2.3 Modeling with linear functions  (Page 3/8)

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the $y\text{-}$ intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004:

• Input: $t,$ years since 2004
• Output: $P(t),$ the town’s population

To predict the population in 2013 $(t=9),$ we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to $t=0,$ giving the point $\left(0,\text{6200}\right).$ Notice that through our clever choice of variable definition, we have “given” ourselves the y -intercept of the function. The year 2009 would correspond to $t=\text{5,}$ giving the point $\left(5,\text{8100}\right).$

The two coordinate pairs are $\left(0,\text{6200}\right)$ and $\left(5,\text{8100}\right).$ Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

We already know the y -intercept of the line, so we can immediately write the equation:

To predict the population in 2013, we evaluate our function at $t=9.$

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set $P(t)=15000$ and solve for $t.$

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.