Check the picture below. So the bridge more or less looks like so.
since we know the vertex, we'll use that, and we also know a point on the parabola as well, namely (16,0).
[tex]\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=8\\ k=2 \end{cases}\implies y=a(x-8)^2+2\qquad (16,0)~~ \begin{cases} x=16\\ y=0 \end{cases}[/tex]
[tex]\bf 0=a(16-8)^2+2\implies -2 = a(8)^2\implies -2=64a \\\\\\ \cfrac{-2}{64}=a\implies \cfrac{-1}{32}=a \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{32}(x-8)^2+2~\hfill[/tex]
