Question

Depreciation

A sports utility vehicle that costs $21,500 new has a book value of $13,600 after 2 years.
(a) Find a linear model for the value of the vehicle.
(b) Find an exponential model for the value of the vehicle.
(c) Find the book values of the vehicle after 1 year and after 4 years using each model.
(d) Use a graphing utility to graph the two models in the same viewing window. Which model depreciates faster in the first 2 years?
(e) Explain the advantages and disadvantages of using eac model to a buyer and to a seller.

Answer 1

a)
`y=-3950\cdot{x}+21500`

b) f(x)
`=21500\cdot{(0.7953)^x}`

c) Year 1: Linear $17550, exponential $17099

Year 4: Linear $5700, exponential $8601

d) Exponential model

e) The linear model depreciates the value quicker than exponential model long term around 4 years

Explanation:

a) At year 0 the price is 21500 and at year 2 the price is 13600

WE can use points (0,21500) and (2,13600)

We can determine the gradient

`m=(13600-21500)/(2-0)=-3950`

We can use the point slow formula:

`y-y_1=m\cdot{(x-x_1)}`

`y-21500=-3950\cdot{(x-0)}`

`y=-3950\cdot{x}+21500`

b) We can use the following equation:

f(x)
`=a\cdot{(1+r)^x}`

f(x) is the depreciation value after amount of time t, a is the new value, r is the rate of depreciation and x is the time.

`13600=21500\cdot{(1+r)^2}`

`r=-0.2047`

The depreciation rate is 20.47% and is negative because it decreases the new value of the car

c) Year 1:

`y=-3950\cdot{1}+21500=17550`

f(x)
`=21500\cdot{(0.7953)^1}=17099`

Year 4:

`y=-3950\cdot{4}+21500=5700`

f(x)
`=21500\cdot{(0.7953)^4}=8601`

d) Year 2

`y=-3950\cdot{2}+21500=13600`

f(x)
`=21500\cdot{(0.7953)^2}=13598`