Question

Use the table to answer the 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)


A.
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.


B. 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.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.


C.
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.


D. 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.

Answer 1

First finish the 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

These values show an increasing exponential curve that reaches about 946 beetles by month 5. The only graph option that correctly shows an increasing curve with an appropriate vertical scale is A.