

function p (z) = , where z is the week of operation and z = 0 represents opening.

a. State the domain of the function. b. Determine any vertical and horizontal asymptotes and intercepts for p (z). c. Graph the function.

SOLUTION:

a. The function is undefined at b(x) = 0. 2z2 + 7z + 5 can be written as (2z + 5)(z+ 1). The zeros of this polynomial

are at 1 and . Since the car wash cannot be open for negative weeks, these zeros do not fall in the domain and

D = {z | z 0,z R}.

b. Though there are vertical asymptotes at x = 1 and , these values of x are not in the domain. Therefore,

there are no vertical asymptotes for this situation.

There is a horizontal asymptote at y = , the ratio of the leading coefficients of the numerator and denominator,

because the degrees of the polynomials are equal.


c. Find and plot points to construct a graph. Since the domain is restricted to real numbers greater than or equal to 0, it is only necessary to show the first and fourth quadrants.

For each function, determine any asymptotes, holes, and intercepts. Then graph the function and state itsdomain.

20.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 0,x R}. There is a vertical asymptote at x = 0. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is , the zero of the numerator. There is no y-intercept because h(x) is undefined for x = 0.


21.h(x) =

SOLUTION:The function is undefined at b(x) = 0.

Thus, .

There is a vertical asymptote at .

There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

Since the polynomial in the numerator has no real zeros, there are no x-intercepts. The y-intercept is becauseh

(0) = .


22.f (x) =

SOLUTION:



The x-intercepts are 5 and 3, the zeros of the numerator. The y-intercept is 5 because f (0) =5. Graph the asymptotes and intercepts. Then find and plot points.

23.g(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 4,x R}. There is a vertical asymptote at x = 4. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.



24.h(x) =

SOLUTION:

The function is undefined at b(x) = 0, so D = {x | x 3, x R}. There is a vertical asymptote at x = 3. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote. The x-intercept is 0, the zero of the numerator. The y-intercept is 0 because h(0) = 0. Graph the asymptote and intercepts. Then find and plot points.

25.g(x) =

SOLUTION:


The function is undefined at b(x) = 0, so D = {x | x 4,x R}. There is a vertical asymptote at x = 4. There is no horizontal asymptote since the degree of the numerator is greater than the degree of the denominator. Since the degree of the numerator is 2 greater than the denominator, there is no oblique asymptote.

The x-intercepts are 2, 1, and 0, the zeros of the numerator. The y-intercept is 0 because g(0) = 0. Graph the asymptote and intercepts. Then find and plot points.

26.f (x) =

SOLUTION:

f(x) can be written as f (x) = or .

The function is undefined at b(x) = 0, so D = {x | x 3, 1, 2, x R}. There are vertical asymptotes at x = 1 and x = 2, the real zeros of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is 7, the zero of the simplified numerator. The y-intercept is becauseh(0) = .

There is a hole at (3, 1) because the original function is undefined when x = 3. Graph the asymptotes and intercepts. Then find and plot points.

27.g(x) =

SOLUTION:

g(x) can be written as g(x) = or .

The function is undefined at b(x) = 0, so D = {x | x 2, 1, or 2, x R}. There is a vertical asymptote at x = 1, the real zero of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator. Since the simplified numerator has no real zeros, there are no x-intercepts. The y-intercept is 1 because g(0) =1.

There are holes at (2, 1) and becausetheoriginalfunctionisundefinedwhenx = 2 and x = 2. Graph the

asymptotes and intercept. Then find and plot points.

28.f (x) =

SOLUTION:

f(x) can be written as f (x) =

The function is undefined at b(x) = 0, so D = {x | x 3, 1, x R}. There is a vertical asymptote at x = 3, the real zero of the simplified denominator. There is a horizontal asymptote at y = 1, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the polynomials are equal.

The x-intercept is 4, the zero of the simplified numerator. The y-intercept is becausef (0) = .

There is a hole at becausetheoriginalfunctionisundefinedwhenx = 1.


