212k views
2 votes
Find the equation of the linear function k that has k ( 8 ) = 18 and that is parallel to g ( t ) = 5/16t − 40

User MJimitater
by
8.6k points

1 Answer

3 votes


\bf g(t)=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{5}{16}}t-40\qquad \impliedby \begin{array}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}

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
\bf k(\stackrel{x}{8})=\stackrel{y}{18}\qquad (8~~,~~18).

so we're really looking for the equation of a line whose slope is 5/16 and runs through (8,18).


\bf (\stackrel{x_1}{8}~,~\stackrel{y_1}{18})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{5}{16} \\\\\\ \begin{array} \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}

User Crackerjack
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories