<PRE><B>Question 333583</B>
your quadratic equation is {{{y=-0.2x^2 + 12x + 11}}}
---
the vertex can be found where  x=-b/(2a) this comes from the quadratic solution for {{{0=-0.2x^2 + 12x + 11}}} and its the front part of the quadratic solution.
--
so vertex (x,y) is where  x=-b/(2a) and substituting this into {{{y=-0.2x^2 + 12x + 11}}}  to get the y part.
--
from {{{y=-0.2x^2 + 12x + 11}}}
a=-0.2, b=12, c=11
--
x=-b/(2a)= -12/(2*(-0.2))=30
y=-0.2*30^2 +12(30)+11 = -180+360+11=191
vertex = (30,191)
===
you could have also worked {{{y=-0.2x^2 + 12x + 11}}}
into the form  {{{y=d*(x-h)^2+k}}} by completing the square and in this form
the vertex would have been (h,k)

</PRE>