Find the equation of the linear function k that has k ( 8 ) = 18 and that is parallel to g ( t ) = 5/16t − 40

[tex]\bf g(t)=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{16}}t-40\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]bearing in mind that parallel lines have the same exact slope, so k(x) will have a slope as the one above, we also know that  [tex]\bf k(\stackrel{x}{8})=\stackrel{y}{18}\qquad (8~~,~~18)[/tex].so we're really looking for the equation of a line whose slope is 5/16 and runs through (8,18).[tex]\bf (\stackrel{x_1}{8}~,~\stackrel{y_1}{18})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{5}{16} \\\\\\ \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}{18}=\stackrel{m}{\cfrac{5}{16}}(x-\stackrel{x_1}{8}) \implies y-18=\cfrac{5}{16}x-\cfrac{5}{2} \\\\\\ y=\cfrac{5}{16}x-\cfrac{5}{2}+18\implies y=\cfrac{5}{16}x+\cfrac{31}{2}[/tex]