Question

Find a linear function h, given h(3) = - 2 and h(- 3) = 16. Then find h(4).

Answer 1

well, from h(3) = -2 we can see the point of (3 , -2) and from h(-3) = 16 we can see the point of (-3 , 16).

To get the equation of any straight line, we simply need two points off of it, let's use those two given.

`(\stackrel{x_1}{3}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{-3}~,~\stackrel{y_2}{16}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{\textit{\large rise}} {\stackrel{y_2}{16}-\stackrel{y1}{(-2)}}}{\underset{\textit{\large run}} {\underset{x_2}{-3}-\underset{x_1}{3}}} \implies \cfrac{ 16 +2 }{ -6 } \implies \cfrac{ 18 }{ -6 } \implies -3`

`\begin{array}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}{(-2)}=\stackrel{m}{-3}(x-\stackrel{x_1}{3}) \implies y +2 = -3 ( x -3) \\\\\\ y+2=-3x+9\implies y=-3x+7\hspace{5em}\boxed{h(x)=-3x+7} \\\\\\ h(4)=-3(4)+7\implies h(4)=-5`