[tex]\bf \boxed{y=a(x-{{ h}})^2+{{ k}}}\\
x=a(y-{{ k}})^2+{{ h}}\qquad\qquad vertex\ ({{ h}},{{ k}})\\\\
-----------------------------\\\\
f(x)=-4x^2\implies
\begin{array}{llll}
y=-4(x-&0)^2+&0\\
&\uparrow &\uparrow \\
&h&k
\end{array}
\\\\\\
\textit{now, you're asked for one with a vertex of -19, -18}\\
\textit{that simply means h = -19, and k= -18}[/tex]
so, just change those h,k coordinates of the center :)
also, we're asked to put it in "standard form", that simply means, expand the squared binomial, and simplify, you'll end up with a trinomial
