Use the order pairs to write a function rule. Give the tule in slope-intercept form.

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to get the equation of any straight line, we simply need two points off of it, so hmmm let's use say (-1 , -1.25) and (8 , -3.5)

$(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1.25})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-3.5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3.5}-\stackrel{y1}{(-1.25)}}}{\underset{run} {\underset{x_2}{8}-\underset{x_1}{(-1)}}}\implies \cfrac{-3.5+1.25}{8+1}\implies \cfrac{-2.25}{9}\implies -\cfrac{~~ \frac{225}{100}~~}{\frac{9}{1}} \\\\\\ -\cfrac{225}{100}\cdot \cfrac{1}{9}\implies -\cfrac{9}{4}\cdot \cfrac{1}{9}\implies -\cfrac{1}{4}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1.25)}=\stackrel{m}{-\cfrac{1}{4}}(x-\stackrel{x_1}{(-1)}) \\\\\\ y+1.25=-\cfrac{1}{4}(x+1)\implies y+\cfrac{5}{4}=-\cfrac{1}{4}x-\cfrac{1}{4} \\\\\\ y=-\cfrac{1}{4}x-\cfrac{1}{4}-\cfrac{5}{4}\implies y=-\cfrac{1}{4}x-\cfrac{6}{4}\implies y=-\cfrac{1}{4}x-\cfrac{3}{2}$