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 |

1 answer

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.