Hi

the vertex form of a parabola,  where(h,k) is the vertex

x2+4x+2y-7=0   OR y = (1/2)x^2 - 2x - 7/2

  2y = -x^2 - 4x - 7

   y = -(1/2)x^2 - 2x - 7/2

Completing the square to put into vertex form

 y = -(1/2)[x^2+4x] - 7/2

 y = -(1/2)[(x+2)^2 - 4] -7/2

 y = -(1/2)(x+2)^2   +2 -7/2

 y = -(1/2)(x+2)^2 - 3/2

Vertex is Pt(-2,-3/2)  This is a maximum point, parabola opens downward -(1/2) < 0



 

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