Question 384955
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Hi
the vertex form of a parabola: {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
f(x) = -5x^2 - 60x - 181   Completing the square
f(x) = -5(x^2 + 12x) - 181
f(x) = -5[(x + 6)^2 -36] - 181
f(x) = -5(x + 6)^2 + 180 - 181
f(x) = -5(x + 6)^2 -1
Vertex is Pt (-6,-1)  Note: parabola opens downward as a < 0
{{{graph( 300, 300,-10,10,-10,10,-5x^2 - 60x - 181) }}}