Gustafson and Hughes - College Algebra 13/e

1. 1/1 points | Previous Answers GHColAlg13 3.4.001. My Notes

This exercise addresses readiness gaps.
Getting Ready exercises appear at the beginning of each exercise set and ensure students have the skills necessary to successfully complete the practice exercises.

Add and simplify the following.

(−8x² + 4x + 1) + (6x² − 8x − 6)

−2x2−4x−5

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2. 2/3 points | Previous Answers GHColAlg13 2.4.040. My Notes

This exercise addresses readiness gaps.
Learn It modules address your students' knowledge gaps with just-in-time instruction. Learn Its provide targeted instruction and practice on that topic using narrative, videos, and tutorialsâall in one place. If the topic is still too challenging, students can choose to continue learning through associated prerequisite Learn Its.

Find the slope and y-intercept and then use them to draw the graph of the line.

y = 2x − 2

slope

y-intercept

(x, y)

=

0,−2

0,0

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3. 0/1 points | Previous Answers WACollMathBootcamp1 11.4.001. My Notes

This exercise addresses readiness gaps.
The College Math Readiness Bootcamp contains a formative student assessment on prerequisite topics needed for success and targeted learning modules for areas where students struggled.

Factor. (If the expression is not factorable using integers, enter PRIME.)

48a⁴ + 8a³ − 320a²

45

Suggested tutorial: Learn It: Factor polynomials completely.

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4. 1/1 points | Previous Answers GHColAlg13 4.5.017.MI. My Notes

This exercise builds problem-solving skills.
Master It tutorials are an optional student-help tool available within select questions for just-in-time support. Students can use the tutorial to guide them through the problem-solving process step-by-step using different numbers.

Find all rational zeros of the polynomial function. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)

P(x) = x³ − 5x² − x + 5

x =

5, 1, −1

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5. 5/5 points | Previous Answers GHColAlg13 4.6.032.MI.SA. My Notes

This exercise builds problem-solving skills.
Stand-Alone Master It Tutorials are embedded, step-by-step tutorials used to help students understand each step required to solve the problem, before inputting their final answer.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Find the domain of the rational function. Do not graph the function.

x + 1

5x² − 24x + 16

Step 1

We can factor the denominator to see what values of x will make the denominator equal to 0. These values are not

are not

in the domain.

Factor the denominator.

f(x)

=

x + 1

5x² − 24x + 16

=

x + 1

$$5x−4

(x − 4)

Step 2

Set each factor in the denominator equal to 0 and solve for x.

5x − 4	=	0	or	x − 4	=	0

x	=	4/5 4/5	or	x	=	4 4

Step 3

Since

4

5

and 4 make the denominator 0, the domain is the set of all real numbers except these values.

We have this domain. (Enter your answer using interval notation.)

$$(−∞,45)∪(45,4)∪(4,∞)

$webMathematica generated answer key$

You have now completed the Master It.

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6. –/5 points GHColAlg13 5.5.104.FI. My Notes

This exercise builds problem-solving skills.
Fix It exercises deepen problem-solving skills by demonstrating a common mistake and asking students to identify the error, "fix it," and then correctly work the problem utilizing problem-solving strategies.

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

The following solution is incorrect. You will be asked to determine which steps are incorrect and fix the solution.

Write the following logarithmic expression as a single logarithm.

9 ln(x) − 2 ln(7) −

1

4

ln(y)

STEP 1:

=

ln(x⁹) − ln(7²) − ln(y^1⁄4)

STEP 2:

=

ln(x⁹) − ln(49) − ln(

4

y

)

STEP 3:

=

ln(x⁹ − 49) − ln(

4

y

)

STEP 4:

=

ln(x⁹ − 49 −

4

y

)

Part 1

Write the following logarithmic expression as a single logarithm.