29.g(x) =

SOLUTION:


The function is undefined at b(x) = 0, so D = {x | x 4, 5, x R}. There is a vertical asymptote at x = 4, the real zero of the simplified denominator. There is a horizontal asymptote at y = 0, because the degree of the denominator is greater than the degree of the numerator.

The x-intercept is , the zero of the simplified numerator.

The y-intercept is becauseg(0) = .

There is a hole at becausetheoriginalfunctionisundefinedwhenx = 5. Graph the asymptotes and

intercepts. Then find and plot points.

30.STATISTICSAnumberx is said to be the harmonic mean of y and z if istheaverageof and .

a. Write an equation for which the solution is the harmonic mean of 30 and 45. b. Find the harmonic mean of 30 and 45.

SOLUTION:

a. Write an equation that represents astheaverageof and .

=

Substitute y = 30 and z = 45.

b. Solve for x for the equation found in part a.

The harmonic mean of 30 and 45 is 36.

31.OPTICSThelensequationis = + , where f is the focal length, di is the distance from the lens to the

image, and do is the distance from the lens to the object. Suppose the object is 32 centimeters from the lens and the

focal length is 8 centimeters.

a. Write a rational equation to model the situation. b. Find the distance from the lens to the image.

SOLUTION:a. Substitute f = 8 and do = 32 into the equation.

b. Solve for di for the equation found in part a.

The distance from the lens to the image is 10 centimeters.

Solve each equation.

32.y + =5

SOLUTION:

y = 3 or y = 2

33. z = 4

SOLUTION:

34. + =1

SOLUTION:

x = 1 or x = 8

35.

SOLUTION:

Because the original equation is not defined when y = 2, you can eliminate this extraneous solution. So,

hasnosolution.

36. + =

SOLUTION:

37.

SOLUTION:

38.

SOLUTION:

x = 4 or x = 5

39.

SOLUTION:

40. + =3

SOLUTION:

x = 7 or x = 1

41.

SOLUTION:

42.WATERThecostperdaytoremovex percent of the salt from seawater at a desalination plant is c(x) = ,

where 0 x < 100. a. Graph the function using a graphing calculator. b. Graph the line y = 8000 and find the intersection with the graph of c(x) to determine what percent of salt can be removed for $8000 per day. c. According to the model, is it feasible for the plant to remove 100% of the salt? Explain your reasoning.

SOLUTION:a.

b. Use the CALC menu to find the intersection of c(x) and the line y = 8000.

For $8000 per day, about 88.9% of salt can be removed. c. No; sample answer: The function is not defined when x = 100. This suggests that it is not financially feasible to remove 100% of the salt at the plant.

Write a rational function for each set of characteristics.43.x-intercepts at x = 0 and x = 4, vertical asymptotes at x = 1 and x = 6, and a horizontal asymptote at y = 0

SOLUTION:Sample answer: For the function to have x-intercepts at x = 0 and x = 4, the zeros of the numerator need to be 0 and

4.Thus, factors of the numerator are x and (x 4). For there to be vertical asymptotes at x = 1 and x = 6, the zeros of the denominator need to be 1 and 6. Thus, factors of the denominator are (x 1) and (x 6). For there to be a horizontal asymptote at y = 0, the degree of the denominator needs to be greater than the degree of the numerator.

Raising the factor (x 6) to the second power will result in an asymptote at y = 0.

f(x) =

44.x-intercepts at x = 2 and x = 3, vertical asymptote at x = 4, and point discontinuity at (5, 0)

SOLUTION:

Sample answer: For the function to have x-intercepts at x = 2 and x = 3, the zeros of the numerator need to be 2 and 3.Thus, factors of the numerator are (x 2) and (x + 3). For there to be a vertical asymptote at x = 4, a zero of the denominator needs to be 4. Thus, a factor of the denominator is (x 4). For there to be point discontinuity at x= 5, (x + 5) must be a factor of both the numerator and denominator. Additionally, for the point discontinuity to be at (5, 0), the numerator must have a zero at x = 5. Thus, the (x + 5) in the numerator must have a power of 2 so that x = 5 remains a zero of the numerator after the function is simplified.

f(x) =

45.TRAVELWhen distance and time are held constant, the average rates, in miles per hour, during a round trip can

be modeled by , where r1 represents the average rate during the first leg of the trip and r2 represents the

average rate during the return trip. a. Find the vertical and horizontal asymptotes of the function, if any. Verify your answer graphically. b. Copy and complete the table shown.

c. Is a domain of r1 > 30 reasonable for this situation? Explain your reasoning.

