Question 706119
graph x=(y+1)^2-4
This is an equation of a parabola that opens right.
Its standard form: (y-k)^2=4p(x-h), (h,k)=(x,y) coordinates of the vertex
For given equation: x=(y+1)^2-4
rewrite: (y+1)^2=(x+4)
vertex: (-4,-1)
y-intercepts
set x=0
(y+1)^2=4
y+1=±√4=±2
y=-1±2
y=-3 and 1
See graph below:
y=-1+(x+4)^.5
{{{ graph( 300, 300, -10, 10, -10, 10,-1+(x+4)^.5,-1-(x+4)^.5 ) }}}