f we plot the provided points, we'd end up with something like the picture below, which is a line, mind you a declining line from left to right, so we can pretty see it has a negative slope, to get the slope we can just pick any two given points and check what it might be as well as get its equation, hmmm let's use the points of (4,-8) and (2,8)[tex](\stackrel{x_1}{4}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{8}-\stackrel{y1}{(-8)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{4}}}\implies \cfrac{8+8}{-2}\implies \cfrac{16}{-2}\implies -8 \\\\\\ \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}{(-8)}=\stackrel{m}{-8}(x-\stackrel{x_1}{4})\implies y =-8x+24[/tex]
so is linear because it has a slope (common difference) of -8,