Let the[tex]x-axis[/tex] presents the time, and the [tex]y-axis [/tex] the gallons of water remaining, so our first point will be (4,800000). We know from the problem that after 4 hours, at 8:00 am, there were 100000 gallons, so our second point will be (8,100000).
To relate those two points we are going to use the point slope formula:
[tex]m= \frac{y_{2}-y_{1} }{x_{2}-x_{1} } [/tex]
where
[tex]x_{2}[/tex] is 8
[tex]x_{1}[/tex] is 4
[tex]y_{1}[/tex] is 800000
[tex]y_{2}[/tex] is 100000
So, [tex]m= \frac{100000-800000}{8-4} [/tex]
[tex]m= \frac{-700000}{4} [/tex]
[tex]m=-175000[/tex]
Now that we have our slope, we are going to use the point-slope formula to find our equation:
[tex]y-y_{1} =m(x-x_{1} )[/tex]
[tex]y-800000=-175000(x-4)[/tex]
[tex]y-800000=-175000x+700000[/tex]
[tex]y=-175000x+1500000[/tex]
The equation that describes the number of gallons of water as a function of the time the field has been irrigated is [tex]y=-175000x+1500000[/tex]
where
[tex]y[/tex] are the remaining gallons of water in the reservoir after [tex]x[/tex] hours
[tex]x[/tex] is the time the field has been irrigated in hours
