Question 979388
The maximum is a vertex,  (4,6) and you can say  {{{y=a(x-4)^2+6}}}.


The given zero should be enough for finding value of a.  Expect to use or find {{{a<0}}}.


{{{y-6=a(x-4)^2}}}
{{{a=(y-6)/(x-4)^2}}}
{{{a=(0-6)/(4+2sqrt(3)-4)^2}}}
{{{a=-6/(2sqrt(3))^2}}}
{{{a=-6/12}}}
{{{a=-1/2}}}


Standard Form function is {{{highlight(f(x)=-(1/2)(x-4)^2+6)}}}.
Simply do the simplification (multiplication...) to put into the general form to identify b and c.
You already know a.