Question 391728
  <pre><font size = 3 color = "indigo"><b>
Hi
the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} 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
{{{drawing(300,300, -6, 6, -6, 6, grid(1),
circle(-2, -1.5,0.3),
circle(0, -3.5,0.3),
graph( 300, 300, -6, 6, -6, 6,-.5x^2 - 2x - 7/2
 ))}}}