Question 633082
The vertex of the quadratic function {{{y = a(x-h)^2 + k }}}is (h,k).
The line of symmetry is {{{x = h}}} and the maximum or minimum value is k.
Therefore, in {{{f(x) = -(x+1)^2-9}}}, the vertex is (-1,-9).
The line of symmetry is {{{highlight(x = -1)}}} and the maximum value is {{{highlight(-9)}}}.
The graph has a maximum value because a < 0.
here's the graph:
{{{drawing(500,500,-18,7,-15,2,grid(1),graph( 500, 500, -18,7,-15,2, -(x+1)^2-9)) }}}