Question

Question 9 (5 points)

Answer 1

to get the equation of any straight line, we simply need two points off of it, let's use those two points from the table in the picture below.

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{13})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-17}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-17}-\stackrel{y1}{13}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-4)}}} \implies \cfrac{-30}{6 +4} \implies \cfrac{ -30 }{ 10 }\implies -3$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{13}=\stackrel{m}{-3}(x-\stackrel{x_1}{(-4)}) \\\\\\ y-13=-3(x+4)\implies y-13=-3x-12\implies y=-3x+1$