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Sagot :
keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above
[tex]y=\stackrel{\stackrel{m}{\downarrow }}{4} x-4\qquad \impliedby \qquad \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 we're really looking for the equation of a line whose slope is 4 and passes through (-2 , 0)
[tex](\stackrel{x_1}{-2}~,~\stackrel{y_1}{0})\qquad \qquad \stackrel{slope}{m}\implies 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}{0}=\stackrel{m}{4}(x-\stackrel{x_1}{(-2)}) \\\\\\ y=4(x+2)\implies y=4x+8[/tex]
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