check the picture below
so hmmm the left-side of the picture, is the focus point and the directrix, notice, the directrix is a horizontal line, that means the parabola is a vertically opening one.
also notice, the focus point is below the directrix, that means the parabola opens downwards
now, the vertex is "p" distance from either the directrix and the focus point, so that means, the vertex is half-way between those two fellows, if you notice the right-side of the picture, that's where the vertex is at.
the distance "p" is clearly just 1unit, however, since the parabola is opening downwards, "p" is negative, thus -1
[tex]\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\
\end{array}
\qquad
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-------------------------------\\\\[/tex]
[tex]\bf \begin{cases}
h=8\\
k=-7\\
p=-1
\end{cases}\implies (x-8)^2=4(-1)(y-(-7))
\\\\\\
(x-8)^2=-4(y+7)\implies \cfrac{(x-8)^2}{-4}=y+7
\\\\\\
-\cfrac{1}{4}(x-8)^2-7=y[/tex]
