keeping in mind that the vertex is "p" distance from either the focus or directrix, that means is half-way between both, check the picture below.
from -1 to -1/2 the distance is just 1/2 and half of that is 1/4.
now, if we go down from the directrix which is at -1/2, by 1/4 of a unit, we'll end up at
[tex] \bf \cfrac{1}{2}+\cfrac{1}{4}\implies \cfrac{2+1}{4}\implies \cfrac{3}{4}\qquad \qquad vertex~\left(\stackrel{h}{2}~,~\stackrel{k}{-\frac{3}{4}} \right) [/tex]
the distance "p" is obviously 1/4, however, since the parabola is opening downwards, is negative, therefore -1/4.
[tex] \bf \textit{parabola vertex form with focus point distance}\\\\\begin{array}{llll}4p(x- h)=(y- k)^2\\\\\boxed{4p(y- k)=(x- h)^2}\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=2\\k=-\frac{3}{4}\\\\p=-\frac{1}{4}\end{cases}\implies 4\left( -\frac{1}{4} \right)\left[y- \left( -\frac{3}{4} \right) \right]=(x-2)^2\\\\\\-1\left( y+\frac{3}{4} \right)=(x-2)^2\implies y+\cfrac{3}{4} =-(x-2)^2\\\\\\y=-(x-2)^2-\cfrac{3}{4} [/tex]
