to get the equation of any straight line, we simply need two points off of it, let's use the provided ones
[tex]f(5)=-1\implies \underline{(\stackrel{x}{5}~~,~~\stackrel{y}{-1})}\hspace{5em}f(0)=-5\implies \underline{(\stackrel{x}{0}~~,~~\stackrel{y}{-5})} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{0}~,~\stackrel{y_2}{-5}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-5}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{5}}} \implies \cfrac{-5 +1}{-5} \implies \cfrac{ -4 }{ -5 } \implies \cfrac{4 }{ 5 }[/tex]
[tex]\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}{(-1)}=\stackrel{m}{ \cfrac{4 }{ 5 }}(x-\stackrel{x_1}{5}) \implies y +1 = \cfrac{4 }{ 5 } ( x -5) \\\\\\ y-1=\cfrac{4 }{ 5 }x-4\implies {\Large \begin{array}{llll} y=\cfrac{4 }{ 5 }x-3 \end{array}}[/tex]
