Check the picture below, so the parabola looks more or less like so, with the vertex half-way from the focus point and the directrix.
Now, the distance from the directrix to the focus point is only 1 unit, so half that, or namely the "p" distance is 1/2 unit, and since the parabola is opening downwards, "p" is negative, so
[tex]\textit{vertical parabola vertex form with focus point distance} \\\\ 4p(y- k)=(x- h)^2 \qquad \begin{cases} \stackrel{vertex}{(h,k)}\qquad \stackrel{focus~point}{(h,k+p)}\qquad \stackrel{directrix}{y=k-p}\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix}\\\\ \stackrel{"p"~is~negative}{op ens~\cap}\qquad \stackrel{"p"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\begin{cases} h=0\\ k=-\frac{1}{2}\\[1em] p=-\frac{1}{2} \end{cases}\implies 4\left( -\frac{1}{2} \right)\left( ~~y-\left( -\frac{1}{2} \right) ~~ \right)~~ = ~~(x-0)^2 \\\\\\ -2\left( y+\frac{1}{2} \right)=x^2\implies y+\cfrac{1}{2} =\cfrac{x^2}{-2}\implies y = -\cfrac{1}{2}x^2-\cfrac{1}{2}[/tex]
now, we could also write that as either
[tex]y = -\cfrac{1}{2}(x^2+1)\qquad or\qquad y = \cfrac{1}{2}(-x^2-1)[/tex]
