Download Lesson 2.5 (19-52 Custom) - MOC-FV

2-5 Rational Functions
19. SALES The business plan for a new car wash projects that profits in thousands of dollars will be modeled by the 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: 2
a. The function is undefined at b(x) = 0. 2z + 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.
31. OPTICS The lens equation is = + , where f is the focal length, d i is the distance from the lens to the
image, and d o 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
d o = 32
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into the equation.
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2-5 Rational Functions
31. OPTICS The lens equation is = + , where f is the focal length, d i is the distance from the lens to the
image, and d o 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 d o = 32 into the equation.
b. Solve for d i for the equation found in part a.
The distance from the lens to the image is 10
centimeters.
Solve each equation.
33. −z = 4
SOLUTION: eSolutions Manual - Powered by Cognero
35. Page 2
2-5 The
Rational
distance Functions
from the lens to the image is 10
centimeters.
Solve each equation.
33. −z = 4
SOLUTION: 35. SOLUTION: Because the original equation is not defined when y = 2, you can eliminate this extraneous solution. So,
has no solution.
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Because the original equation is not defined when y = 2, you can eliminate this extraneous solution. So,
has no solution.
2-5 Rational Functions
37. SOLUTION: 39. SOLUTION: 41. SOLUTION: eSolutions Manual - Powered by Cognero
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2-5 Rational Functions
41. SOLUTION: 42. WATER The cost per day to remove x 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.
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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
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2-5 Rational Functions
42. WATER The cost per day to remove x 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 Page
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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.
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) =
2-5 Rational Functions
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. TRAVEL When 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 =
or r2 = 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.
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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.
=
2-5 f(x)
Rational
Functions
45. TRAVEL When 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 =
or r2 = 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.
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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
2-5 holds
Rational
Functions
true about
r as r approaches 30 from the right.
2
1
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) =
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47. .
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function that can represent the graph is f (x) =
2-5 A
Rational
Functions
.
47. SOLUTION: Sample answer: Since there are vertical asymptotes at x = −4 and x = 3, (x + 4) and (x − 3) are factors of the
denominator. Since x = 5 is a x-intercept, (x − 5) is a factor of the numerator. Because there is a horizontal
asymptote at y = 2, the degrees of the polynomials in the numerator and denominator are equal and the ratio of their
leading coefficients is 2. 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 (−1, −3) into the equation and solve for a.
Substitute a = 2 into the original equation.
A function that can represent the graph is f (x) =
.
52. CHEMISTRY When a 60% acetic acid solution is added to 10 liters of a 20% acetic acid solution in a 100-liter
tank, the concentration of the total solution changes.
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2-5 A
Rational
Functions
function that
can represent the graph is f (x) =
.
52. CHEMISTRY When a 60% acetic acid solution is added to 10 liters of a 20% acetic acid solution in a 100-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.
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