well for a line, to get its slope all we need is two points, so let's use (-6, 5) and (0, 1), and get the equation of it.
[tex]\bf (\stackrel{x_1}{-6}~,~\stackrel{y_1}{5})\qquad
(\stackrel{x_2}{0}~,~\stackrel{y_2}{1})
\\\\\\
slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-5}{0-(-6)}\implies \cfrac{1-5}{0+6}\implies -\cfrac{5}{6}[/tex]
[tex]\bf \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-5=-\cfrac{5}{6}[x-(-6)]\implies y-5=-\cfrac{5}{6}(x+6)
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\textit{to get the x-intercept, we set y = 0, solve for \underline{x}}
\\\\\\
0-5=-\cfrac{5}{6}(x+6)\implies -30=5x+30\implies -60=5x
\\\\\\
\cfrac{-60}{5}=x\implies -12=x~\hfill \boxed{\stackrel{x-intercept}{(-12,0)}}[/tex]
now, where's the y-intercept of that line? well, to get the y-intercept, we set x = 0 and solve for "y"....hmmmm wait a second, notice (0, 1), x = 0, y = 1, that's the y-intercept already.
