Question 379726
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Hi
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y= -1/3(x+6)^2+3  
vertex is Pt(-6,3) negative 'a' of -(1/3) tells us the parabola opens downward
y intercept (when x = 0)
y = -1/3(6)^2 + 3
y = -12 + 3 = -9  Pt(0,-9)
x intercepts (when y = 0)
0 = -1/3(x+6)^2+3  
0 = -(x+6)^2 + 9
(x+6)^2 = 9
(x+6) = ±3
x +6 = 3    x = -3  Pt(-3,0)
x +6 = -3   x = -9  Pt(-9,0)
{{{drawing(300,300,-10,10,-10,10,blue(line(-6,10,-6,-10)),grid(1),graph(300,300,-10,10,-10,10, -x^2/3 -4x -9  ))}}}