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 $

1 answer

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.

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

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

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

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

  5. 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.