so, to get the equation of a straight line, we simply can use two points it passes through, now, from the picture, we can say hmmm it passes through (10, 300) and (15, 375), so let's use those two points.
[tex]\bf (\stackrel{x_1}{10}~,~\stackrel{y_1}{300})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{375}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{375-300}{15-10}\implies \cfrac{75}{5}\implies 15 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-300=15(x-10)\implies y-300=15x-150[/tex]
[tex]\bf y=\underset{constant}{\stackrel{coefficient}{\stackrel{\downarrow }{15}x+\underset{\uparrow }{150}}}\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
what does the constant represent?
it represents as you see there, the y-intercept or where the graph touches the y-axis, at y = 150 and x = 0, at (0, 150).
at 0 linear feet, the cost is $150, meaning without even putting a foot of fencing, he's getting charged $150, which is most likely a fixed cost the installer is charging, usually for tooling and other equipment he has to use.
what does the coefficient represent?
well, is the slope, or average rate of change, in this case 15, or 15/1.
meaning for every 1 linear foot added, the charge is 15, or each linear foot of fencing has a cost of $15.