SOLUTION:a. There is a vertical asymptote at the real zero of the denominator r1 = 30. There is a horizontal asymptote at r2 =

orr2 = 30, the ratio of the leading coefficients of the numerator and denominator, because the degrees of the

polynomials are equal.

b. Evaluate the function for each value of r1 to determine r2.

c. No; sample answer: As r1 approaches infinity, r2 approaches 30. This suggests that the average speeds reached

during the first leg of the trip have no bounds. Being able to reach an infinite speed is not reasonable. The same

holds true about r2 as r1 approaches 30 from the right.

Use your knowledge of asymptotes and the provided points to express the function represented by each graph.

46.

SOLUTION:



f(x) = .




47.

SOLUTION:



f(x) = .




Use the intersection feature of a graphing calculator to solve each equation.

48. = 8

SOLUTION:


graphs.


about 10.05.

49. = 1

SOLUTION:


graphs.


about 2.65.

50. = 2

SOLUTION:


graphs.


about 0.90.

51. = 6

SOLUTION:

Graph y = andy = 6 using a graphing calculator. Use the CALC menu to find the intersection of

the graphs.

The intersections occur at x 3.87, x1.20,andx3.70.Thus,thesolutionsto =6areabout

3.87, about 1.20, and about 3.70.

52.CHEMISTRYWhena60%aceticacidsolutionisaddedto10litersofa20%aceticacidsolutionina100-liter tank, the concentration of the total solution changes.

a. Show that the concentration of the solution is f (a) = , where a is the volume of the 60% solution.

b. Find the relevant domain of f (a) and the vertical or horizontal asymptotes, if any. c. Explain the significance of any domain restrictions or asymptotes. d. Disregarding domain restrictions, are there any additional asymptotes of the function? Explain.

SOLUTION:a. Sample answer: The concentration of the total solution is the sum of the amount of acetic acid in the original 10

liters and the amount in the a liters of the 60% solution, divided by the total amount of solution or .

Multiplying both the numerator and the denominator by 5 gives or .

b. Since a cannot be negative and the maximum capacity of the tank is 100 liters, the relevant domain of a is 0 a 90. There is a horizontal asymptote at y = 0.6, the ratio of the leading coefficients of the numerator and denominator,because the degrees of the polynomials are equal. c. Sample answer: Because the tank already has 10 liters of solution in it and it will only hold a total of 100 liters, the amount of solution added must be less than or equal to 90 liters. It is also impossible to add negative amounts of solution, so the amount added must be greater than or equal to 0. As you add more of the 60% solution, the concentration of the total solution will get closer to 60%, but because the solution already in the tank has a lower concentration, the concentration of the total solution can never reach 60%. Therefore, there is a horizontal asymptote at y = 0.6.

d. Yes; sample answer: The function is not defined at a = 10, but because the value is not in the relevant domain, the asymptote does not pertain to the function. If there were no domain restrictions, there would be a vertical

asymptote at a = 10.

53.MULTIPLEREPRESENTATIONS In this problem, you will investigate asymptotes of rational functions. a. TABULARCopyandcompletethetablebelow.Determinethehorizontalasymptoteofeachfunctionalgebraically.

b. GRAPHICAL Graph each function and its horizontal asymptote from part a. c. TABULAR Copy and complete the table below. Use the Rational Zero Function to help you find the real zeros of the numerator of each function.

d. VERBAL Make a conjecture about the behavior of the graph of a rational function when the degree of the denominator is greater than the degree of the numerator and the numerator has at least one real zero.

