Question 946128
Let's call the middle of the road (0,0).
The other point of the road is (32,-0.04). 
It's negative because it's lower than the center to allow for drainage.
So in vertex form,
{{{y=ax^2}}}
Using the point,
{{{-0.4=a(32)^2}}}
So then,
{{{a=-0.4/32^2}}}
{{{a=-(4/10)(1/1024)}}}
{{{a=-1/2560}}}
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.
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{{{y=-(x^2/2560)}}}