Financial Algebra 1st Edition Chapter 5 Automobile Ownership
Page 246 Problem 1 Answer
Given: A car sells for D dollars and totally depreciates after T years.
Here we will define the intercepts of the straight line depreciation equation.
Let x represent the time in years. The minimum x−value is 0 years, the purchase year of the car.
Because the car totally depreciates after T years, the maximum x−value will be T.
In a straight line depreciation equation, the intercepts are (0, maximum car value) and (maximum lifespan, 0 )
Let y represent the value of the car at any time during its lifetime.
The minimum y−value is zero dollars and the maximum y−value is the pur− chase price of D dollars.
Knowing this information, you can identify the intercepts as (0,D) and (T,0).
The intercepts are (0,D) and (T,0).
Page 246 Problem 2 Answer
Given: A car is purchased for D dollars and totally depreciates after T years.
Here By using this given data we will find the slope of the depreciation line.
Two points determine a line, so you only need two points to determine the slope of a line.
Let the coordinates of the y−intercept be the first point.
That is, (x1,y1)=(0,D). Let the coordinates of the x−intercept be the second point. That is, (x2,y2)=(T,0).
Using the slope ration we get y2−y1/x2−x1
Put all the values 0−D/T−0=−D/T
The slope of the depreciation line is −D/T
Page 247 Problem 3 Answer
Given: A car that was purchased for $22,000 and totally depreciates after 11 years.
Here by using this data we will plot the graph of the depreciation line.
The general form for the equation of a straight line is
y=mx+b
Where in represents the slape of the line and b represents the y−intercept.
The two intercepts of the given line are (0,22,000) and (11,0)
Thus the slope is 0−22,000/11−0=−2000
And the Y-intercept is 22,000
Thus equation of line is written as y=−2,000x+22,000
Graph is
The equation of line is y=−2,000x+22,000 and the graph is
Page 248 Problem 4 Answer
Given: A car sells for $18,495 dollars and straight line depreciates to zero after 9 years.
Here by using this data we will find the equation of the line.
The intercepts are (0,18,495) and (9,0).
Thus slope is y2−y1/x2−x1
=0−18,495/9−0
=−18,495/9
=−2,055
The y-intercept of the line is 18,495
Thus the equation of line is y=−2,055x+18,495
Because x represents years, it is necessary to convert W months into years by dividing by 12.
Thus the value of car after W months is y=−2,300(W/12)+27,600
The value of the car is =−2,300(W/12)+27,600
Page 248 Problem 5 Answer
Given : Value of car is D dollars
The straight line depreciation equation for a car is y=−4,000x+D
Here we will find the value of x with respect to the value of the car after the car’s value decreased by 25%.
The original value of the car is y intercept of the equation.
We must determine the actual value of the car after it has dropped by 25%. This can be done
As we can find 25% of the original value of the car and then subtract that amount from the original value
0.25×D=0.25D
D−0.25D=0.75D
The value is 0.75D
We are trying to determine a length of time. Solve the depreciation equation for x.
Y/0.75D/0.25D
0.25D/4000
=−4,000x+D
=−4,000x+D
=4,000x
=x
The length of the time is represented as 0.25D/4000=x
Page 249 Problem 6 Answer
Given;The expense function be altered so that it refl ects a more accurate amount spent over time
To find ; What effect might that have on the graphs?
Let x represent time in months and y represent dollars.
Celine’s expense function is the sum of her monthly payments over this time period and her initial down payment.
Expense function y=560x+4,000
The time, x, is in months rather than years. Express Celine’s depreciation function in terms of months as well.
Celine’s car totally depreciates after 10 years, or 120 months.
To determine her monthly depreciation amount, divide the original car value by 120.
33,600/120
=280
Celine’s car depreciates 280 per month. To calculate the slope of the depreciation equation, use the intercepts (0,33,600) and 120,0.
Slope0−33,600/120−0
=−33,600/120
=−280
Depreciation function y=−280x+33,600
Using a graphing calculator, the coordinates of the intersection point, rounded to the nearest hundredth, are 35.24,23,733.33.
