Question 749162
Since you have a maximum value, your variable, 'a', is negative.  Use standard form.


{{{y=a(x+3)^2+9}}}
but you would use the zeros to help find the value for 'a'.  Right in the middle of -6 and 0 is -3, which is why you have (x+3) as part of the standard form.  Vertex is (-3,9).


The zeros give two more equations which may help in finding 'a'.
(-6,0) gives us {{{0=a(-6+3)^2+9}}}
(0,0) gives us {{{0=a(0+3)^2+9}}}.
What can you do with that?  Can YOU finish this?