To model the value of the car as it depreciates over time, we can use an exponential decay function. The general form for such a function is:
\[ f(x) = V_0 \cdot (1 - r)^x \]
where:
- \( f(x) \) is the value of the car after \( x \) years,
- \( V_0 \) is the initial value of the car,
- \( r \) is the rate of depreciation (as a decimal),
- \( x \) is the number of years.
In this case, the initial value \( V_0 = 40000 \) and the rate of depreciation \( r = 0.15 \).
Thus, the function becomes:
\[ f(x) = 40000 \cdot (1 - 0.15)^x \]
Which simplifies to:
\[ f(x) = 40000 \cdot (0.85)^x \]
Now we can fill in the table for various values of \( x \):
Table of Values:
-
For \( x = 0 \): \[ f(0) = 40000 \cdot (0.85)^0 = 40000 \cdot 1 = 40000 \]
-
For \( x = 1 \): \[ f(1) = 40000 \cdot (0.85)^1 = 40000 \cdot 0.85 = 34000 \]
-
For \( x = 3 \): \[ f(3) = 40000 \cdot (0.85)^3 \approx 40000 \cdot 0.614125 = 24565 \]
-
For \( x = 5 \): \[ f(5) = 40000 \cdot (0.85)^5 \approx 40000 \cdot 0.443705 = 17748.20 \]
-
For \( x = 8 \): \[ f(8) = 40000 \cdot (0.85)^8 \approx 40000 \cdot 0.272490 = 108996 \]
Putting it all together, the filled table is:
\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 0 & 40000 \ 1 & 34000 \ 3 & 24565 \ 5 & 17748.20 \ 8 & 10899.60 \ \hline \end{array} \]
So the final answers are:
- f(x) = 40000(0.85)^x
- The table values filled in would be:
- 0: $40000
- 1: $34000
- 3: $24565
- 5: $17748.20
- 8: $10899.60