Let's solve each part of this problem step-by-step.

Problem Statement Summary:

The problem involves a 2014 luxury sedan originally sold for $38,180. After 2 years, its value depreciates to $23,719. You are required to:

1. Find a linear model for the value $V(t)$ of the sedan.
2. Find an exponential model for $V(t)$.
3. Graph both models to compare them.

Step 1: Linear Model

In a linear model, the value $V(t)$ can be represented as: $V(t) = mt + b$ where:

• $t$ is the time in years,
• $m$ is the slope (rate of change of value per year),
• $b$ is the initial value (value when $t = 0$).

Calculate the Slope $m$

Given data points:

• $(0, 38180)$ (initial value at $t = 0$)
• $(2, 23719)$ (value after 2 years)

The slope $m$ is calculated as: $m = \frac{V(2) - V(0)}{2 - 0} = \frac{23719 - 38180}{2} = \frac{-14461}{2} = -7230.5$

Determine the Linear Model Equation

Now, we can write the linear equation as: $V(t) = -7230.5t + 38180$

Step 2: Exponential Model

In an exponential model, the value $V(t)$ can be represented as: $V(t) = V_0 e^{kt}$ where:

• $V_0$ is the initial value at $t = 0$,
• $k$ is the decay constant.

Using the data points, we can solve for $k$.

Since $V(0) = 38180$ and $V(2) = 23719$, we can set up the equation: $23719 = 38180 \cdot e^{2k}$

Solving for $k$:

1. Divide both sides by 38180: $\frac{23719}{38180} = e^{2k}$

2. Take the natural logarithm of both sides: $\ln\left(\frac{23719}{38180}\right) = 2k$

3. Calculate the left side: $k = \frac{\ln(23719 / 38180)}{2} \approx -0.2282$

Determine the Exponential Model Equation

Now, we can write the exponential equation as: $V(t) = 38180 \cdot e^{-0.2282t}$

Step 3: Graphing Both Models

For graphing both models, we would use the equations derived:

1. Linear Model: $V(t) = -7230.5t + 38180$
2. Exponential Model: $V(t) = 38180 \cdot e^{-0.2282t}$

These equations should be graphed over the same time range to visually compare the depreciation of the sedan's value according to both models.

Would you like a further breakdown on any specific part or a graph illustrating these models? Here are some questions that expand on this:

1. How does depreciation differ between linear and exponential models over longer periods?
2. What are typical real-world examples where linear vs. exponential models might apply better?
3. How would a different depreciation rate (slope or decay constant) affect the model?
4. How can you calculate the rate of depreciation as a percentage for each year?
5. How would you modify the models if you had a third data point?

Tip: In exponential decay models, $k$ (the decay constant) helps determine how quickly the value approaches zero. Smaller values of $k$ mean slower decay.