Question

What do I do for this because I don’t think my Equation

Answer 1

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

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{16}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{16}-\stackrel{y1}{12}}}{\underset{\textit{\large run}} {\underset{x_2}{-4}-\underset{x_1}{(-6)}}} \implies \cfrac{4}{-4 +6} \implies \cfrac{ 4 }{ 2 } \implies 2`

`\begin{array} \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}{12}=\stackrel{m}{ 2}(x-\stackrel{x_1}{(-6)}) \implies y -12 = 2 ( x +6) \\\\\\ y-12=2x+12\implies {\Large \begin{array}{llll} y=2x+24 \end{array}}`