by **Integr88** » Thu 07 April, 2011 20:25

we have find the function. Because it is liner, we have to represent it as y=mx+b

we have:

at 2003, 6400

at 2007, 13168

so `x_1=2003`, `y_1=6400`

and `x_2=2007`, `y_2=13168`

to get slope m we have to use:

`m=(y2-y1)/(x2-x1)` sub in the values `m=(13168-6400)/(2007-2003)=6768/4=1692`

so the slope m=1692

when know the initial data that we have at 2003 which is the begining, we can set x=0, y=6400 we had 6400. Using this information

y=mx+b

`6400=(1692)(0)+b`

`6400=b` so we found b=6400

our linear equation is y=1692t+6400