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Write a linear function f with values f(-3)=5 and f(6)=11
f(x)=

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to get the equation of any straight line, we simply need two points off of it, so let's use the provided ones


f(-3)=5\implies \begin{cases} x=-3\\ y=5 \end{cases}\hspace{5em}f(6)=11\implies \begin{cases} x=6\\ y=11 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{11}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{11}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-3)}}} \implies \cfrac{6}{6 +3} \implies \cfrac{ 6 }{ 9 } \implies \cfrac{2 }{ 3 }


\begin{array}c \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}{5}=\stackrel{m}{ \cfrac{2 }{ 3 }}(x-\stackrel{x_1}{(-3)}) \implies y -5 = \cfrac{2 }{ 3 } ( x +3) \\\\\\ y-5=\cfrac{2}{3}x+2\implies {\Large \begin{array}{llll} y=\cfrac{2}{3}x+7 \end{array}}

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