SOLUTION:a. Since the degree of the denominator is greater than the degree of the numerator, f (x), h(x), and g(x) will have horizontal asymptotes at y = 0.

b. Use a graphing calculator to graph f (x), h(x), and g(x).

c. c. To find the real zeros of the numerator of f (x), use the Rational Zero Theorem and factoring. Because the leading coefficient is 1, the possible rational zeros are the integer

factors of the constant term 4. Therefore, the possible rational zeros of f are 1, 2, and 4. Factoring x2 5x + 4 results in (x 4)(x 1). Thus, the real zeros of the numerator of f (x) are 4 and 1. To find the real zeros of the numerator of h(x), use the Rational Zero Theorem and synthetic division. Because the

leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 12. Therefore, the possible rational zeros of f are 1, 2, 3, 4, 6, and 12. By using synthetic division, it can be determined that x = 3 is a rational zero.

The remaing quadratic factor (x2 + 4) yields no real zeros. Thus, the real zero of the numerator of h(x) is 3.

To find the real zeros of the numerator of g(x), use the Rational Zero Theorem and factoring. Because the leading

coefficient is 1, the possible rational zeros are the integer factors of the constant term 1. Therefore, the possible

rational zeros of f are 1. Factoring x4 1 results in (x2 1)(x2 + 1) or (x + 1)(x 1)(x2 + 1). The quadratic factor

(x2 + 1) yields no real zeros. Thus, the real zeros of the numerator of g(x) are 1 and 1.

d. Sample answer: When the degree of the numerator is less than the degree of the denominator and the numerator has at least on real zero, the graph of the function will have y = 0 as an asymptote and will intersect the asymptote atthe real zeros of the numerator.

54.REASONING Given f (x) = , will f (x) sometimes, always, or never have a horizontal asymptote at

y = 1 if a, b, c, d, e, and f are constants with a0andd0?Explain.

SOLUTION:Sometimes; sample answer: When a = d, the function will have a horizontal asymptote at y = 1. When ad, the function will not have a horizontal asymptote at y = 1.

55.PREWRITE Design a lesson plan to teach the graphing rational functions topics in this lesson. Make a plan that addresses purpose, audience, a controlling idea, logical sequence, and time frame for completion.


56.CHALLENGE Write a rational function that has vertical asymptotes at x = 2 and x = 3 and an oblique asymptotey = 3x.

SOLUTION:

Sample answer: Since there are vertical asymptotes at x = 2 and x = 3, (x + 2) and (x 3) are factors of the denominator. Since there is an oblique asymptote y = 3x, the degree of the numerator is exactly 1 greater than the degree of the denominator. Additionally, when the numerator is divided by the denominator, the quotient polynomial q

(x) is q(x) = 3x. So, the function can be written as f (x) = , where a(x) is the numerator of the

function. We know that =3x + , where r(x) is the remainder. We can write the

denominator of f (x) as x2 x 6 and can use the equation to solve for a(x).

The degree of the remainder has to be less than the degree of the denominator, so it will either be 1 or 0. Thus, the

sum of 18x + r(x) cannot be determined, but the first two terms of a(x) must be 3x3 3x

2. Substitute this

expression for a(x) and use long division to verify that the quotient is 3x.

The quotient is 3x. There is a remainder but it has no bearing on quotient, which is the oblique asymptote. Thus, a

function that has vertical asymptotes at x = 2 and x = 3 and an oblique asymptote y = 3x is f (x) = .

57.Writing in Math Use words, graphs, tables, and equations to show how to graph a rational function.


58.CHALLENGE Solve for k so that the rational equation has exactly one extraneous solution and one real solution.

= +

SOLUTION:

Let k = 3 so that the denominator factors to (x 1)(x 3).

The solutions are x = 3 and . Because the original equation is not defined when x = 3, this is the extraneous

solution. Thus, when k = 3, the rational equation has exactly one extraneous solution and one real solution.

59.Writing in Math Explain why all of the test intervals must be used in order to get an accurate graph of a rational function.

SOLUTION:Sample answer: The test intervals are used to determine the location of points on the graph. Because many rational functions are not continuous, one interval may include y-values that are vastly different than the next interval. Therefore, at least one, and preferably more than one, point is needed for every interval in order to sketch a reasonably accurate graph of the function.

