Question

the 2012 jeep liberty limited sport 2wd has one of the largest depreciation values of any car. suppose a 2012 jeep liberty limited sport 2wd sold for $26,600.00 in 2012 and its value depreciates by $3,000.00 per year. assume the depreciation is linear.
a) let y be the value of the 2012 jeep liberty limited sport 2wd and t be the number of years since 2012. find the equation that represents the price of the car.
y=
b) how much is the car worth in 1 year?
$
c) how much is the car worth in 2 years?
$
d) in what year will the value of the car reach $0.00? round up to the nearest whole year.

Explanation:

Step1: Determine the linear - equation form

The general form of a linear equation is $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. In the context of depreciation, the initial value of the car ($b$) is the price in 2012, which is $26600$, and the rate of depreciation ($m$) is $- 3000$ (negative because the value is decreasing). So the equation is $y=-3000t + 26600$.

Step2: Find the value of the car after 1 year

Substitute $t = 1$ into the equation $y=-3000t + 26600$. Then $y=-3000\times1 + 26600=23600$.

Step3: Find the value of the car after 2 years

Substitute $t = 2$ into the equation $y=-3000t + 26600$. Then $y=-3000\times2+26600 = 20600$.

Step4: Find the year when the car's value is 0

Set $y = 0$ in the equation $y=-3000t + 26600$. Then $0=-3000t + 26600$. Solve for $t$:
\[

$$\begin{align*} 3000t&=26600\\ t&=\frac{26600}{3000}\approx8.87 \end{align*}$$

\]
Round up to the nearest whole - year, so $t = 9$. Since $t$ is the number of years since 2012, the year is $2012 + 9=2021$.

Answer:

a) $y=-3000t + 26600$
b) $23600$
c) $20600$
d) $2021$