In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the population continues to change linearly: A.) Find a formula for the moose population, P , in terms of t , the years since 1990. P ( t ) = B.) What does your model predict the moose population to be in 2008?
<u>Answer:</u>
Model predict the moose population to be in 2008 is 1860
<u>Solution:</u>
Let us consider this problem on graph, taking the x axis to be years since 1990 and y on the graph be the number of moose
We have two points on the graph: (2, 4260) and (6, 3660)
Using the slope formula we can find the slope :
![\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%5Cmathrm%7Bm%7D%3D%5Cfrac%7By_%7B2%7D-y_%7B1%7D%7D%7Bx_%7B2%7D-x_%7B1%7D%7D)
![m=\frac{(3660-4260)}{(6-2)}](https://tex.z-dn.net/?f=m%3D%5Cfrac%7B%283660-4260%29%7D%7B%286-2%29%7D)
m = -150
Now that we have the slope of the line, we can use the point-slope formula to find the equation for the line as follows:-
![\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-4260=-150(x-2)} \\\\ {y-4260=-150 x+300} \\\\ {y=-150 x+4560}\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bl%7D%7By-y_%7B1%7D%3Dm%5Cleft%28x-x_%7B1%7D%5Cright%29%7D%20%5C%5C%5C%5C%20%7By-4260%3D-150%28x-2%29%7D%20%5C%5C%5C%5C%20%7By-4260%3D-150%20x%2B300%7D%20%5C%5C%5C%5C%20%7By%3D-150%20x%2B4560%7D%5Cend%7Barray%7D)
Since the formula that is being sought is supposed to be interms of P and t, we will replace y with P and x with t
P(t) = -150t + 4560
Number of years from 1990 to 2008 is
2008 - 1990 = 18
So, population in 2008 will be
P(18) = -150 (18) + 4560
P(18) = -2700 + 4560
P(18) = 1860
Thus the model to predict the moose population in 2008 is found