Originally Posted by

**Unenlightened** a)

The fact that the model is LINEAR tells you that the depreciation is a constant amount every year.

So if after two years the car is worth 15,300, that means that over the two years we have lost 23,500 - 15,300 = 8,200, which is 4100 per year.

So 4100 is the amount you lose each year.

So at any given year t, the value V of your car will have depreciated by 4100*t.

So your model is V(t) (the value after t years) = -4100*t (the yearly depreciation by the number of years) + 23,500 (the initial value).

b)

For the exponential model, we know it satisfies

at time 0

23,500 = 23,500e^(k*0) ... which isn't much use, but important to understand,

and more usefully at time 2

15,300 = 23,500e^(k*2)

So we need to solve for k:

taking the natural log of both sides:

ln (15,300/23,500) = 2k

-0.429147593 = 2k

k = -0.214573796

So your exponential model is given by

V=23,500e^(-0.215* t )

c) The answer should be more or less given in part a there...