Belinda is thinking about buying a car for \$45,000. The table below shows the projected value of two different cars for three years.

\begin{tabular}{|c|c|c|c|}
\hline
Number of years & 1 & 2 & 3 \\
\hline
Car 1 (value in dollars) & 40,500 & 36,450 & 32,805 \\
\hline
Car 2 (value in dollars) & 42,000 & 39,000 & 36,000 \\
\hline
\end{tabular}

Part A: What type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? Explain your answer. (2 points)

Part B: Write one function for each car to describe the value of the car $ f(x) $ in dollars, after $ x $ years. (1 point)

Part C: Belinda wants to purchase a car that would have the greatest value in 13 years. Will there be any significant difference in the value of either car after 13 years? Explain your answer, and show the value of each car after 13 years. (4 points)

Certainly! Let's tackle each part individually with detailed explanations:

Part A:
To determine the type of function, we need to analyze the given values for each car year by year.

- Car 1:
- Year 1: \$40,500
- Year 2: \$36,450
- Year 3: \$32,805

For Car 1, the value decrement isn't constant. Instead, it seems to decrease by a fixed percentage each year. This type of behavior is characteristic of an exponential decay function.

- Car 2:
- Year 1: \$42,000
- Year 2: \$39,000
- Year 3: \$36,000

For Car 2, the difference in value between years is \$3,000 each time. This indicates a constant decrease, which is characteristic of a linear decay function.

Part B:
Next, we will establish the functions that describe the value of each car after $x$ years.

- Car 1:
Since Car 1 is decreasing by 10% annually, we can express it as:
[tex]\[ f_1(x) = 45000 \times (0.9)^x \][/tex]
Here, 45,000 is the initial value, and 0.9 represents the remaining value after a 10% decrease each year.

- Car 2:
Since Car 2 decreases by \$3,000 each year, we can express it as:
[tex]\[ f_2(x) = 45000 - 3000 \times x \][/tex]
Here, 45,000 is the initial value, and the term $3000 \times x$ represents a linear decrease of \$3,000 per year.

Part C:
To find out which car will have the greater value after 13 years and whether there's a significant difference in their values, we use our functions:

- For Car 1 after 13 years:
[tex]\[ f_1(13) = 45000 \times (0.9)^{13} \approx 11438.40 \][/tex]

- For Car 2 after 13 years:
[tex]\[ f_2(13) = 45000 - 3000 \times 13 = 45000 - 39000 = 6000 \][/tex]

Now, comparing these values, Car 1’s value is approximately $\[tex]$11,438.40$ and Car 2’s value is $\$[/tex]6,000$.

To determine the difference:
[tex]\[ \text{Difference} = |11438.40 - 6000| = 5438.40 \][/tex]

Therefore, after 13 years, Car 1 will have a value of approximately $\[tex]$11,438.40$ and Car 2 will have a value of \$[/tex]6,000. The difference in their values will be around \$5,438.40, which is significant.

Conclusion:
Belinda should consider buying Car 1 since it retains more value after 13 years compared to Car 2. There is indeed a significant difference in their values, with Car 1 being worth approximately \$5,438.40 more than Car 2 after 13 years.