Question
3. 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 $
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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 $
Skip to navigation
Answers
GPT-4o mini
### Part A: Exponential Function
To represent the value of the car after x years, we start with the initial value of the car, which is $40,000. The car depreciates at a rate of 15% per year, which means it retains 85% of its value each year.
The exponential function can be formulated as follows:
\[
f(x) = 40000(0.85)^x
\]
### Part B: Values of the Car After x Years
To fill out the table for the values of the car after x years, we can calculate \(f(x)\) for each value 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 \(x = 3\):**
\[
f(3) = 40000(0.85)^3 \approx 40000(0.614125) \approx 24565.01
\]
4. **When \(x = 5\):**
\[
f(5) = 40000(0.85)^5 \approx 40000(0.443705) \approx 17748.20
\]
5. **When \(x = 8\):**
\[
f(8) = 40000(0.85)^8 \approx 40000(0.272490) \approx 10899.61
\]
### Completed Table
| x | f(x) |
|-----|----------|
| 0 | 40000.00 |
| 1 | 34000.00 |
| 3 | 24565.01 |
| 5 | 17748.20 |
| 8 | 10899.61 |
To represent the value of the car after x years, we start with the initial value of the car, which is $40,000. The car depreciates at a rate of 15% per year, which means it retains 85% of its value each year.
The exponential function can be formulated as follows:
\[
f(x) = 40000(0.85)^x
\]
### Part B: Values of the Car After x Years
To fill out the table for the values of the car after x years, we can calculate \(f(x)\) for each value 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 \(x = 3\):**
\[
f(3) = 40000(0.85)^3 \approx 40000(0.614125) \approx 24565.01
\]
4. **When \(x = 5\):**
\[
f(5) = 40000(0.85)^5 \approx 40000(0.443705) \approx 17748.20
\]
5. **When \(x = 8\):**
\[
f(8) = 40000(0.85)^8 \approx 40000(0.272490) \approx 10899.61
\]
### Completed Table
| x | f(x) |
|-----|----------|
| 0 | 40000.00 |
| 1 | 34000.00 |
| 3 | 24565.01 |
| 5 | 17748.20 |
| 8 | 10899.61 |