Answer
$ y-9=-\displaystyle \frac{11}{3}(x+1) \qquad$ ... point-slope form
$y=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... slope-intercept form
$f(x)=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... function notation
Work Step by Step
$(x_{1},y_{1})=(1,9) \; (x_{2},y_{2})=(4,-2)$
$m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-9}{4-1}=\frac{-11}{3}=-\frac{11}{3}$.
So with $m=-\displaystyle \frac{11}{3}$ and $(x_{1},y_{1})=(1,9)$, we write the point-slope form
$y-y_{1}=m(x-x_{1})$
$ y-9=-\displaystyle \frac{11}{3}(x-1) \qquad$ ... point-slope form
Simplify to slope-intercept form, $ y=mx+b$
... distribute
$ y-9=-\displaystyle \frac{11}{3}x+\frac{11}{3} \qquad$ ...add $9$
$ y=-\displaystyle \frac{11}{3}x+\frac{38}{3} \qquad$ ... is the slope-intercept form
For function notation, replace $y$ with $f(x)$.
