Question 1008550: In the year 2000, Anna bought a new car for $26,000. In 2005, she was told that the value of her car was $15,000 due to depreciation. She is told that the value of her car depreciates linearly.
a) Find a function V(t) which gives the value of the car t years after the year 2000
b) In 2008, Anna is told that she will be given $7,000 for her car if she decides to trade it in for a new car. Use the function from part (a) above to determine the value of her car in 2008.
c) Is the $7,000 value fair based on what she was told about the linear depreciation?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! year 2000 car is bought for 26,000.
year 2005, value of car is 15,000.
let x = 0 represent the year 2000
then x = 5 represents the year 2005
then x = 8 represents the year 2008
y = mx + b is the slope intercdept form of a straight line.
y = value of the car.
x = year
m is the slope of the change in the value of y divided by the change in the value of x.
b is the y-intercept which is the value of y when the value of x is 0.
when x = 0, y = 26,000
when x = 5, y = 15,000
the slope is equal to (y2-y1)/(x2-x1)
y2 = 15,000
y1 = 26,000
x2 = 5
x1 = 0
slope = (15,000 - 26,000) / (5 - 0) = -11,000 / 5 = -2,200.
this means that the car is losing value by 2,2000 per year.
y = mx + b becomes y = -2,200 * x + b
b is the value of y when x is equal to 0 which is the value of the car when it was brand new.
b is therefore equal to 26,000 and the equation is therefore equal to:
y = -2,200 * x + 26,000
in year 0, the car is worth 26,000.
in year 5, the car is worth 26,000 - 2,200 * 5 = 15,000.
in year 8, the car is worth 26,000 - 2,200 * 8 = 8,400.
the offer of 7,000 is 1,400 dollar below the remaining value of the car based on straight line depreciation.
you can see this graphically below:
the first picture shows x = 0 and x = 5
the value at x = 0 is 26,000
the vlaue at x = 5 is 15,000
this is based on loss of value of 2,200 per year based on straight line depreciation.
the second picture shows x = 5 and x = 8.
the value at x = 5 is 15,000
the value at x = 8 is 8,400
you can also see the offer of 7,000 at x = 8.
it is below the value of 8,400.
the difference is 1,400.
the red line is the value of the car in year x.
the blue line is the value of the offer in year x.
the actual year is 2000 + x.
when x is 0, the year is 2000
when x is 5, the year is 2005
when x is 8, the year is 2008
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