Exponential decay occurs when the quantity of something decreases at a rate that is proportional to its current value. For example, a population of animals might decay as their habitat is slowly eroded by human encroachment, or the amount of radiation that is emitted from a radioactive substance could decrease as time passes.

To write the exponential decay function for the value of a car, we can use the formula:

\[ V(t) = V_0 \cdot e^{-kt} \]

where:

• \( V(t) \) is the value of the car at time \( t \).
• \( V_0 \) is the initial value (cost of the car).
• \( k \) is the decay constant.
• \( t \) is the time in years.

However, for simplicity and based on the values provided in the problem, we can also represent the decay using the decay factor, expressed as:

\[ V(t) = V_0 \cdot (1 - r)^t \]

where:

• \( r \) is the decay rate in decimal form (for example: 12% becomes 0.12).
• \( (1 - r) \) is the factor by which the value reduces each year.

Car A:

• Initial Cost: \( V_0 = 8,750 \)
• Decay Factor: 12% (or 0.12 as a decimal)

The decay factor (for one year) is \( (1 - r) = (1 - 0.12) = 0.88 \).

Thus, the exponential decay function for Car A can be expressed as:

\[ V_A(t) = 8750 \cdot (0.88)^t \]

Car B:

• Initial Cost: \( V_0 = 9,995 \)
• Decay Factor: 18% (or 0.18 as a decimal)

The decay factor (for one year) is \( (1 - r) = (1 - 0.18) = 0.82 \).

Thus, the exponential decay function for Car B can be expressed as:

\[ V_B(t) = 9995 \cdot (0.82)^t \]

Explanation of the Value of \( b \):

In these functions, the base value \( b \) represents the decay factor per year, which is determined by the percentage decrease in value of the car:

• For Car A, with a 12% decay factor, it retains 88% of its value each year (or decays by multiplying \( V_0 \) by 0.88).
• For Car B, with an 18% decay factor, it retains 82% of its value each year (or decays by multiplying \( V_0 \) by 0.82).

Thus:

• For Car A, \( b = 0.88 \).
• For Car B, \( b = 0.82 \).

These decay functions help predict how the value of each car will decrease over time based on their initial prices and decay rates.

2. Use the Desmos graphing utility to create two graphs; one for each function that you wrote in question 1. Explain what each axis represents and why you choose the scale you did. You may also plot these graphs by hand, if you prefer.

To graph the exponential decay functions for both Car A and Car B using a graphing utility like Desmos or by hand, you would follow these steps:

Functions to Graph

1. Car A: \[ V_A(t) = 8750 \cdot (0.88)^t \]

2. Car B: \[ V_B(t) = 9995 \cdot (0.82)^t \]

Axes Representation

X-Axis (Horizontal):

• Label: Time (in years)
• Scale: You can choose a scale that goes from 0 to, for example, 10 or 15 years. This gives a good range to see how the values of the cars decrease over a realistic time frame. A mark at every 1-year increment would be helpful for a clear representation of the decay over time.

Y-Axis (Vertical):

• Label: Value of the Car (in dollars)
• Scale: Setting the scale from approximately $0 to about $10,000 would allow both functions to be clearly visible within the initial values and their decay. Each major divide could represent increments of $1,000, which keeps the graph uncluttered and easy to read.

Steps to Create the Graph on Desmos:

1. Open Desmos: Access the graphing utility.
2. Input Functions: Enter the two functions:
   • \( y = 8750 \cdot (0.88)^x \) for Car A
   • \( y = 9995 \cdot (0.82)^x \) for Car B
3. Adjust View: Make sure the window is set to show an appropriate portion of both axes, particularly focusing on the first 10-15 years and the values starting from $0 to about $10,000.
4. Label: You can add labels for each curve to distinguish which line represents which car.

Explanation of the Graph:

When the graph is generated, you should observe the following:

• Both curves start at their respective initial values (Car A starts at $8,750, and Car B begins at $9,995).
• As time progresses (moving along the X-axis), both curves show a downward trend that reflects the exponential decay in value over time. The curves you see should trend downward, demonstrating how the car values decrease due to their respective decay rates.

Interpretation of the Graph:

• Steeper Decay: You will notice that Car B (with the higher decay factor of 18%) will decrease in value faster compared to Car A (with a 12% decay factor). This means that after the same period, Car B will have a lower value than Car A.
• Long-Term Trend: Over the long run, both cars decline towards zero, but the rate at which they do so varies based on the decay factors.

This visual representation provides a clear illustration of how the initial investments in both cars will lose value over time due to depreciation. It helps emphasize the importance of considering decay rates when making purchasing decisions related to vehicles.

3. Look at your graphs from question 2. Explain the value of each y-intercept in context of the scenario.

The y-intercept of a graph represents the value of the function at \( t = 0 \), which corresponds to the initial value or cost of the cars in this scenario. Let's analyze the y-intercepts for both Car A and Car B from their respective functions.

Y-Intercept Values

1. Car A:

   • Function: \( V_A(t) = 8750 \cdot (0.88)^t \)
   • At \( t = 0 \): \[ V_A(0) = 8750 \cdot (0.88)^0 = 8750 \cdot 1 = 8750 \]
   • Y-Intercept Value: $8,750

2. Car B:

   • Function: \( V_B(t) = 9995 \cdot (0.82)^t \)
   • At \( t = 0 \): \[ V_B(0) = 9995 \cdot (0.82)^0 = 9995 \cdot 1 = 9995 \]
   • Y-Intercept Value: $9,995

Explanation of the Y-Intercept Values in Context

• Car A Y-Intercept ($8,750$): This value represents the initial cost of Car A. It indicates that if Rhyley were to purchase this car, the amount of money she would need to invest upfront is $8,750. This cost is also important to consider because it reflects how much value Rhyley and her parents will put into the car at the beginning.

