We will consider x on the graph to be years since 1990 and y on the graph to be the number of Moose.

We have two points on the graph: (4, 4290) and (8, 3850)

Using the slope formula (m = [y_{2} - y_{1}]/[x_{2} - x_{1}]), we can find the slope of this line:

m = (y_{2} - y_{1})/(x_{2} - x_{1}) Slope formula

m = (3850 - 4290)/(8 - 4) Substitution

m = -440/4 Solve the subtraction problems in the parenthesis

m = -110 Reduce the fraction

Now that we have the slope of the line, we can use the point-slope formula (y - y_{1} = m[x - x_{1}]) to find the equation for the line. I will use the first point, but either point will produce the same answer.

y - y_{1} = m(x - x_{1}) Point-slop e formula

y - 4290 = -110(x - 4) Substitution

y - 4290 = -110x + 440 Distribute the -110

y - 4290 + 4290 = -110x + 440 + 4290 Add 4290 to each side

y = -110x + 4730 Simplify

Since the formula that is being sought is supposed to be interms of P and t, I will replace y with P and x with t.

P(t) = -110t + 4730

To predict the population in 2006, we need to determine the years from 1990 to 2006.

2006 - 1990 = 16

P(16) = -110 (16) + 4730

P(16) = -1760 + 4730

P(16) = 2970