I require help graphing the parabola y^2=-4x. Also, what will the focus be?
TIA
=============================================
300C,
standard from is x = a [( y-k)^2] + h
If a<0, it opens to left horizontally.
where, Focus point is { [ h+ (1/4a)], k }
Rewrite your question to standard from and get
x = (- 1/4) [ ( y-0)^2 ] + 0
where you see a = -1/4, k=0, h=0
Now Focust point is ([ h+ (1/4a)] ,0 )
Solve [ h+ 1/(4a) ] by substitue a = -1/4 and get -1
so Focusto points =( -1, 0 )