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,  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)



 

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