Question 417091
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Hi
Using the vertex form of a parabola, {{{y=a(x-h)^2 +k}}} where(h,k) is the vertex
the graph of the quadratic 
y = -2x^2 + 3   Vertex is Pt(0,3)  
translated down 4 units  Vertex would be Pt(0,-1)
y = -2x^2 - 1
{{{drawing(300,300,   -6, 6, -6, 6,  grid(1),
circle(0, 3,0.3),
circle(0, -1,0.3),
graph( 300, 300, -6, 6, -6, 6,0,-2x^2 + 3 ,-2x^2 -1 ))}}}