Video: US-SAT04S4-Q20-107180930192

The number of flies in a forest is increasing such that it doubles every year. Which of the following graphs could model the number of flies in the forest as a function of time? Note: in each graph below, 𝑂 represents the coordinate (0, 0).

03:44

Video Transcript

The number of flies in a forest is increasing such that it doubles every year. Which of the following graphs could model the number of flies in the forest as a function of time? Note: in each graph below, 𝑂 represents the coordinate zero, zero.

Before we consider the graphs, let’s think about the information we were given. We were told that the number of flies in the forest is increasing and that each of these graphs are a function of time. This means as we move along the 𝑥-axis to the right. As time passes along the 𝑥-axis, the number of flies should go up. It should increase as time goes along. This means we’re looking for some kind of positive correlation. This also means that the horizontal line shown in option C is not possible. The graph in model C is saying that the number of flies stays the same, is constant as time passes. It also means that option D is not possible because it represents a negative correlation. It is a decrease in the number of flies as the years go by.

This leaves us with A or B. Both A and B represent a positive correlation. And that means we’ll need to look more specifically for the one that is doubling each year. One way to do this is to make a table and examine a few of the points on each of these graphs. Because graph B says that 𝑦 equals 𝑥, we know that our points are one, one; two, two; and three, three. So we can fill those in. The points on graph A are not quite as clear. So instead of using the graph, we can use this function 𝑦 equals 0.6 times two to the 𝑥 power. In year one, we have 0.6 times two to the first power. Two to the first power is two. 0.6 times two is 1.2. In year two, we have 0.6 times two squared flies. Two squared is four. 0.6 times four is 2.4. And finally, in year three, 0.6 times two cubed equals 4.8.

Back to this statement, we know that the number of flies should double every year. From year one to year two in graph A, we went from 1.2 flies to 2.4 flies. And on graph B, we went from one fly to two flies. Both times from year one to year two, they did double. Now we need to check from year two to year three. 2.4 times two does equal 4.8. But this times-two rule does not work for graph B. In graph B, what’s actually happening is that the number of flies are just increasing by one every year. And that means graph B cannot fit the model for the number of flies doubling every year.

So we can circle graph A. It’s also important to note that we could’ve chosen graph A based on the fact that it has two to the 𝑥 power. This two to the 𝑥 power is a function for doubling.

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