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).
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...