to get the equation of any straight line, we simply need two points off of it, so hmm let's use the two points you see in the picture below from the provided table.
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{5}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{5}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{1}}}\implies \cfrac{5+1}{3}\implies 2 \\\\\\ \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}{(-1)}=\stackrel{m}{2}(x-\stackrel{x_1}{1}) \\\\\\ y+1=2x-2\implies y=2x-3[/tex]
