to get the equation of any straight line, we simply need two points off of it, let's use those points in the graph
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{2}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-4)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{0}}}\implies \cfrac{2+4}{4}\implies \cfrac{6}{4}\implies \cfrac{3}{2}[/tex]
[tex]\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-\stackrel{y_1}{(-4)}=\stackrel{m}{\cfrac{3}{2}}(x-\stackrel{x_1}{0}) \\\\\\ y+4=\cfrac{3}{2}x\implies y=\cfrac{3}{2}x-4[/tex]
