to get the equation of any straight line, we simply need two points off of it, so for the equation in the table, let'use the points in the picture below
[tex](\stackrel{x_1}{1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{2}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{2}-\stackrel{y1}{(-2)}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{2+2}{2}\implies 2 \\\\\\ \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}{(-2)}=\stackrel{m}{2}(x-\stackrel{x_1}{1})\implies y+2=2x-2[/tex]
[tex]y=2x\stackrel{\stackrel{b}{\downarrow }}{-4}\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
so the equation in the table has a y-intercept of -4.
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]y+\cfrac{1}{4}x=2\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{1}{4}} x+2\qquad \impliedby \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}[/tex]
so hmmm we're really looking for an equation whose slope is -1/4 and with a y-intercept greater than -4, hmmm let's recall something, that on the negative side of the number line, the closer to 0, the greater.
that said, on the negative side of the number line -10 is much larger than -1,000,000, because -10 is closer than -1,000,000 to 0, and so on.
so hmmm what's greater than -4? hmm well, heck let's use -2
[tex]y = -\cfrac{1}{4}x - 2[/tex]
