Answer
$$\eqalign{
& {\text{Vertex }}\left( {0,0} \right) \cr
& \,\,\,\,\,\,\,{\text{Focus: }}\left( {0,p} \right):\,\,\left( {0, - \frac{1}{2}} \right) \cr
& \,\,\,\,\,\,\,{\text{Directrix:}}\,\,y = - p:\,\,\,\,\,y = \frac{1}{2} \cr
& \,\,\,\,\,\,{\text{Vertical axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\,\,\,\,\left( { - \infty ,0} \right] \cr} $$
Work Step by Step
$$\eqalign{
& {x^2} + 2y = 0 \cr
& {x^2} = - 2y \cr
& {\text{The equation is written in the form }}{x^2} = 4py \cr
& {x^2} = - 2y \to \,\,\,\,\,\, - 2 = 4p\,\,\,\,\,\,\,\,p = - \frac{1}{2} \cr
& {\text{With: }} \cr
& \,\,\,\,\,\,\,{\text{Vertex }}\left( {0,0} \right) \cr
& \,\,\,\,\,\,\,{\text{Focus: }}\left( {0,p} \right):\,\,\left( {0, - \frac{1}{2}} \right) \cr
& \,\,\,\,\,\,\,{\text{Directrix:}}\,\,y = - p:\,\,\,\,\,y = \frac{1}{2} \cr
& \,\,\,\,\,\,{\text{Vertical axis of symmetry}} \cr
& \,\,\,\,\,\,\,{\text{Domain: }}\left( { - \infty ,\infty } \right) \cr
& \,\,\,\,\,\,\,{\text{Range:}}\,\,\,\,\left( { - \infty ,0} \right] \cr} $$
