Answer
$ y+3=-2(x-0) \qquad$ ... point-slope form
$y=-2x -3 \qquad$ ... slope-intercept form
$f(x)=-2x -3 \qquad$ ... function notation
Work Step by Step
$(x_{1},y_{1})=(-4,5) \; (x_{2},y_{2})=(0,-3)\quad$ (the y-intercept)
$m=\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-3-5}{0-(-4)}=\frac{-8}{4}=-2$.
So with $m=-2$ and $(x_{1},y_{1})=(0,-3)$, we write the point-slope form
$y-y_{1}=m(x-x_{1})$
$ y+3=-2(x-0) \qquad$ ... point-slope form
Simplify to slope-intercept form, $ y=mx+b$
$y+3=-2x$ $\qquad$ ... add (-3)
$ y=-2x -3 \qquad$ ... is the slope-intercept form
For function notation, replace $y$ with $f(x)$.
