Question 389254
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Hi,         
Find the x-intercepts for the parabola y=-2x^2-4x+6
  -2x^2-4x+6 = 0
  -2(x^2+2x-3)=0 
 (x+3)(x-1)=0 Note:SUM of the inner product(3x) and the outer product(-x) = 2x
 (x+3)= 0
  x = -3
(x-1)=0
  x = 1
please note: Vertex is Pt(-1,8) 
as square completed for y=-2x^2-4x+6  gives y= -2(x+1)^2+ 8
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
{{{drawing(300,300, -10,10,-10,10, grid(1),
circle(-3, 0,0.3),
circle(1, 0,0.3),
circle(-1, 8,0.4),
graph( 300, 300, -10,10,-10,10,  -2x^2-4x+6))}}}