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 | x ≠ –5, 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 Graph the asymptotes and intercepts. Then find and plot points.
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 function is undefined at b(x) = 0, so D = {x | x –3, 8, x R}.
There are vertical asymptotes at x = 8 and 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-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}.
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.
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.
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:
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 its domain.
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 because h (0) = .
Graph the asymptotes and intercept. Then find and plot points.
22.
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 are vertical asymptotes at x =−3 and x =−1.
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−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.
The x-intercept is−7, the zero of the numerator. The y-intercept is because . Graph the asymptotes and intercepts. Then find and plot points.
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 because h(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 because the original function is undefined when x =−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 because f (0) = . There is a hole at because the original function is undefined when x = 1.
Graph the asymptotes and intercepts. Then find and plot points.
29.
g(x) =SOLUTION:
g(x) can be written as g(x) = .
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 because g(0) = .
There is a hole at because the original function is undefined when x = 5. Graph the asymptotes and intercepts. Then find and plot points.
30.
STATISTICS A number x is said to be the harmonic mean of y and z if is the average of 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 as the average of 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.
OPTICS The lens equation is = + , 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 + = 5SOLUTION:
y= 3 or y = 2
33.
−z = 4SOLUTION:
34.
+ = 1SOLUTION:
x= 1 or x = 8
35.
SOLUTION:
Because the original equation is not defined when y = 2, you can eliminate this extraneous solution. So, has no solution.
36.
+ =SOLUTION:
37.
SOLUTION:
38.
SOLUTION:
x= 4 or x =−5
39.
SOLUTION:
40.
+ = 3SOLUTION:
x= 7 or x = 1
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.
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 = 0SOLUTION:
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.
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.
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:
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) = .
Use the intersection feature of a graphing calculator to solve each equation.
48.
= 8SOLUTION:
Graph y = and y = 8 using a graphing calculator. Use the CALC menu to find the intersection of the graphs.
The intersections occur at x ≈−1.59 and x ≈ 10.05. Thus, the solutions to = 8 are about−1.59 and about 10.05.
49.
= 1SOLUTION:
Graph y = and y = 1 using a graphing calculator. Use the CALC menu to find the intersection of the graphs.
The intersections occur at x ≈−2.65 and x ≈ 2.56. Thus, the solutions to = 1 are about−2.65 and about 2.65.
50.
= 2SOLUTION:
Graph y = and y = 2 using a graphing calculator. Use the CALC menu to find the intersection of the graphs.
The intersections occur at x ≈−0.98 and x ≈ 0.90. Thus, the solutions to = 2 are about−0.98 and about 0.90.
51.
= 6SOLUTION:
Graph y = and y = 6 using a graphing calculator. Use the CALC menu to find the intersection of the graphs.
The intersections occur at x ≈−3.87, x ≈ 1.20, and x ≈ 3.70. Thus, the solutions to = 6 are about
−3.87, about 1.20, and about 3.70.
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.
53.
MULTIPLE REPRESENTATIONS In this problem, you will investigate asymptotes of rational functions.a. TABULAR Copy and complete the table below. Determine the horizontal asymptote of each function algebraically.
b. GRAPHICALGraph each function and its horizontal asymptote from part a.
c. TABULARCopy and complete the table below. Use the Rational Zero Function to help you find the real zeros of the numerator of each function.
d. VERBALMake 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 at the 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 a ≠ 0 and d ≠ 0? Explain.SOLUTION:
Sometimes; sample answer: When a = d, the function will have a horizontal asymptote at y = 1. When a ≠ d, the function will not have a horizontal asymptote at y = 1.