Question

[PLEASE ANSWER] Precalculus Question (has 2 parts)

Answer 1

part I


`\bf \begin{array}lllllll x&2&\boxed{4}&6&8&\boxed{10}&12&14\\\\ y&1&\boxed{6}&11&16&\boxed{21}&26&31 \end{array}\implies \begin{array}{llll} (4,6)\\\\ (10,21) \end{array}\\\\ -------------------------------\\\\`





part II


`\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad \begin{array}{llll} \textit{plug in the values for } \begin{cases} y_1=6\\ x_1=4\\ m=\boxed{?} \end{cases}\\ \end{array}\\ \left. \qquad \right. \uparrow\\ \textit{point-slope form}`


now, the form above is the point-slope form, and that'd be the equation, you can leave it like so, or you can solve for "y", and put it in slope-intercept, same equation though.