Question

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: 41, 27, 18, 12

Answer 1

To write an exponential function for the given insect population data, we can use the general form of an exponential decay function:

\[ y = a \cdot b^x \]

Where:
- \( y \) is the population at time \( x \),
- \( a \) is the initial population (at \( x = 0 \)),
- \( b \) is the decay factor (the base of the exponential function).

Given the data points:
- At \( x = 0 \), \( y = 41 \)
- At \( x = 1 \), \( y = 27 \)
- At \( x = 2 \), \( y = 18 \)
- At \( x = 3 \), \( y = 12 \)

From this, we find that the initial population \( a = 41 \).

Next, we can find the decay factor \( b \). To find \( b \), we can use the values for \( y \) at \( x = 0 \) and \( x = 1 \):

\[
27 = 41 \cdot b^1
\]

Solving for \( b \):

\[
b = \frac{27}{41} \approx 0.6585
\]

Now, to check the consistency with the other data points:

- For \( x = 2 \):

\[
y = 41 \cdot b^2 \approx 41 \cdot (0.6585)^2 \approx 41 \cdot 0.4346 \approx 17.82 \quad \text{(which is close to 18)}
\]

- For \( x = 3 \):

\[
y = 41 \cdot b^3 \approx 41 \cdot (0.6585)^3 \approx 41 \cdot 0.2868 \approx 11.76 \quad \text{(which is close to 12)}
\]

Now, we can summarize the results into a table and confirm:

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

The exponential function that models this scenario is:

\[
y = 41 \cdot \left(\frac{27}{41}\right)^x
\]

This function describes the population of the insect species over the given time periods.