Question

A car was purchased for $25,000. Research shows that the car has an average yearly depreciation rate of 18.5%. Create a function that will determine the value V(t), of the car t years after purchase. Determine to the nearest cent, how much the car will depreciate from year 3 to year 4

Answer 1

Answer:

V(t) = 25000 * (0.815)^t

The depreciation from year 3 to year 4 was $2503.71

Step-by-step explanation:

We can model V(t) as an exponencial function:

V(t) = Vo * (1+r)^t

Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.

We have that Vo = 25000, r = -18.5% = -0.185, so:

V(t) = 25000 * (1-0.185)^t

V(t) = 25000 * (0.815)^t

In year 3, we have:

V(3) = 25000 * (0.815)^3 = 13533.58

In year 4, we have:

V(4) = 25000 * (0.815)^4 = 11029.87

The depreciation from year 3 to year 4 was:

V(3) - V(4) = 13533.58 - 11029.87 = $2503.71

Answer 2

The depreciation from year 3 to year 4 was $2503.71

Exponential function

The standard exponential function is expressed as:

$y=a(1\pm r)^t$

r is the rate

t is the time

Given the following parameters

a = 25000,

r = -18.5% = -0.185,

Substitute into the formula

y = 25000 * (1-0.185)^t

y = 25000 * (0.815)^t

In year 3, we have:

y(3) = 25000 * (0.815)^3 = 13533.58

In year 4, we have:

y(4) = 25000 * (0.815)^4 = 11029.87

The depreciation from year 3 to year 4 was:

y(3) - y(4) = 13533.58 - 11029.87 = $2503.71

The depreciation from year 3 to year 4 was $2503.71

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