List all the possible rational zeros of each function. Then determine which, if any, are zeros.

60.f (x) = x3 + 2x2 5x 6

SOLUTION:

Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 6. Therefore, the possible rational zeros of g are 1, 2, 3, and 6. By using synthetic division, it can be determined that x = 1 is a rational zero.

Because (x + 1) is a factor of f (x), we can use the quotient to write a factored form of f (x) as f (x) = (x + 1)(x2 + x

6). Factoring the quadratic expression yields f (x) = (x + 1)(x + 3)(x 2). Thus, the rational zeros of f are 1, 3, and 2.

61.f (x) = x3 2x2 + x + 18

SOLUTION:Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 18. Therefore, the possible rational zeros of g are 1, 2, 3, 6, 9, and 18. By using synthetic division, it can be determined that x = 2 is a rational zero.

Because (x + 2) is a factor of f (x), we can use the quotient to write a factored form of f (x) as f (x) = (x + 2)(x2 4x

+ 9). The quadratic expression (x2 4x + 9) yields no rational zeros. Thus, the rational zero of f is 2.

62.f (x) = x4 5x3 + 9x2 7x + 2

SOLUTION:Because the leading coefficient is 1, the possible rational zeros are the integer factors of the constant term 2. Therefore, the possible rational zeros of g are 1 and 2. By using synthetic division, it can be determined that x = 1 is a rational zero.

By using synthetic division on the depressed polynomial, it can be determined that x = 2 is a rational zero.

Because (x 1) and (x 2) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x 1)(x 2)(x2 2x + 1). Factoring the quadratic expression yields f (x) = (x 1)(x 2)(x 1)(x 1) or f (x) =

(x 1)3(x 2). Thus, the rational zeros of f are 1 (multiplicity: 3) and 2.

Use the Factor Theorem to determine if the binomials given are factors of f (x). Use the binomials that are factors to write a factored form of f (x).

63.f (x) = x4 2x3 13x2 + 14x + 24; x 3, x 2

SOLUTION:

Use synthetic division to test each factor, (x 3) and (x 2).

Because the remainder when f (x) is divided by (x 3) is 72, (x 3) is not a factor. Test the second factor, (x 2).

Because the remainder when f (x) is divided by (x 2) is 0, (x 2) is a factor. Because (x 2) is a factor of f (x), we can use the quotient of f (x)(x 2) to write a factored form of f (x) as f (x)

= (x 2)(x3 13x 12).

64.f (x) = 2x4 5x3 11x2 4x; x 4, 2x 1

SOLUTION:

Use synthetic division to test each factor, (x 4) and (2x 1).

Because the remainder when f (x) is divided by (x 4) is 0, (x 4) is a factor.

For (2x 1), use the depressed polynomial 2x3 + 3x

2 + x and rewrite the division expression so that the divisor is of

the form x c.


Because the remainder when the depressed polynomial is divided by (2x 1) is , (2x 1) is not a factor.

Because (x 4) is a factor of f (x), we can use the quotient of f (x)(x 4) to write a factored form of f (x) as f (x)

= (x 4)(2x3 + 3x2 + x). Factoring the cubic expression 2x3 + 3x2 + x results in x(2x2 + 3x + 1) or x(2x + 1)(x + 1).

Thus, f (x) = (x 4)(2x3 + 3x

2 + x) or f (x) = x(2x + 1)(x + 1)(x 4).

65.f (x) = 6x4 + 59x3 + 138x2 45x 50; 3x 2, x 5

SOLUTION:




with the depressed polynomial 2x3 + 21x

2 + 60x + 25.

Because the remainder when the depressed polynomial is divided by (x 5) is 1100, (x 5) is not a factor of f (x). Because (3x 2) is a factor of f (x), we can use the quotient of f (x)(3x 2) to write a factored form of f (x) as f

(x) = (3x 2)(2x3 + 21x2 + 60x + 25).

Synthetic division can be used to factor 2x3 + 21x

2 + 60x + 25. By using synthetic division, it can be determined that

x = 5 is a rational zero.

