Question 486883
When the equation is in the form 
{{{ f(x) = ax^2 + bx + c }}}, the vertex 
is at {{{ x = -b/(2a) }}}
{{{ f(x)= (1/2)*x^2 + 6x + 11}}}
{{{ -b/(2a) = -6/(2*(1/2)) }}}
{{{ -b/(2a) = -6 }}}
Now put this value of {{{ x }}} into the
equation to find {{{ y }}}, or {{{ f(x) }}}.
{{{ f(x)= (1/2)*(-6)^2 + 6*(-6) + 11}}}
{{{ f(x) = 18 - 36 + 11 }}}
{{{ f(x) = -7 }}}
The vertex is at (-6, -7)
Here is a plot:
{{{ graph( 400, 400, -14, 6, -10, 10, (1/2)*x^2 + 6x + 11) }}}