Question

3. Exponential Functions in Context(1 point)
Markus bought a car for $40,000
and it decreases in value by 15%
every year.

Part A: Write an exponential function that represents the value of the car at the end of every year. Use the variable x
in your equation. Do not include a comma in your equation. Simplify the expression in the parenthesis.

f(x)=


Part B: Fill out the table that represents the value of the car after x
years. Do not include a comma in your response. Write your answer to the nearest cent, write ".00" if there are no cents.

x
f(x)

0 $
1 $
3 $
5 $
8 $
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page 3 of 3

Answer 1

To represent the depreciation in value of Markus's car using an exponential function, we can start with the following information:

- The initial value of the car is $40,000.
- The car depreciates by 15% each year, which means it retains 85% of its value each year.

### Part A
The exponential function can be expressed as:

\[
f(x) = 40000 \times (0.85)^x
\]

This simplifies to:

\[
f(x) = 40000(0.85)^x
\]

### Part B
Now, we need to calculate the value of the car after x years using the function we derived above.

Let's calculate the values for different years (0, 1, 3, 5, and 8):

1. For \( x = 0 \):
\[
f(0) = 40000(0.85)^0 = 40000 \times 1 = 40000
\]

2. For \( x = 1 \):
\[
f(1) = 40000(0.85)^1 = 40000 \times 0.85 = 34000
\]

3. For \( x = 3 \):
\[
f(3) = 40000(0.85)^3 \approx 40000 \times 0.614125 = 24565
\]

4. For \( x = 5 \):
\[
f(5) = 40000(0.85)^5 \approx 40000 \times 0.443705 = 17748.20
\]

5. For \( x = 8 \):
\[
f(8) = 40000(0.85)^8 \approx 40000 \times 0.272490 = 10899.60
\]

### Completed Table

| x | f(x) |
|---|---------|
| 0 | $40000.00 |
| 1 | $34000.00 |
| 3 | $24565.00 |
| 5 | $17748.20 |
| 8 | $10899.60 |

So, the final answers are:

- **Part A**: \( f(x) = 40000(0.85)^x \)
- **Part B**:
- \( f(0) = 40000.00 \)
- \( f(1) = 34000.00 \)
- \( f(3) = 24565.00 \)
- \( f(5) = 17748.20 \)
- \( f(8) = 10899.60 \)