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Sagot :
to get the equation of any straight line, we simply need two points off of it, let's use the provided points in the picture above.
[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{(-3)}}}{\underset{run} {\underset{x_2}{1}-\underset{x_1}{0}}} \implies \cfrac{1 +3}{1 +0}\implies 4 \\\\\\ \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}{(-3)}=\stackrel{m}{4}(x-\stackrel{x_1}{0}) \\\\\\ y+3=4x\implies y=4x-3[/tex]
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