let's notice something, the parabola is a vertical one, so the squared variable is the x, and is opening downwards, meaning the x² will have a negative coefficient.
the distance from the vertex to the directrix/focus is the amount of "p" units, let's see in the graph, the distance from the vertex to the directrix is 2, and since the parabola is opening downwards, "p" is a negative 2, p = -2. The vertex is of course at (0, 2).
[tex] \bf \textit{parabola vertex form with focus point distance}
\\\\
4p(y- k)=(x- h)^2
\qquad
\begin{array}{llll}
vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\begin{cases}
h=0\\
k=2\\
p=-2
\end{cases}\implies 4(-2)(y-2)=(x-0)^2\implies -8(y-2)=x^2
\\\\\\
y-2=\cfrac{x^2}{-8}\implies \blacktriangleright y=-\cfrac{1}{8}x^2+2 \blacktriangleleft [/tex]
