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Josh Strater · Answer 1 · 2018-03-02T23:46:50+0000

part A)

$\bf \begin{array}{ccllll} hours(x)&velocity(y)\\ -----&-----\\ 1&6\\ 3&2 \end{array}\\\\ -------------------------------\\\\$

$\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ 1}}\quad ,&{{ 6}})\quad % (c,d) &({{ 3}}\quad ,&{{ 2}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{2-6}{3-1}\implies \cfrac{-4}{2}\implies -2$

$\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-6=-2(x-1)\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form} \\\\\\ y-6=-2x+2\implies y=-2x+2+6\implies \boxed{y=-2x+8}$

part B)

well, we know y = -2x+8.... so.. what's the runner's velocity after 5hours? well, x = 5, thus y = -2(5) +8 ---> y = -2

to graph it, well, is a LINEar equation, meaning the graph is a LINE, and to graph a line, all you need is two points, and by now, you have more than two.. so graph it away.

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