To solve this problem, let's break it down into parts.
Part A: Exponential Function
Markus originally bought the car for $40,000, and it depreciates by 15% each year. This means that after one year, the car is worth 85% of its previous year's value (since 100% - 15% = 85%).
The exponential function that represents the value of the car after \( x \) years can be written as:
\[ f(x) = 40000 \times (0.85)^x \]
Part B: Value of the Car After \( x \) Years
Now let's fill in the table with the values after \( x \) years using the function derived in Part A.
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For \( x = 0 \): \[ f(0) = 40000 \times (0.85)^0 = 40000 \times 1 = 40000.00 \]
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For \( x = 1 \): \[ f(1) = 40000 \times (0.85)^1 = 40000 \times 0.85 = 34000.00 \]
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For \( x = 3 \): \[ f(3) = 40000 \times (0.85)^3 = 40000 \times 0.614125 = 24565.00 \]
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For \( x = 5 \): \[ f(5) = 40000 \times (0.85)^5 = 40000 \times 0.443705 = 17748.20 \]
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For \( x = 8 \): \[ f(8) = 40000 \times (0.85)^8 = 40000 \times 0.232630 = 9305.19 \]
Summary Table
Here is the completed table of the car's value after \( x \) years:
| x | f(x) | |-----|------------| | 0 | $40000.00 | | 1 | $34000.00 | | 3 | $24565.00 | | 5 | $17748.20 | | 8 | $9305.19 |
Make sure to round the final values to the nearest cent as required.