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

1 answer

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.