To determine the correct graph for the population of beetles increasing at a rate of 22% per month, we first need to calculate the values of the function \( f(x) = 350(1.22)^x \) for \( x = 0, 1, 2, 3, 4, 5 \).
Let's compute these values:
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For \( x = 0 \): \[ f(0) = 350(1.22)^0 = 350 \times 1 = 350 \]
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For \( x = 1 \): \[ f(1) = 350(1.22)^1 = 350 \times 1.22 \approx 427 \]
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For \( x = 2 \): \[ f(2) = 350(1.22)^2 = 350 \times (1.22 \times 1.22) \approx 521.54 \]
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For \( x = 3 \): \[ f(3) = 350(1.22)^3 = 350 \times (1.22 \times 1.22 \times 1.22) \approx 636.51 \]
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For \( x = 4 \): \[ f(4) = 350(1.22)^4 \approx 774.76 \]
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For \( x = 5 \): \[ f(5) = 350(1.22)^5 \approx 930.95 \]
Now, summarizing these computed values:
- \( f(0) = 350 \)
- \( f(1) \approx 427 \)
- \( f(2) \approx 521.54 \)
- \( f(3) \approx 636.51 \)
- \( f(4) \approx 774.76 \)
- \( f(5) \approx 930.95 \)
Now that we have the values, we can plot the points:
- \( (0, 350) \)
- \( (1, 427) \)
- \( (2, 521.54) \)
- \( (3, 636.51) \)
- \( (4, 774.76) \)
- \( (5, 930.95) \)
Considering the responses provided for different graphs, we should look for a graph that shows an increasing trend that goes beyond 350 and continues to rise sharply as \( x \) increases.
Based on the calculations, the graph that fits best with our observations is:
- An increasing curve with an arrow on top plotted in a coordinate plane (the third image).
This graph corresponds to an exponential growth pattern (which is exhibited by our calculations) and will accurately represent the situation of the increasing population of beetles with time.