This is a quadratic equation, with a vertex at the top.
(x+2)^2 = -8(y-1)(x+2)2=−8(y−1) --> Expand the left hand side
x^2+4x+4 = -8(y-1)x2+4x+4=−8(y−1) --> Divide both sides by -1/8
-1/8x^2-1/2x-1/2= y-1−18x2−12x−12=y−1 --> Add one to both sides
-1/8x^2-1/2x-1/2+1= y−18x2−12x−12+1=y --> Simplify
y=-1/8x^2-1/2x+1/2y=−18x2−12x+12
Apply the equation for the xx-value of the vertex of a parabola. For a quadratic of the form y=ax^2+bx+cy=ax2+bx+c, the xx-value of the vertex is
x=-b/(2a)x=−b2a
For b=-1/2b=−12 and a=-1/8a=−18, we have vertex x=-2x=−2. Plugging x=-2x=−2 into y=-1/8(-2)^2-1/2(-2)+1/2y=−18(−2)2−12(−2)+12 gives y=1y=1. So the vertex is (-2,1)(−2,1).
Graphically,
graph{-1/8x^2-.5x+.5 [-10, 10, -5, 5]}
