Question

This table tracks the number of beetles (350) over time as it increases at a rate of 25% per month, modeled by an exponential equation.
x f(x)
0 350
1
2
3
4
5
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.(1 point)
Responses

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.
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Image with description An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 2000 in increments of 250.

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
▶Click for Long Desc
Image with description A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.

An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.
▶Click for Long Desc
Image with description An increasing curve with an arrow on the top is plotted on a coordinate plane. The x-axis for months ranges from 0 to 2,000 in increments of 200. The y-axis for Beetles ranges from 0 to 10 in increments of 1.

A decreasing curve with an arrow at the end is plotted on a coordinate plane. The x-axis for months ranges from 0 to 10 in increments of 1. The y-axis for Beetles ranges from 0 to 350 in increments of 50.
▶Click for Long Desc

Answer 1

Finished table (using f(x)=350(1.22)^x):

- x = 0 → 350
- x = 1 → 350(1.22) = 427
- x = 2 → 350(1.22)^2 ≈ 520.94
- x = 3 → 350(1.22)^3 ≈ 635.55
- x = 4 → 350(1.22)^4 ≈ 775.37
- x = 5 → 350(1.22)^5 ≈ 945.95

The correct graph is the first one (an increasing curve with y up to 2000), because the population grows exponentially (increasing) and reaches about 946 beetles by month 5, which fits that scale.