Question

Mike bought a car for $40,000 and it decreases in value every year by 15%
Write an exponential function to represent the value of the car at the end of every year. Use the variable x but no commas
f(x)=____
Then fill in the table
x f(x)
0 $__
1 $___
3 $___
5 $___
8 $ ___

Answer 1

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