well, let's grab a couple of points off the line hmmmm let's see, the lines runs through (0, 4) and also (3,5), so let's use those to get its slope and thus its function.
[tex]\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) ~\hfill slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-4}{3-0}\implies \cfrac{1}{3}[/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-4=\cfrac{1}{3}(x-0)\implies y-4=\cfrac{1}{3}x \\\\\\ y=\cfrac{1}{3}x+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]