An SUV is purchased new for $\$ 31,500$.
Part 1 of 4
(a) Write a linear function of the form $y=m t+b$ to represent the value $y$ of the vehicle $t$ years after purchase. Assume that the vehicle is depreciated by $\$ 3150$ per year.
The model for the value of the car $t$ years after purchase is $y=-3150 t+31500$.
Alternate Answer:
\[
y=-3150 t+31,500
\]
Part 2 of 4
(b) Suppose that the vehicle is depreciated so that it holds only $90 \%$ of its value from the previous year. Write an exponential function of the form $y=V_{0} b^{t}$, where $V_{0}$ is the initial value and $t$ is the number of years after purchase.
The model for the value of the vehicle $t$ years after purchase is $y=31500(.90)^{t}$
Alternate Answer:
\[
y=31,500(0.9)^{t}
\]
Part: $2 / 4$
Part 3 of 4
(c) To the nearest dollar, determine the value of the vehicle after $5 \mathrm{yr}$ and after $10 \mathrm{yr}$ using the linear model. According to the linear model, the vehicle will be worth $\$ \square$ after 5 years, and $\$ \square$ after 10 years,
Answer
\(\boxed{\text{According to the linear model, the vehicle will be worth $15750 after 5 years, and $0 after 10 years.}}\)
Step 1 ：The linear function to represent the value of the vehicle t years after purchase, assuming that the vehicle is depreciated by $3150 per year, is given by \(y=-3150t+31500\).
Step 2 ：The exponential function to represent the value of the vehicle t years after purchase, assuming that the vehicle holds only 90% of its value from the previous year, is given by \(y=31500(0.9)^{t}\).
Step 3 ：To find the value of the vehicle after 5 years and 10 years using the linear model, we substitute t=5 and t=10 into the linear model \(y=-3150t+31500\).
Step 4 ：Substituting t=5 into the linear model gives a value of $15750 for the vehicle after 5 years.
Step 5 ：Substituting t=10 into the linear model gives a value of $0 for the vehicle after 10 years.
Step 6 ：\(\boxed{\text{According to the linear model, the vehicle will be worth $15750 after 5 years, and $0 after 10 years.}}\)