This means that after a little more than 35 months, both your expenses and the car’s value are the same In the region before the intersection point, the expenses are lower than the value of the car.
The region after the intersection point indicates a period of time that the value of the car is less than what you have invested in it.
Hence we conclude that after a little more than 35 months, both your expenses and the car’s value are the same.
In the region before the intersection point, the expenses are lower than the value of the car.
The region after the intersection point indicates a period of time that the value of the car is less than what you have invested in it.
Page 250 Problem 7 Answer
We have to find how might those words apply to what we have learned.
We will explain it by two statement which are stated below.
Computers have improved over the years in their tech while becoming more affordable for the masses.
Cars too have improved but the cost for a car has increased over the years.
Therefore, Unlike computers and other tech which has improved over the years and dropped in expenses (becoming more accessible to the public), cars have improved but simultaneously increased in cost.
Page 250 Problem 8 Answer
Given : Delia purchased a new car for $25,350. This make and model straight line depreciates to zero after 13 years.
We have to identify the coordinates of the x – and y -intercepts for the depreciation equation.
We will find it.
Here,x is the number of years and y is the price.
Now,The x intercept has y zero.
The model depreciates to 0 after 13 years, which implies that when x=13
⇒y=0
i.e. (13,0)
Now,The y-intercept has x zero.
At time x=0, Delia purchases the car and thus the price is $25,350.
i.e. (0,25350)
Therefore, the coordinates of the x – and y -intercepts for the depreciation equation is(13,0),(0,25350)
Page 250 Problem 9 Answer
Given: Delia purchased a new car for $25,350.
This make and model straight line depreciates to zero after 13 years.
We have to find the slope of the depreciation equation.
We will use above formula.
Here,We have, x – and y-intercept is (13,0),(0,25350)
So,y2−y1/x2−x1
=25350−0/0−13
=−1,950
Therefore,−1,950 is the slope of the depreciation equation.
Page 250 Problem 10 Answer
Given: Delia purchased a new car for $25,350. This make and model straight line depreciates to zero after 13 years.
We have to write the straight line depreciation equation that models this situation.
We will use above information.
Here,
We have,
m=−1,950
b=25350
So,y=mx+b
y=−1950x+25350
Therefore,
y=−1950x+25350
Page 250 Problem 11 Answer
Given: Delia purchased a new car for $25,350. This make and model straight line depreciates to zero after 13 years.
We have to draw the graph of the straight line depreciation equation.
We will put different values of x to get the values of y and then draw a graph.
Here,We have ,y=−1950x+25350
We will put different values of x to get the values of y in the above equation.
So, the graph is :
Therefore,
Given : Vince purchased a used car for $11,200. This make and model used car straight line depreciates to zero after 7 years.
We have to identify the coordinates of the x-and y-intercepts for the depreciation equation.
We will find it.
Here,x is the number of years and y is the price.
Now,The x -intercept has y zero.
After 7 years, the price is depreciated to 0 .This implies that when x=7
⇒y=0
i.e. (0,11200)
Noe, The y-intercept has x zero.
When x=0, Vince purchase the car and then the price was$11,200.
i.e. (7,0)
Therefore, the coordinates of the x – and y -intercepts for the depreciation equation is(0,11200),(7,0)
Page 250 Problem 13 Answer
Given : Vince purchased a used car for$11,200. This make and model used car straight line depreciates to zero after 7 years.
We have to find the slope of the depreciation equation.
We will use above formula.
Here,We have, x- and y-intercept is (7,0),(0,11200)
So, y2−y1/x2−x1
=11200−0/0−7
=−1,600
Therefore, The slope of the depreciation equation is−1,600
Page 250 Problem 14 Answer
Given: Vince purchased a used car for $11,200. This make and model used car straight line depreciates to zero after 7 years.
We have to write the straight line depreciation equation that models this situation.
We will use above information.
Here,We have,m=11200
y=−1600
So, y=mx+b
y=−1,600x+11,200
Therefore,
y=−1,600x+11,200
Page 250 Problem 15 Answer
Given: Vince purchased a used car for $11,200. This make and model used car straight line depreciates to zero after 7 years.
