well, we know it's a line, and to get the equation of any line all we need is two points, let's pick two from the table hmmmm say let's use (-3 , -7) and (3 , 11)
[tex](\stackrel{x_1}{-3}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{11}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{11}-\stackrel{y1}{(-7)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-3)}}}\implies \cfrac{11+7}{3+3}\implies \cfrac{18}{6}\implies 3[/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}{(-7)}=\stackrel{m}{3}(x-\stackrel{x_1}{(-3)}) \\\\\\ y+7=3(x+3)\implies y+7=3x+9\implies y = 3x+2[/tex]
