Question 364579
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Hi,
*Note: vertex form of a parabola, 
{{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
vertex is (-2,8)
y = a(x+2)^2 + 8
y -intercept is 0. (0,0) a point on the parabola
0 = 4a + 8
-2 = a
y = -2(x+2) + 8
{{{graph( 300, 300,-10,10,-10,10,-2(x+2)^2 +8) }}}