We have to Draw the graph of the straight line depreciation equation.
So,
We get,
Page 250 Problem 16 Answer
Given: Examine the straight line depreciation graph for a car.
To find: At what price was the car purchased?
The price at purchase is the price at time 0 and is thus the y-intercept (intersection of the graph with the y axis).
28000
The price of the car at purchase is at time 0 and as per the graph it is 28000.
Page 250 Problem 17 Answer
Given: Examine the straight line depreciation graph for a car.
To find: After how many years does the car totally depreciate?
The car totally depreciates if the value becomes zero and thus at the x-intercept (intersection of the graph with the x-axis).
So, in 10 years does the car totally depreciate
So, in 10 years does the car totally depreciate
Page 250 Problem 18 Answer
Given: Examine the straight line depreciation graph for a car.
We have to Write the equation of the straight line depreciation graph shown.
x − and y−intercept found in previous exercises (0,28000) and (10,0)
The slope can be determined with y2−y1/x2−x1
y2−y1/x2−x1
=0−28000/10−0
=−2,800
The equation of the straight line depreciation is y=mx+b with m the slope and b the y-intercept
y=−2,800x+28,000
We get,
y=−2,800x+28,000
Given: The straight line depreciation equation for a luxury car is y=−3,400x+85,000
To find: What is the original price of the car?
The price at purchase is the price at time 0 and is thus the y-intercept (x is zero).
y=−3,400×0+85,000=85,000
We get,
y=−3,400×0+85,000=85,000
Page 250 Problem 19 Answer
Given: The straight line depreciation equation for a luxury car is y=−3,400x+85,000
To find: How much value does the car lose p
The yearly depreciation is given by the slope (number in front of the x in the straight line depreciation equation) $3,400
The slope of the graph gives the yearly depreciation. So as per the equation, it is $3,400.
Page 250 Problem 20 Answer
Given: The straight-line depreciation equation for a luxury car is y=−3,400x+85,000
To find: How many years will it take for the car to totally depreciate?
The car totally depreciates if the value becomes zero and thus at the x-intercept ( y is zero).
0=−3,400x+85,000
Add 3,400x to both sides of the equation
3,400x=85,000
Divide both sides of the equation by 3400
x=25
25 years
It will take 25 years for the car to totally depreciate.
Page 250 Exercise 1 Answer
Given: The straight line depreciation equation for a motorcycle is y=−2,150x+17,200
To find: What is the original price of the motorcycle?
The price at purchase is the price at time 0 and is thus the y-intercept ( x is zero).
u=−2.150×0+17.200=17.200
We get,
u=−2.150×0+17.200=17.200
Page 250 Exercise 2 Answer
The straight-line depreciation equation for a motorcycle is y=−2,150x+17,200
To find: How much value does the motorcycle lose per year.
The yearly depreciation is given by the slope (number in front of the x in the straight line depreciation equation) $2,150
The motorcycle loses its value by$2,150 per year.
Page 250 Exercise 3 Answer
Given: The straight line depreciation equation for a motorcycle is y=−2,150x+17,200
To find: How many years will it take for the motorcycle to totally depreciate.
The car totally depreciates if the value becomes zero and thus at the x-intercept (y is zero).
0=−2,150x+17,200
Add 2,150x to both sides of the equation
2,150x=17,200
Divide both sides of the equation by 2,150
x=8
It would take8 years for the motorcycle to totally depreciate.
Page 250 Exercise 4 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000
To find: What is the car worth after 5 years?
Replace x with 5 in the straight line depreciation equation.
y=−2,750×5+22,000
=8,250
We get,y=8250
Page 250 Exercise 5 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000
To find: What is the car worth after 8 years
Replace x with 8 in the straight line depreciation equation.
y=−2,750×8+22,000=0
The worth of the car after 8 years is $0.
Page 250 Exercise 6 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000
Suppose that A represents a length of time in years when the car still has value.
