An insect population after `x` months can be modeled by the graph at left. Complete the table below and write the exponential function that describes this scenario. x: 0, 1, 2, 3. y: 12, 18, 27, 41

1 answer

To model the insect population using an exponential function, we can start from the information given:

  • x: 0, 1, 2, 3
  • y: 12, 18, 27, 41

We can observe that the growth appears to be accelerating, which suggests an exponential growth pattern.

The general form of the exponential function is:

\[ y = ab^x \]

where:

  • \( y \) is the population at time \( x \),
  • \( a \) is the initial population (when \( x = 0 \)),
  • \( b \) is the growth factor (how much the population multiplies each time period).

From the data:

  1. When \( x = 0 \), \( y = 12 \). Therefore, \( a = 12 \).

So, we can start with:

\[ y = 12b^x \]

Next, we can find \( b \) using the data points we have. Let's use the point when \( x = 1 \):

\[ y = 18 \] \[ 18 = 12b^1 \] \[ b = \frac{18}{12} = \frac{3}{2} = 1.5 \]

Now, let's check if this value of \( b \) works with the other points.

  • For \( x = 2 \):

\[ y = 12(1.5^2) \] \[ y = 12 \cdot 2.25 = 27 \] (This matches the given data)

  • For \( x = 3 \):

\[ y = 12(1.5^3) \] \[ y = 12 \cdot 3.375 = 40.5 \] (This is approximately 41, which is within rounding error)

The model closely fits the data, so we can conclude that:

The exponential function that describes the insect population over time is:

\[ y = 12(1.5^x) \]

And the completed table reflects the values already provided:

| x | y | |---|----| | 0 | 12 | | 1 | 18 | | 2 | 27 | | 3 | 41 |