Answer:
![V(t) = 35000(0.92)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2035000%280.92%29%5E%7Bt%7D)
Decay function
The value of the car in 2015 is $19,525.
Step-by-step explanation:
A exponential value function has the following format:
![V(t) = V(0)(1+r)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%20V%280%29%281%2Br%29%5E%7Bt%7D)
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](https://tex.z-dn.net/?f=V%280%29%20%3D%2035%2C000)
If the car depreciates in value approximately 8% each year
Depreciates, then r is negative. So ![r = -0.08](https://tex.z-dn.net/?f=r%20%3D%20-0.08)
Then
![V(t) = V(0)(1+r)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%20V%280%29%281%2Br%29%5E%7Bt%7D)
![V(t) = 35000(1-0.08)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2035000%281-0.08%29%5E%7Bt%7D)
![V(t) = 35000(0.92)^{t}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2035000%280.92%29%5E%7Bt%7D)
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}](https://tex.z-dn.net/?f=V%28t%29%20%3D%2035000%280.92%29%5E%7Bt%7D)
![V(7) = 35000(0.92)^{7}](https://tex.z-dn.net/?f=V%287%29%20%3D%2035000%280.92%29%5E%7B7%7D)
![V(7) = 19525](https://tex.z-dn.net/?f=V%287%29%20%3D%2019525)
The value of the car in 2015 is $19,525.