Question

Find the rates of change in population for both parks between 2009 and 2014, and determine which park showed faster population growthduring those years.

Answer 1

The table and graph given show the information about two national parks.

It is required to find the rate of change of the population of the two parks between 2009 and 2014, and then determine the park that shows faster population growth during those years.

Notice that the variable x represents the number of years after 2005.

Hence, the years 2009 and 2014 represent, x=4 and x=9, respectively.

The formula for the rate of change of a function f(x) between x=a and x=b is given by the formula:

`(f(b)-f(a))/(b-a)`

Calculate the rate of change for park A:

Substitute a=4 and b=9 into the formula:

`(f(9)-f(4))/(9-4)`

Substitute f(9)=2870 and f(4)=2370 from the table for park A into the expression for the rate of change:

`(2870-2370)/(9-4)=(500)/(5)=100\text{ swallows per year}`

Calculate the rate of change for park B:

`(f(9)-f(4))/(9-4)`

Substitute f(9)=2800 and f(4)=2200 from the graph of park B:

`(2800-2200)/(9-4)=(600)/(5)=120\text{ swallows per year}`

Notice that the rate of change in population for park B is higher than that of park A.

Hence, park B showed faster population growth during those years.

Answers:

Rate of change for park A= 100 swallows per year.

Rate of change for park B= 120 swallows per year.

Park B showed faster population growth during those years.