We have to Write an algebraic expression to represent the value of the car after A yea
Replace x with A in the straight line depreciation equation.
y=−2,750A+22,000
We get,y=−2,750A+22,000
Page 251 Exercise 7 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000
To find: What is the car worth after 5 years?
Replace x with 5 in the straight line depreciation equation.
y=−2,750×5+22,000
=8,250
We get,y=8250
Page 251 Exercise 8 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000
To find: What is the car worth after 8
Replace x with 8 in the straight line depreciation equation.
y=−2,750×8+22,000=0
The worth of the car after 8 years is $0.
Page 251 Exercise 9 Answer
Given: The straight line depreciation equation for a car is y=−2,750x+22,000 Suppose that A represents a length of time in years when the car still has value.
We have to Write an algebraic expression to represent the value of the car after A years.
Replace x with A in the straight line depreciation equation.
y=−2,750A+22,000
We get,
y=−2,750A+22,000
Page 251 Exercise 10 Answer
Given: The graph of a straight line depreciation equation is shown.
We have to Use the graph to approximate the value of the car after 4 years.
Find the vertical line with 4 years at the bottom and determine the intersection between the vertical line and the blue line.
Determine the horizontal line that also goes through this point and read the value of this line on the y-axis.
$12,800
As per the graph, the approximate value of the car after 4 years is $12,800.
Page 251 Exercise 11 Answer
Given: The graph of a straight line depreciation equation is shown.
we have to Use the graph to approximate the value of the car after 5 years.
Find the vertical line with 5 years at the bottom and determine the intersection between the vertical line and the blue line.
Determine the horizontal line that also goes through this point and read the value of this line on the y-axis.
So, we get $9,600
We get, $9,600
Page 251 Exercise 12 Answer
Given: The graph of a straight line depreciation equation
We have to Use the graph to approximate when the car will be worth half its original value.
The original value is the value at the intersection of the blue graph and the y-axis.
$25,600 Half of the original value is then the original value divided by 2
$25,600÷2=$12,800
Determine the horizontal line at the value $12,800, find its intersection with the blue line.
Find the vertical line that also goes through this intersection and determine the number of years of this vertical line.
4 years
We get,4 years
Page 251 Exercise 13 Answer
Given: A car is originally worth $34,450. It takes 13 years for this car to totally depreciate.
We have to write the straight line depreciation equation for this situation.
x – and y-intercept
(0,34450)and(13,0)
The slope can be determined with y2−y1/x2−x1
y2−y1 /x2−x1
=0−34450/13−0
=−2,650
The equation of the straight line depreciation is y=mx+b with m the slope and b the y-intercept
y=−2,650x+34,450
We get,
y=−2,650x+34,450
Page 251 Exercise 14 Answer
Given: A car is originally worth $34,450. It takes 13 years for this car to totally depreciate.
We have to find that how long will it take for the car to be worth half its value
Equation found in previous exercise
y=−2,650x+34,450
Half of its value is the original value divided by 2.
34,450/2
=−2,650x+34,450
Subtract 34,450 from both sides of the equation−17,225=−2,650x
Divide both sides of the equation by −2,650/6.5=x
It will take 6.5 years for the car to be half its value.
Page 251 Exercise 15 Answer
Given: A car is originally worth $34,450. It takes 13 years for this car to totally depreciate.
We have to find that how long will it take for the car to be worth Round your answer to the nearest tenth of a year.$10,000?
Equation found in previous exercise
y=−2,650x+34,450
The value is$10,000
10,000=−2,650x+34,450
Subtract 34,450 from both sides of the equation−24,450=−2,650x
Divide both sides of the equation by −2,650/9.2=x
It will take 9.2 years for the car to be worth it.
Page 251 Exercise 16 Answer
Given: The original price of a car is entered into spreadsheet cell A1 and the length of time it takes to totally depreciate is entered into cell B1.
We have to write the spreadsheet formula that calculates the amount that the car depreciates each year.
x-and y-intercept (0,A1) and (B1,0)
The slope can be determined with y2−y1/x2−x1
y2−y1/x2−x1
=0−A1/B1−0
=−A1/B1
We get,−A1/B1
Page 251 Exercise 17 Answer
Given: The original price of a car is entered into spreadsheet cell Al and the length of time it takes to totally depreciate is entered into cell B1.