9 ln(x) − 2 ln(7) −

1

4

ln(y)

STEP 1:

=

ln(x⁹) − ln(7²) − ln(y^1⁄4)

STEP 2:

=

ln(x⁹) − ln(49) − ln(

4

y

)

STEP 3:

=

ln(x⁹ − 49) − ln(

4

y

)

STEP 4:

=

ln(x⁹ − 49 −

4

y

)

For the above worked-out solution, choose the step in which the first error is made.

STEP 1 STEP 2 correct

STEP 3 STEP 4

The first error was made in Step 3, so Steps 3 and 4 are incorrect.

9 ln(x) − 2 ln(7) −

1

4

ln(y)

STEP 1:

=

ln(x⁹) − ln(7²) − ln(y^1⁄4)

STEP 2:

=

ln(x⁹) − ln(49) − ln(

4

y

)

STEP 3:

=

ln(x⁹ − 49) − ln(

4

y

)

STEP 4:

=

ln(x⁹ − 49 −

4

y

)

Select the error that is made in Step 3 of this solution.

When rewriting

ln(x⁹) − ln(49)

as a single logarithm, the arguments

x⁹ and 49 were .

The arguments should have been instead.

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7. –/9 points GHColAlg13 5.GA.001. My Notes

This exercise builds problem-solving skills.
Group Activities at the end of each chapter enable students to explore topics in greater depth and apply the mathematics to real life. These activities and applications can also be assigned individually.

Group Activity: Installment Loans

What Are Fixed Installment Loans?

A fixed installment loan is a loan that is repaid in equal payments. Sometimes part of the cost is paid at the time of purchase. This amount is the down payment.

Real-World Example of Installment Loans

Installment loans allow you the ability to purchase an item and use it now. This is called installment purchasing, and being able to use the item is an advantage. The disadvantage is that interest is paid on the amount borrowed. A common example is purchasing an automobile.

Group Activity

Suppose you graduate from college, obtain an amazing job, get married, and then purchase a new vehicle. Use the given information below to answer four questions.
- Cost of automobile: $29,500
- Down payment: $5,000
- Monthly payment: $550.25
- Loan term: 48 months
1. What is the amount financed (in dollars)? Note that the amount financed is the price of the car minus the down payment.
  
  $
2. What is the total amount (in dollars) of all monthly payments?
  
  $
3. What is the total installment price (in dollars)? Note that the total installment price is the sum of all the monthly payments plus the down payment.
  
  $
4. What was the financial charge (in dollars)? Note that the financial charge is the total installment price minus the purchase price of the automobile.
  
  $
  
  Explain what it represents.
  
  The financial charge represents the .
Prior to making an automobile purchase, it is often helpful to know in advance what the monthly payment will be. The monthly payment can be determined using the formula shown below, where M is the monthly payment, P is the principal value of the loan, r is the APR (annual percentage rate) in decimal form, and n is the total number of payments.

M =
P
r
12
1 −
1 +
r
12
−n
1. Using the internet, identify the MSRP (Manufacturer's Suggested Retail Price) of an automobile you would be interested in purchasing (in dollars). A suggested website would be www.edmunds.com.
  
  $
2. Use the MSRP from part (2a) and the given information below to determine the monthly payment (in dollars) for the automobile. Compare your calculated monthly payment with an online loan calculator. A suggested loan calculator is found at www.bankrate.com. (Round your answer to two decimal places.)
  - APR is 3.19%
  - 48 months
  - Trade-in value of your current automobile is $4,650
  M = $
3. Use the rounded value you found in part (2b) to find the total installment price (in dollars). (Round your answer to two decimal places.)
  
  $
4. Use the rounded value you found in part (2b) to find the financial charge (in dollars). (Round your answer to two decimal places.)
  
  $

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8. 0/1 points | Previous Answers GHColAlg13 5.3.026. My Notes

This exercise develops conceptual understanding.
Read It links are available as a learning tool under each question so students can quickly jump to the corresponding section of the eTextbook.

Write the equation in logarithmic form.

3⁻⁴ =

1

81

67

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9. 1/1 points | Previous Answers GHColAlg13 6.3.034. My Notes

This exercise develops conceptual understanding.
Students get just-in-time learning support with Watch It videos that contain narrated and closed-captioned videos walking students through the proper steps to solve a similar problem.

