Question

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated that the population will be 1,000. After 3 years, it is estimated that the population will be 640. a. Write a function to model this scenario. b. Create a graph to show the bee population over the next 10 years. c. Identify the key features of the function. Identify the x- and y-intercepts. Determine the maximum, the minimum, whether the function is increasing or decreasing, the rate of change of the function over the interval [0, 10], and any asymptotes.

Answer 1

To model this situation, we are going to use the decay formula:
`A=Pe^(rt)`
where

`A` is the final pupolation

`P` is the initial population

`e` is the Euler's constant

`r` is the decay rate

`t` is the time in years

A. We know for our problem that the initial population is 1,250, so
`P=1250`; we also know that after a year the population is 1000, so
`A=1000` and
`t=1`. Lets replace those values in our formula to find
`r`:

`A=Pe^(rt)`

`1000=1250e^(r)`

`e^(r)= (1000)/(1250)`

`e^(r)= (4)/(5)`

`ln(e^(r))=ln( (4)/(5) )`

`r=ln( (4)/(5) )`

`r=-02231`

Now that we have
`r`, we can write a function to model this scenario:

`A(t)=1250e^(-0.2231t)`.

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:

`A(0)=1250e^{(-0.2231)(0)`

`A_(0)=1250e^(0)`

`A_(0)=1250`
- Over the interval [0,10], the function will have a minimum at t=10:

`A(10)=1250e^{(-0.2231)(10)`

`A_(10)=134.28`

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula:
`m= (A(0)-A(10))/(10-0)`
where

`m` is the rate of change

`A(10)` is the function evaluated at 10

`A(0)` is the function evaluated at 0
We know from previous calculations that
`A(10)=134.28` and
`A(0)=1250`, so lets replace those values in our formula to find
`m`:

`m= (134.28-1250)/(10-0)`

`m= (-1115.72)/(10)`

`m=-111.572`
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.