Markus bought a car for $40,000

Question

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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Answer 1

Part A:

The value of the car after each year can be modeled using an exponential decay function. Since the car loses 15% of its value each year, it retains 85% of its value after one year.

To express this mathematically:

• The initial value of the car when x = 0 is $40,000.
• After each year, the value is multiplied by 0.85 (which represents 85% retention of value).

Thus, the exponential function can be written as:

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

Part B:

Now, we can calculate the value of the car for specific years \( x \):

Calculating for Different Values of x:

1. When \( x = 0 \): \[ f(0) = 40000(0.85)^0 = 40000(1) = 40000.00 \]

2. When \( x = 1 \): \[ f(1) = 40000(0.85)^1 = 40000(0.85) = 34000.00 \]

3. When \( x = 3 \): \[ f(3) = 40000(0.85)^3 = 40000(0.614125) \approx 24565.00 \]

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

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

Final Table:

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

This completes the answers for both parts A and B.