Find A − B.

A =

2

0

−3

7

2

−2

−7

, B =

−4

3

6

−2

1

0

2

4

−1

A − B =

Need Help?

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10. –/6 points GHColAlg13 3.2.EI.002. My Notes

This exercise develops conceptual understanding.
Explore It exercises engage students with interactive learning modules that include video and explorations. The module is also available to use for studying in the eTextbook

Review the Explore It, then use it to complete the exercise below.

Click here to access the Explore It in a new window.

Select Parent Function 3 to investigate shifting the graph of the parent function

y = |x|.

(a)

The graph of

y = |x|

has its vertex at the origin. What horizontal and vertical shifts are needed to shift the vertex to the point (5, 3).

horizontal shift vertical shift

(b)

In the equation of the modified function from part a, the amount of the horizontal shift is the absolute value bars, and the amount of the vertical shift is the absolute value bars.

(c)

For the modified function

y = |x + 4| − 6,

the graph is shifted 4 units and 6 units.

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11. –/3 points GHColAlg13 4.1.094.EP. My Notes

This exercise develops conceptual understanding.
Expanded Problems enhance student understanding by going beyond a basic exercise and asking students to solve each step of the problem in addition to their final answer.

A city's transit authority serves 154,000 commuters daily when the fare is $1.90. Market research has determined that every penny decrease in the fare will result in 1,100 new riders.

Let x be the number of penny decreases. Represent the revenue with a quadratic function, in terms of x.

R(x) =

How many penny decreases are needed to maximize revenue?

penny decreases

What fare will maximize revenue (in dollars)?

$

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12. –/72 points GHColAlg13 4.LAC.001. My Notes

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This exercise develops conceptual understanding.
Looking Ahead to Calculus interactive modules prepare STEM students for future success by previewing calculus concepts they will encounter.

Looking Ahead to Calculus: Using the Derivative to Analyze the Graph of a Function

Introduction
You have already learned how to sketch the graph of a polynomial function using information about the end behavior, symmetry, the intercepts, and a sign chart for where the graph is above and below the x-axis. In calculus, the derivative of a function provides us with even more information to accurately sketch the graph of a function.

The derivative of a function represents the slope of the tangent lineA tangent line is a line that intersects a curve at only one point. The word "tangent" is derived from the Latin word "tangens," which means "touching." to the curve y = f(x) at the point (x, f(x)). The derivative of a function is denoted f ′(x) (read as f prime of x).
The x y-coordinate plane is given. A curve and a line are given.

The curve labeled y = f(x) begins in the first quadrant, goes up and right becoming less steep, passes through the point labeled P, changes direction at an unlabeled point up and right of the point P, goes down and right becoming more steep, and exits the window in the first quadrant.

The line labeled t enters the window in the first quadrant, goes up and right, touches the curve at the point P, and exits the window in the first quadrant.
In the figure above, the slope of the tangent line t is the derivative of f(x) at the point P.

So, the derivative will tell us the direction in which the curve proceeds at each point. In other words, it tells us where the curve is increasing or decreasing. The derivative will also tell us when the curve changes directions, and thus the location of any local maximum or minimum values.

We focus on polynomial functions in this lab, but, in calculus, the concepts extend to other types of functions as well.
Interactive Reading
Consider the graph of y = f(x) below. A few tangent lines are also graphed at various points on the curve.
The x y-coordinate plane is given. A curve and seven short tangent line segments are given.

The curve begins in the first quadrant at a point labeled A, goes up and right becoming less steep, changes direction at a point labeled B, goes down and right, changes direction at a point labeled C, goes up and right becoming more steep, and ends at a point labeled D.

There are three tangent line segments that touch the curve between points A and B.

There are two tangent line segments that touch the curve between points B and C.

There are two tangent line segments that touch the curve between points C and D.
Between points A and B, the tangent lines have positive slope. On this same interval, the graph of f(x) is .