The spreadsheet user is instructed to enter a length of time in years that is within the car’s lifetime in cell C1.
We have to write the spreadsheet formula that will calculate the car’s value after that period of time.
The equation of the straight line depreciation is y=mx+b with m the slope and b the y-intercept
y=−A1/B1
x+A1
Replace x with C1
y=−A1/B1×C1+A1
We get,−A1/B1×C1+A1
Page 251 Exercise 18 Answer
Given: Winnie purchased a new car for $54,000. She has determined that it straight line depreciates to zero over 10 years.
When she purchased the car, she made an $8,000 down payment and financed the rest with a 4-year loan at 4.875%.
You can use the monthly payment formula from the last chapter to determine the monthly payment to the nearest cent.
We have to Create an expense and depreciation function.
Expense function:
Purchase price=$54,000
Down payment= $8,000
r= Interest rate =4.875%=0.04875
t= Number of years =4
The principal (amount loaned) is the purchase price decreased by the down payment:
p= Principal
= Purchase price − Down payment
=$54,000−$8,000
=$46,000
Suppose the first determine the monthly payment using the monthly payment loan formula:
M=p(r/12)(1+r/12)12t
(1+r/12)12t−1
=46000(0.04875/12)(1+0.04875/12)12×4
(1+0.04875/12)12×4−1
≈1,056.74
Suppose x represent the number of months and let y represent the total expense.
The total expense is the total of the monthly payments and the down payment, while the total of the monthly payments is the product of the number of months and the monthly payment.
y= Total expense = Total monthly payments + Down payment = Number of months × Monthly payment+ Down payment
=x×1,056.74+8,000
=1,056.74x+8,000
Depreciation function
Let x represent the number of months and let y represent the value of the car.
General depreciation function:
y=ax+b
with a the slope and b the y-intercept.
When x=0, the value of the car is $54,000 (purchase price).
This then implies that (0,54000) needs to be a point on the depreciation function and thus they-intercept is $54,000.
b=y-intercept =54,000
After 10 years (or 10×12=120 months), the car has depreciated to 0 and thus (120,0) needs to be a point on the depreciation function as well.
We can then calculate the slope using y2−y1/x2−x1 with (x1,y1)=(0,54000) and (x2,y2)=(120,0).
a=y2−y1/x2−x1
=0−54000/120−0
=−54000/120
=−450
We finally determine the depreciation function by replacing a by −450 and replacing b by 54,000 in the general depreciation function:
y=ax+b=−450x+54,000
Expense function: y = 1, 056.74x + 8, 000
Depreciation function: y = −450x + 54, 000
Page 251 Exercise 19 Answer
Given: Winnie purchased a new car for $54,000. She has determined that it straight line depreciates to zero over 10 years.
When she purchased the car, she made an $8,000 down payment and financed the rest with a 4-year loan at 4.875%.
You can use the monthly payment formula from the last chapter to determine the monthly payment to the nearest cent.
We have to draw the Graph these functions on the same axes.
Expense function
Purchase price =$54,000
Down payment=$8,000
r= Interest rate =4.875%=0.04875
t= Number of years =4
The principal (amount loaned) is the purchase price decreased by the down payment:
p= Principal
= Purchase price − Down payment
=$54,000−$8,000
=$46,000
Let us first determine the monthly payment using the monthly payment loan formula:
M=p(r/12)(1+r/12)12t
(1+r/12)12t−1
=46000(0.04875/12)(1+0.04875/12)12×4
(1+0.04875/12)12×4−1
≈1,056.74
Let x represent the number of months and let y represent the total expense.
The total expense is the total of the monthly payments and the down payment, while the total of the monthly payments is the product of the number of months and the monthly payment.
y= Total expense = Total monthly payments + Down payment = Number of months × Monthly payment + Down payment
=x×1,056.74+8,000
=1,056.74x+8,000
Depreciation function
Suppose x represent the number of months and let y represent the value of the car.
