to get the slope of a line all we need is two points off of it, so let's get two points from that table
[tex]\bf ~\hspace{4em}\downarrow ~\hspace{1.5em}\downarrow \\ \begin{array}{|ccccc|} \cline{1-5} x&0&1&2&3\\ \cline{1-5} y&-\frac{1}{3}&\frac{1}{3}&1&\frac{5}{3} \\ \cline{1-5} \end{array}\qquad \qquad (\stackrel{x_1}{1}~,~\stackrel{y_1}{\frac{1}{3}})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{\frac{5}{3}})[/tex]
[tex]\bf \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{\frac{5}{3}}-\stackrel{y1}{\frac{1}{3}}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{1}}}\implies \cfrac{~~\frac{4}{3}~~}{2}\implies \cfrac{~~\frac{4}{3}~~}{\frac{2}{1}}\implies \cfrac{\stackrel{2}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}{3}\cdot \cfrac{1}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies \cfrac{2}{3}[/tex]