Between points B and C, the tangent lines have slope. On this same interval, the graph of f(x) is decreasing.

Between points C and D, the tangent lines have positive slope. On this same interval, the graph of f(x) is .

The graph of f(x) has a local (relative) maximum at point . If we were to draw a tangent line through this point, it would be a horizontal line and the slope would be .

The graph of f(x) has a local (relative) minimum at point . If we were to draw a tangent line through this point, it would be a horizontal line and the slope would be .

From the statements above, we can draw the following conclusions regarding the information that f ′(x) tells us about f(x).
Summary of Information from the First Derivative of a Function

Suppose f(x) is a polynomial function.
- If the derivative is positive (f ′(x) > 0) on an interval, then f is increasing on that interval.
- If the derivative is negative (f ′(x) < 0) on an interval, then f is decreasing on that interval.
- If f(x) has a turning point (local maximum or minimum) at c, then f ′(c) = 0.
- If f ′(x) changes from positive to negative at c, then f has a local maximum at c. If f ′(x) changes from negative to positive at c, then f has a local minimum at c.
So, to sketch the graph of a function, we will use the derivative and a sign chart to determine the intervals of increasing and decreasing, as well as the location of any turning points. This information can be combined with what you already know about the end behavior, symmetry, and intercepts of a function to obtain an accurate sketch.

Interactive Example

For the polynomial function

f(x) = 3x⁴ − 4x³ − 12x² + 5,

the derivative is

f ′(x) = 12x³ − 12x² − 24x.

Use a sign chart for f ′(x) to determine the open intervals where f(x) is increasing and where it is decreasing. Also, identify any local maximum and minimum values of f(x). Then, sketch the graph of f(x).

Solution

First, we must set the derivative equal to zero and solve for any critical numbersA critical number of a function f is a number c in the domain of f such that either f ′(c) = 0 or f ′(c) does not exist.. These values will help us construct our sign chart.

We factor f ′(x) and then solve for x.

12x³ − 12x² − 24x = 0

12x(x + 1)

= 0

x = 0, x = −1, or x =

Now, we create a sign chart. The three critical numbers divide the domain of f(x) into four intervals.

In each interval, we choose a test point and evaluate the derivative at the test point. If the sign of f ′(x) is positive, we know that f(x) is increasing, and if the sign of f ′(x) is negative, we know that f(x) is decreasing.

Interval

Test Point

Sign of f ′(x)

Behavior of f(x)

(−∞, −1)

f ′(−2) = −96

−

decreasing

(−1, 0)

f ′

−

1

2

=

15

2

(0, 2)

f ′(1) =

(2, ∞)

f ′(3) =

+

increasing

From the sign chart, we see that f(x) is increasing on the intervals

and (2, ∞). We see that f(x) is decreasing on the intervals (−∞, −1) and

.

Now, we determine the behavior at the turning points.

Because the sign of f ′(x) changes from negative to positive at x = −1, we know f(x) has a local at x = −1. The value of this local is
f(−1) = 0.
Because the sign of f ′(x) changes from positive to negative at x = 0, we know f(x) has a local at x = 0. The value of this local is f(0) = .
Because the sign of f ′(x) changes from negative to positive at x = 2, we know f(x) has a local at x = 2. The value of this local is
f(2) = −27.

To graph

y = f(x),

we combine the information above with some additional information that we can determine using college algebra techniques.

Since the leading term of f(x) is
3x⁴,
we know that the graph on the left and on the right.
Since f(x) ≠ f(−x), and f(x) ≠ − f(x), the function is . Thus, f(x) is not symmetric with respect to the y-axis nor the origin.

If additional points are still needed to help graph the function, we can create a table of values. You may also choose to look for the x and y intercept using the techniques you have learned in your college algebra course. The graph of

y = f(x)

is shown below.

