Question

X y

-1 -10
3 14
Complete the slope-intercept form of the linear equation that represents the relationship in the table.

Answer 1

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

`\begin{array} \cline{1-2} x&y\\ \cline{1-2} -1&-10\\ 3&14\\ \cline{1-2} \end{array}\hspace{5em} (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-10})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{14}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{14}-\stackrel{y1}{(-10)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{(-1)}}} \implies \cfrac{14 +10}{3 +1}\implies \cfrac{24}{4}\implies 6`

`\begin{array}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}{(-10)}=\stackrel{m}{6}(x-\stackrel{x_1}{(-1)}) \\\\\\ y+10=6(x+1)\implies y+10=6x+6\implies y=6x-4`