Question

Mark bought a brand new car for $35,000 in 2008. If the car depreciates in value approximately 8% each year, write an exponential function to model the situation. Then, find the value of the car in 2015. Is this considered growth or decay?

Answer 1

`V(t) = 35000(0.92)^(t)`

Decay function

The value of the car in 2015 is $19,525.

Explanation:

A exponential value function has the following format:

`V(t) = V(0)(1+r)^(t)`

In which V(t) is the value after t years, V(0) is the initial value and 1+r is the yearly variation rate.

If 1+r>1, the function is a growth function.

If 1-r<1, the function is a decay function.

Mark bought a brand new car for $35,000 in 2008.

This means that
`V(0) = 35,000`

If the car depreciates in value approximately 8% each year

Depreciates, then r is negative. So
`r = -0.08`

Then

`V(t) = V(0)(1+r)^(t)`

`V(t) = 35000(1-0.08)^(t)`

`V(t) = 35000(0.92)^(t)`

0.92 < 1, so decay function.

Then, find the value of the car in 2015.

2015 is 2015-2008 = 7 years after 2008. So this is V(7).

`V(t) = 35000(0.92)^(t)`

`V(7) = 35000(0.92)^(7)`

`V(7) = 19525`

The value of the car in 2015 is $19,525.