The x y-coordinate plane is given. The curve enters the window in the second quadrant, goes down and right becoming less steep, changes direction at the point (−1, 0), goes up and right becoming more steep, passes through the approximate point (−0.55, 2.3), goes up and right becoming less steep, changes direction at the point (0, 5), goes down and right becoming more steep, crosses the x-axis at approximately x = 0.61, passes through the approximate point (1.22, −13.4), goes down and right becoming less steep, changes direction at the point (2, −27), goes up and right becoming more steep, crosses the x-axis at approximately x = 2.72, and exits the window in the first quadrant.

Practice 1

Consider the equation below.

f(x) = x³ − 6x² − 15x + 6

The derivative is

f ′(x) = 3x² − 12x − 15 = 3(x + 1)(x − 5).

Complete the sign chart to determine the intervals on which f is increasing or decreasing.

Interval	Test Point	Sign of f ′(x)	Behavior of f(x)
	f ′(−2) =
	f ′(0) =
	f ′(8) =

Find the interval(s) on which f is increasing. (Enter your answer using interval notation.)

Find the interval(s) on which f is decreasing. (Enter your answer using interval notation.)

Find the coordinates of the points where the local minima and maxima occur.

local minimum

(x, y)

=

local maximum

(x, y)

=

Determine the end behavior of f(x).

The graph of f(x) on the left and on the right.

Determine if the graph of f(x) has any symmetries.

about the y-axis about the origin no symmetries

Use the information to sketch the graph of f(x).

The x y-coordinate plane is given. The curve enters the window in the third quadrant, goes up and right becoming less steep, crosses the x-axis at approximately x = −2.2, changes direction at the point (−1, 14), goes down and right becoming more steep, crosses the y-axis at y = 6, crosses the x-axis at approximately x = 0.4, passes through the point (2, −40), goes down and right becoming less steep, changes direction at the point (5, −94), goes up and right becoming more steep, crosses the x-axis at approximately x = 7.8, and exits the window in the first quadrant.

The x y-coordinate plane is given. The curve enters the window in the second quadrant, goes down and right becoming less steep, crosses the x-axis at approximately x = −2.2, changes direction at the point (−1, −14), goes up and right becoming more steep, crosses the y-axis at y = −6, crosses the x-axis at approximately x = 0.4, passes through the point (2, 40), goes up and right becoming less steep, changes direction at the point (5, 94), goes down and right becoming more steep, crosses the x-axis at approximately x = 7.8, and exits the window in the fourth quadrant.

Practice 2

Consider the equation below.

f(x) = x⁴ − 8x² + 4

The derivative is

f ′(x) = 4x³ − 16x = 4x(x − 2)(x + 2).

Complete the sign chart to determine the intervals on which f is increasing or decreasing.

Interval	Test Point	Sign of f ′(x)	Behavior of f(x)
	f ′(−3) =
	f ′(−1) =
	f ′(1) =
	f ′(3) =

Find the interval(s) on which f is increasing. (Enter your answer using interval notation.)

Find the interval(s) on which f is decreasing. (Enter your answer using interval notation.)

Find the coordinates of the points where the local minima and maxima occur.

local minimum (smaller x-value)

(x, y)

=

local minimum (larger x-value)

(x, y)

=

local maximum

(x, y)

=

Determine the end behavior of f(x).

The graph of f(x) on the left and on the right.

Determine if the graph of f(x) has any symmetries.

about the y-axis about the origin no symmetries

Use the information to sketch the graph of f(x).

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Gustafson and Hughes - College Algebra 13/e (Homework)

Current Score : 10 / 108

Due : Sunday, January 27, 2030 00:00 EST

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Group Activity: Installment Loans

Looking Ahead to Calculus: Using the Derivative to Analyze the Graph of a Function

Introduction

Interactive Reading

Interactive Example

Practice 1

Practice 2

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Gustafson and Hughes - College Algebra 13/e (Homework)

Current Score : 10 / 108

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

Instructions

Assignment Submission

Assignment Scoring

WebAssign Graphing Tool

Graph LayersToggle Open/Closed

Group Activity: Installment Loans

Looking Ahead to Calculus: Using the Derivative to Analyze the Graph of a Function

Introduction

Interactive Reading

Interactive Example

Practice 1

Practice 2