This is a quadratic equation, with a vertex at the top.

#(x+2)^2 = -8(y-1)# --> Expand the left hand side

#x^2+4x+4 = -8(y-1)# --> Divide both sides by -1/8

#-1/8x^2-1/2x-1/2= y-1# --> Add one to both sides

#-1/8x^2-1/2x-1/2+1= y# --> Simplify

#y=-1/8x^2-1/2x+1/2#

Apply the equation for the #x#-value of the vertex of a parabola. For a quadratic of the form #y=ax^2+bx+c#, the #x#-value of the vertex is

#x=-b/(2a)#

For #b=-1/2# and #a=-1/8#, we have vertex #x=-2#. Plugging #x=-2# into #y=-1/8(-2)^2-1/2(-2)+1/2# gives #y=1#. So the vertex is #(-2,1)#.

Graphically,

graph{-1/8x^2-.5x+.5 [-10, 10, -5, 5]}