The remaining quadratic expression (2x2 + 11x + 5) can be written as (x + 5)(2x + 1). Thus, f (x) = (3x 2) (2x3 +

21x2 + 60x + 25) or f (x) = (3x 2)(2x + 1)(x + 5)

2.

66.f (x) = 4x4 3x3 12x2 + 17x 6; 4x 3; x 1

SOLUTION:




with the depressed polynomial x3 3x + 2.

Because the remainder when the depressed polynomial is divided by (x 1) is 0, (x 1) is a factor of f (x). Because (4x 3) and (x 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f

(x) = (4x 3)(x 1)(x2 + x 2). Factoring the quadratic expression yields f (x) = (4x 3)(x 1)(x 1)(x + 2) or (4x

3)(x 1)2(x + 2).

67.f (x) = 4x5 + 15x4 + 12x3 4x2; x + 2, 4x + 1

SOLUTION:Use synthetic division to test each factor, (x + 2) and (4x + 1).

Because the remainder when f (x) is divided by (x + 2) is 0, (x + 2) is a factor of f (x).

For (4x + 1), use the depressed polynomial 4x4 + 7x

3 2x2 and rewrite the division expression so that the divisor is

of the form x c.


Because the remainder when the depressed polynomial is divided by (4x + 1) is , (4x + 1) is not a factor.

Because (x + 2) is a factor of f (x), we can use the quotient of f (x)(x + 2) to write a factored form of f (x) as f (x)

= (x + 2)(4x4 + 7x

3 2x2). The factor (4x4 + 7x3 2x2) can be written as x2(x + 2)(4x 1). Thus, f (x) = (x + 2)

(4x4 + 7x

3 2x

2) or f (x) = x

2(x + 2)

2(4x 1).

68.f (x) = 4x5 8x4 5x3 + 10x2 + x 2; x + 1, x 1

SOLUTION:

Use synthetic division to test each factor, (x + 1) and (x 1).

Because the remainder when f (x) is divided by (x + 1) is 0, (x + 1) is a factor. Test the second factor, (x 1), with

the depressed polynomial 4x4 12x

3 + 7x

2 + 3x 2.

Because the remainder when the depressed polynomial is divided by (x 1) is 0, (x 1) is a factor of f (x). Because (x + 1) and (x 1) are factors of f (x), we can use the final quotient to write a factored form of f (x) as f (x)

= (x + 1)(x 1)(4x3 8x2 x + 2). The factor (4x3 8x2 x + 2) can be written as (x 2)(2x + 1)(2x 1). Thus, f

(x) = (x + 1)(x 1) (4x3 8x

2 x + 2) or f (x) = (x 2)(2x 1)(2x+ 1)(x + 1)(x 1).

Graph each function.

69.f (x) = (x + 7)2

SOLUTION:

The graph of f (x) = (x + 7)2 is the graph of y = x

2 translated 7 units to the left.

70.f (x) = (x 4)3

SOLUTION:

The graph of f (x) = (x 4)3 is the graph of y = x

3 translated 4 units to the right.

71.f (x) = x4 5

SOLUTION:

This is an even-degree function, so its graph is similar to the graph of y = x2. The graph of f (x) = x

4 5 is the graph

of y = x4 translated 5 units down.

72.RETAIL Sara is shopping at a store that offers $10 cash back for every $50 spent. Let andh(x) =

10x, where x is the amount of money Sara spends. a. If Sara spends money at the store, is the cash back bonus represented by f [h(x)] or h[f (x)]? Explain your reasoning. b. Determine the cash back bonus if Sara spends $312.68 at the store.

SOLUTION:

a. representstheamountoftimes$50wasspent.h(x) = 10x represents the cash back bonus. The

cash back bonus is found after calculating the amount of times $50 was spent. So, the cash back bonus is represented by h[f (x)]. b.

The cash back bonus is $60.

73.INTERIOR DESIGN Adrienne Herr is an interior designer. She has been asked to locate an oriental rug for a ne

2-5 rational functions - montville township public schools · find the domain of each function and...

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