Question 854644
Find the vertex, focus, and directrix of the parabola. Graph the equation. 
y^2 - 2y = 8x - 1
complete the square:
(y^2-2y+1) = 8x - 1+1
(y-1)^2=8x
This is an equation of a parabola that opens right.
Its basic form of equation: (y-k)^2=4p(x-h)^2, (h,k)=coordinates of the vertex
For given parabola:
vertex: (0,1)
axis of symmetry: y=1
4p=8
p=2
focus: (2,1) (p-distance to the right of vertex on the axis of symmetry)
directrix: x=-2 (p-distance to the left of vertex on the axis of symmetry)

see  graph below:
y=(8x)^.5+1
{{{ graph( 300, 300, -10, 10, -10, 10,(8x)^.5+1,-(8x)^.5+1) }}}