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4. The equation [tex]f(t) = 24,500 \cdot (0.88)^t[/tex] represents the value of a car, in dollars, [tex]t[/tex] years after it was purchased.

a. What do the numbers 24,500 and 0.88 mean?

b. What does [tex]f(9)[/tex] represent?

c. Sketch a graph that represents the function [tex]f[/tex] and shows [tex]f(0)[/tex], [tex]f(1)[/tex], and [tex]f(2)[/tex].

Answer :

Certainly! Let's break down the solution to the questions step-by-step.

### Part a: Understanding the Numbers

- 24,500: This number represents the initial value of the car in dollars. This means when the car was first purchased, it was valued at [tex]$24,500.

- 0.88: This number represents the rate of depreciation of the car per year. A depreciation rate of 0.88 indicates that each year, the car's value is 88% of its value from the previous year. In other words, the car loses 12% of its value each year.

### Part b: Calculating \( f(9) \)

The function \( f(t) = 24,500 \cdot (0.88)^t \) is used to determine the value of the car after \( t \) years. To find \( f(9) \), we calculate the car's value 9 years after purchase:

\[
f(9) = 24,500 \cdot (0.88)^9
\]

From the calculation (done previously), the value of the car after 9 years is approximately \$[/tex]7,753.72.

### Part c: Values for [tex]\( f(0) \)[/tex], [tex]\( f(1) \)[/tex], and [tex]\( f(2) \)[/tex]

Let's find the initial value of the car and its value after 1 and 2 years:

- [tex]\( f(0) \)[/tex]: This calculates the value at the time of purchase (initial value).
[tex]\[
f(0) = 24,500 \cdot (0.88)^0 = 24,500 \times 1 = 24,500
\][/tex]

- [tex]\( f(1) \)[/tex]: This calculates the value 1 year after purchase.
[tex]\[
f(1) = 24,500 \cdot (0.88)^1 = 24,500 \times 0.88 = 21,560
\][/tex]

- [tex]\( f(2) \)[/tex]: This calculates the value 2 years after purchase.
[tex]\[
f(2) = 24,500 \cdot (0.88)^2 = 24,500 \times 0.7744 \approx 18,788.80
\][/tex]

Now, we would typically sketch a graph of this exponential decay function. On the graph, plot points corresponding to these values: [tex]\( f(0) = 24,500 \)[/tex], [tex]\( f(1) = 21,560 \)[/tex], and [tex]\( f(2) \approx 18,788.80 \)[/tex]. The graph will show a decreasing trend, getting closer to zero but never quite reaching it, as the car depreciates over time.

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