2 part algebra question

Question

In 1993, the moose population in a park was measured to be 4720. By 1996, the population was measured again to be 5200. If the population continues to change linearly:

Find a formula for the moose population, P, in terms of t, the years since 1990.

P(t)=__________________  

What does your model predict the moose population to be in 2004?____________

Answer 1

(3, 4720) abd (6,5200) are two points on a line

the line has slope

= (5200-4720)/(6-3) = 480/3 = 160 = (y-4720)/(x-3)

y = 160x - 480 +4720

P(t) = 160t + 4240

P(14) = 160(14) + 4240 = 2240 + 4240 =6480 moose in 2004

Answer 2

The key is that you are to assume the population varies linearly. That means you will be able to model the population with a line using either the slope-intercept form: y = mx + b or point-slope form: y - y₁ = m(x - x₁)

Using years after 1990, the 1993 population of 4720 is the point (3, 4720) and the 1996 population of 5200 is the point (6, 5200).

First calculate the slope (m) using the slope formula:

where x₁ = 3, y₁ = 4720, x₂ = 6, and y₂ = 5200

m = (5200 - 4720)/(6 - 3) = 160

Now, use the point-slope form:

y - y₁ = m(x - x₁)

where m = 160, x₁ = 3, and y₁ = 4720:

y - 4720 = 160(x - 3)

y - 4720 = 160x - 480

y = 160x + 4240

In our case, the "x" is the time (or should be "t") and the "y" is the population (or should be P(t)), so:

P(t) = 160t + 4240

Since 2004 is 14 years after 1990, plug in t = 14 to get your estimate of the population in 2004