Question 652637
graph the horizontal parabola and give domain and range: 
-1/2x=(y+3)^2
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This is an equation of a parabola that opens leftwards.
Its standard form: (y-k)^2=-4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given problem:
(y+3)^2=1/2x
vertex: (0,-3)
Domain: (-∞, 0]
Range: (-∞,∞)
x-intercept:
set y=0
3^2=x/2
x=18
axis of symmetry: y=-3
I don't have the means to graph this parabola, but you should be able to do it with the information provided.