A vehicle depreciates by 22% each year after it is first purchased.

Question

A vehicle depreciates by 22% each year after it is first purchased.

Question 1

Part A

What is the base of the exponential decay function that models this situation?

Part B

Enter the exponential decay function that can be used to model the value of the vehicle.

Enter the correct answer in the box, using x to represent the time in years. Do not include the dollar sign in your answer. 

Part C

How much will the vehicle be worth 8 years after it was purchased?

Enter the correct answer in the box. Round to the nearest cent.

Answer 1

A =Pe^rx

A =30,000e^.22x

x=8 years, r= -22%=.22

A =30,000e^-.22(8)

A = 30,000e^-1.76

A = $5,161.25 after 8 years

which is depreciation continuously which is most realistic

or just once a year, then

after one year

A =.78(30K= 23,400

after two years

A =.78(23.4K) = 18,252

3 years

A = .78(18.252K =14,236.56

4 years

A = .78(14,236.56) = 11,104.52

5 years

A = .78(11,104.52 = 8,661.38

6 years

A =.78(8,661.38) = 6755.98

7 years

A = .78(6,755.98) = 5269.67

8 years

A =.78(5269.67= $4,110,34

formual for annual compounding is

A =P(1+r)x where x = number of years, r = rate of interest or rate of decay if negative

A = 30,000(1-.22)8

= 30K(.78)^8

= 30K(.137011)

= $4,110.34 after 8 years

where the car suddenly loses 22% value on the 365th day or 366th if a leap year

while cars typically lose major value in just one second when driven off the lot

the annual once a year depreciation is not realistic, but it may be the answer you want

but if you randomly guess one or the other, you have a 50% chance of the right answer

simple interest or decay

would mean

.22(30,000) = loss in value of $6,600 yearly

after 8 years that's 30,00-8(6600) = 30,000-52800 = -$22,800 which is not the answer you want, and totally unrealistic