Question 390961
  <pre><font size = 3 color = "indigo"><b>
Hi
y=2x^2+4x-6  |determining the x-intercepts (when y=0)
 2x^2+4x-6 = 0
factoring
(2x -2)(x+3)=0 Note:SUM of the inner product(-2x) and the outer product(6x) = 4x
 (2x -2)=0  x = 1  x-intercept at Pt(1,0)
 (x-3)=0    x= -3    x-intercept at Pt(-3,0)

Note: the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
y=2x^2+4x-6   |completing the square to put into vertex form
y = 2(x^2+2)-6
y= 2[(x+1)^2 -1] -6
y= 2(x+1)^2 -8   Vertex is at Pt(-1,-8)  parabola opens upward (2>0)
{{{drawing(300,300, -10,10,-10,10, grid(1),
circle(1, 0,0.3),
circle(-3, 0,0.3),
circle(-1, -8,0.3),
graph( 300, 300, -10,10,-10,10,2x^2+4x-6  ))}}}