Question
A certain population of beetles is increasing at a rate of 22% per month. If there are currently 350 beetles, the equation: f (x) = 350 (1.22)^x, can be used to model the situation. Finish the chart and plot the points to determine which graph correctly displays the situation.
0 | 350
1 |
2 |
3 |
4 |
5 |
0 | 350
1 |
2 |
3 |
4 |
5 |
Answers
GPT-4o mini
To complete the chart based on the given exponential growth model \( f(x) = 350 (1.22)^x \), we need to calculate the number of beetles for each month \( x \) from 0 to 5.
1. **For \( x = 0 \):**
\[
f(0) = 350 (1.22)^0 = 350 \times 1 = 350
\]
2. **For \( x = 1 \):**
\[
f(1) = 350 (1.22)^1 = 350 \times 1.22 \approx 427
\]
3. **For \( x = 2 \):**
\[
f(2) = 350 (1.22)^2 = 350 \times 1.4884 \approx 520
\]
4. **For \( x = 3 \):**
\[
f(3) = 350 (1.22)^3 = 350 \times 1.8110 \approx 633.5
\]
5. **For \( x = 4 \):**
\[
f(4) = 350 (1.22)^4 = 350 \times 2.20536 \approx 771.9
\]
6. **For \( x = 5 \):**
\[
f(5) = 350 (1.22)^5 = 350 \times 2.68832 \approx 941.9
\]
Now, we can fill in the chart:
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
0 & 350 \\
1 & 427 \\
2 & 520 \\
3 & 633.5 \\
4 & 771.9 \\
5 & 941.9 \\
\hline
\end{array}
\]
### Summary of the Completed Chart:
- At \( x = 0 \), there are 350 beetles.
- At \( x = 1 \), there are approximately 427 beetles.
- At \( x = 2 \), there are approximately 520 beetles.
- At \( x = 3 \), there are approximately 633.5 beetles.
- At \( x = 4 \), there are approximately 771.9 beetles.
- At \( x = 5 \), there are approximately 941.9 beetles.
### Plotting Points:
To plot these points on a graph, you would mark the coordinates:
- (0, 350)
- (1, 427)
- (2, 520)
- (3, 633.5)
- (4, 771.9)
- (5, 941.9)
### Graph Characteristics:
You should look for a graph that shows exponential growth, starting at (0, 350) and steeply increasing as \( x \) increases. The curve should rise more sharply as it moves to the right, which is characteristic of exponential functions.
1. **For \( x = 0 \):**
\[
f(0) = 350 (1.22)^0 = 350 \times 1 = 350
\]
2. **For \( x = 1 \):**
\[
f(1) = 350 (1.22)^1 = 350 \times 1.22 \approx 427
\]
3. **For \( x = 2 \):**
\[
f(2) = 350 (1.22)^2 = 350 \times 1.4884 \approx 520
\]
4. **For \( x = 3 \):**
\[
f(3) = 350 (1.22)^3 = 350 \times 1.8110 \approx 633.5
\]
5. **For \( x = 4 \):**
\[
f(4) = 350 (1.22)^4 = 350 \times 2.20536 \approx 771.9
\]
6. **For \( x = 5 \):**
\[
f(5) = 350 (1.22)^5 = 350 \times 2.68832 \approx 941.9
\]
Now, we can fill in the chart:
\[
\begin{array}{|c|c|}
\hline
x & f(x) \\
\hline
0 & 350 \\
1 & 427 \\
2 & 520 \\
3 & 633.5 \\
4 & 771.9 \\
5 & 941.9 \\
\hline
\end{array}
\]
### Summary of the Completed Chart:
- At \( x = 0 \), there are 350 beetles.
- At \( x = 1 \), there are approximately 427 beetles.
- At \( x = 2 \), there are approximately 520 beetles.
- At \( x = 3 \), there are approximately 633.5 beetles.
- At \( x = 4 \), there are approximately 771.9 beetles.
- At \( x = 5 \), there are approximately 941.9 beetles.
### Plotting Points:
To plot these points on a graph, you would mark the coordinates:
- (0, 350)
- (1, 427)
- (2, 520)
- (3, 633.5)
- (4, 771.9)
- (5, 941.9)
### Graph Characteristics:
You should look for a graph that shows exponential growth, starting at (0, 350) and steeply increasing as \( x \) increases. The curve should rise more sharply as it moves to the right, which is characteristic of exponential functions.