[tex]\bf \begin{array}{|c|c|l} \cline{1-2} \stackrel{\textit{years after 2000}}{x}&\stackrel{\textit{millions of volunteers}}{y}\\ \cline{1-2} 4&6.1\\ 10&11.1 \\ \cline{1-2} \end{array} \qquad \qquad (\stackrel{x_1}{4}~,~\stackrel{y_1}{6.1})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{11.1})[/tex]
[tex]\bf slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{11.1-6.1}{10-4}\implies \cfrac{5}{6} \\\\\\ \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-6.1=\cfrac{5}{6}(x-4)\implies y-6.1=\cfrac{5}{6}x-\cfrac{10}{3} \\\\\\ y=\cfrac{5}{6}x-\cfrac{10}{3}+6.1\implies y=\cfrac{5}{6}x-\cfrac{10}{3}+\cfrac{61}{10}\implies \blacktriangleright y=\cfrac{5}{6}x+\cfrac{83}{30} \blacktriangleleft[/tex]
now, for 2009, that's 9 years after 2000 so x = 9, and 2014, well clearly x = 14, so
[tex]\bf \stackrel{x = 9}{y=\cfrac{5}{6}(9)+\cfrac{83}{30}}\implies y=\cfrac{15}{2}+\cfrac{83}{30}\implies y=\cfrac{154}{15}\implies y=10\frac{4}{15} \\\\\\ \stackrel{x = 14}{y=\cfrac{5}{6}(14)+\cfrac{83}{30}}\implies y=\cfrac{35}{3}+\cfrac{83}{30}\implies y=\cfrac{433}{30}\implies y=14\frac{13}{30}[/tex]