• Car B Y-Intercept ($9,995$): Similarly, this amount signifies the initial cost of Car B. Rhyley would need to spend $9,995 to purchase this car initially. This higher price signifies that, even though both cars aim to provide similar features, Car B is more expensive right from the beginning.

Contextual Implications

• The y-intercepts illustrate the initial investment required for each vehicle, which is crucial for Rhyley and her parents in determining how much combined money they would need for the car purchase.
• These values also set the stage for understanding the depreciation of each vehicle over time. The higher initial cost of Car B ($9,995) not only indicates a higher financial commitment but may also reflect differences in quality, mileage, and age of the car.
• When considering which car to purchase, Rhyley should also weigh future depreciation against the initial costs. Although Car B is more expensive upfront, it may also provide a better long-term value based on factors like its lower mileage and higher decay factor.

In conclusion, the y-intercepts of the graphs provide the critical initial cost information needed for analyzing the financial implications of purchasing each car in Rhyley's scenario.

5. Rhyley is unsure how long she will own the car she chooses. The graph below shows the exponential decay functions for Car A (in green) and Car B (in purple). Interpret the value of the marked point in relation to the value of each car as time passes.

Without a visual of the graph, I can provide a general interpretation based on typical behavior of exponential decay functions and how specific points on such a graph might be interpreted in context.

Interpreting a Marked Point on the Graph

Assuming that the marked point on the graph corresponds to a specific time \( t \) (in years) and the respective values for each car at that point in time, here's how to interpret this point:

1. Identifying Time and Value:
   • Let's say the marked point is at \( t = x \) years.
   • For Car A, its value at that point would be \( V_A(x) = 8750 \cdot (0.88)^x \).
   • For Car B, its value at that point would be \( V_B(x) = 9995 \cdot (0.82)^x \).

Interpretation:

1. Current Value:

   • The marked point indicates the current values of both cars after \( x \) years of ownership. This provides Rhyley an idea of how much each car will depreciate in value over that period.

2. Comparison Between Cars:

   • By evaluating the values of Car A and Car B at \( t = x \):
     • If the value of Car A is higher than that of Car B, it implies that despite its lower initial cost and slower depreciation rate, it holds its value better over that period.
     • Conversely, if Car B has a higher value at the marked point, it suggests that the higher initial investment may be justified if its performance and durability lead to less depreciation.

3. Decision-Making Insight:

   • The marked point provides practical insights for Rhyley regarding the future resale value or trade-in value of the car at year \( x \). If Rhyley plans to keep the car for a specific duration, knowing the value at that time allows her to assess the potential financial impact of her purchase.
   • For instance, if she sees that after \( 5 \) years, Car A is worth $4,000 and Car B is worth $3,500, she might consider that Car A has been a better investment in that timeframe, even if it initially cost less.

Summary:

In conclusion, the marked point on the graph, indicating the value of both Car A and Car B at a specific time, serves as a crucial data point for Rhyley. It not only reflects the depreciation behaviors of the cars but also aids her in making an informed decision about which vehicle might be a better investment based on her plans for ownership duration. If you have specific values associated with the marked point or a description of its position, please share that information so I can provide a more tailored interpretation!

6. After some consideration, Rhyley’s best guess is that she will own the car she chooses for 5 years. Explain two different ways you could determine the value of each car in the fifth year Rhyley owns it.

To determine the value of each car after 5 years of ownership, you can use several methods:

Method 1: Substituting Directly into the Exponential Decay Function

The most straightforward way is to substitute \( t = 5 \) directly into the exponential decay functions for Car A and Car B.

1. Car A Function: \[ V_A(t) = 8750 \cdot (0.88)^t \] \[ V_A(5) = 8750 \cdot (0.88)^5 \]

   • Calculate \( (0.88)^5 \) and then multiply by $8,750 to get the value for Car A after 5 years.

2. Car B Function: \[ V_B(t) = 9995 \cdot (0.82)^t \] \[ V_B(5) = 9995 \cdot (0.82)^5 \]

   • Similarly, calculate \( (0.82)^5 \) and then multiply by $9,995 to find the value for Car B after 5 years.

Method 2: Using a Graphing Utility

Another way to determine the values after 5 years is to use a graphing utility (like Desmos or a graphing calculator) to visualize the functions and find the corresponding values at \( t = 5 \).

1. Graph the Functions:

   • Enter the two functions into the graphing utility:
     • \( y = 8750 \cdot (0.88)^x \) (for Car A)
     • \( y = 9995 \cdot (0.82)^x \) (for Car B)

2. Find the Point at \( t = 5 \):

   • Use the graphing utility's features to locate the point on the graph where \( x = 5 \). The y-coordinate at this point will give you the value for each car after 5 years.
   • Some graphing utilities have a "table" feature or allow you to click on the graph to find the specific y-value at a selected x-value.

Summary of Both Methods

• Method 1 involves mathematical calculation using the exponential decay functions directly, which provides an exact numerical answer.
• Method 2 involves a visual approach using graphing technology, allowing for quick verification and understanding of how the values change over time through a graphical representation. This method can also highlight trends and comparisons between the two cars more effectively.

Both methods ultimately achieve the same result, enabling Rhyley to understand how much each car is expected to be worth after 5 years of ownership.