First, let's divide both sides by 8 to get:
#x = (y-2)^2/8#
#(y - 2)^2 >= 0# for all values of #y#, but will equal #0# when #y = 2#.
So the parabola has its vertex to the left on the #y# axis at #(0, 2)# and its axis is a horizontal line parallel to the #x# axis with formula #y = 2#. It is symmetric about this axis. The means that for every point #(x, y)# that lies on the parabola, so does the point #(x, (4-y))#.
The lower 'arm' of the parabola will intersect the #x# axis when #y = 0#. So substitute #y = 0# in the equation of the parabola to derive #x# ...
#x = (y - 2)^2/8 = (0 - 2)^2 / 8 = (-2)^2/8 = 4/8 = 1/2#
In other words, the parabola intersects the #x# axis at #(1/2, 0)#.
If you want to know any more points the parabola passes through, just choose a value of #y# and substitute it into the formula for #x# to get the corresponding value of #x#.