General depreciation function:
y=ax+b with a the slope and b the y-intercept.
When x=0, the value of the car is $54,000 (purchase price). This then implies that (0,54000) needs to be a point on the depreciation function and thus the y-intercept is $54,000.
b=y−intercept =54,000
After 10 years (or 10×12=120 months), the car has depreciated to 0 and thus (120,0) needs to be a point on the depreciation function as well.
We can then calculate the slope using y2−y1/x2−x1 with (x1,y1)=(0,54000) and (x2,y2)=(120,0).
a=y2−y1/x2−x1
=0−54000/120−0
=−54000/120
=−450
We finally determine the depreciation function by replacing a by −450 and replacing b
by 54,000 in the general depreciation function:
y=ax+b=−450x+54,000
So, the graph is
We get,
Page 251 Exercise 20 Answer
Given: Winnie purchased a new car for $54,000. She has determined that it straight line depreciates to zero over 10 years.
When she purchased the car, she made an $8,000 down payment and financed the rest with a 4-year loan at 4.875%.
You can use the monthly payment formula from the last chapter to determine the monthly payment to the nearest cent.
We have to Interpret the region before, at, and after the intersection point in light of the context of this situation.
Expense function
Purchase price =$54,000
Down payment =$8,000
r= Interest rate =4.875%=0.04875
t= Number of years =4
The principal (amount loaned) is the purchase price decreased by the down payment:
p= Principal
= Purchase price − Down payment
=$54,000−$8,000
=$46,000
Let us first determine the monthly payment using the monthly payment loan formula:
M=P(T/12)(1+r/12)12t
(1+r/12)12t−1
=46000(0.04875/12)(1+0.04875/12)12×4
(1+0.04875/12)12×4−1
≈1,056.74
Let x represent the number of months and let y represent the total expense.
The total expense is the total of the monthly payments and the down payment, while the total of the monthly payments is the product of the number of months and the monthly payment
y= Total expense =Total monthly payments + Down payment = Number of months × Monthly payment + Down payment
=x×1,056.74+8,000
=1,056.74x+8,000
Depreciation function
Let x represent the number of months and let y represent the value of the car.
General depreciation function: y=ax+b with a the slope and b the y-intercept.
When x=0, the value of the car is$54,000 (purchase price). This then implies that (0,54000) needs to be a point on the depreciation function and thus they-intercept is $54,000.
b=y-intercept =54,000
After 10 years (or 10×12=120 months), the car has depreciated to 0 and thus (120,0) needs to be a point on the depreciation function as well.
We can then calculate the slope using y1−y1/x2−x1 with (x1,y1)=(0,54000) and (x2,y2)=(120,0).
a=y2−y1/x2−x1
=0−54000/120−0
=−54000/120
=−450
We finally determine the depreciation function by replacing a by −450 and replacing b by 54,000 in the general depreciation function:
y=ax+b=−450x+54,000
Graph is:
Before Intersection
Before the intersection, we note that the depreciation function is above the expenses function, which means that the value of the car exceeds the expenses made
At Intersection Let us first determine the intersection, which is the value of x for which the expense function and the depreciation function are equal :
1,056.74x+8,000=−450x+54,000
Subtract 8,000 from each side:
1,056.74x=−450x+46,000
Add 450x to each side:
1,506.74x=46,000
Divide each side by 1,506.74 :
x=46,000
1,506.74
≈30.5
Thus the intersection occurs at x=30.5. Let us determine the corresponding expense/depreciation by evaluating the expense function at 30.5 (which should be roughly the same as the depreciation function evaluate at 30.5):
y=1,056.74(30.5)+8,000≈40,261.73
This then implies that the intersection has x=30.5 and y=40,261.73, which means that: after 30.5 months, the depreciation and the expenses are both equal to$40,261.73.
After intersection
After the intersection, we note that the depreciation function is below the expenses func
Before intersection: The value of the car exceeds the expenses made.
At intersection: After30.5 months, the depreciation and the expenses are both equal to $40,261.73.
After intersection: The expenses made exceed the value of the car.