First, let's divide both sides by 8 to get:
x = (y-2)^2/8x=(y−2)28
(y - 2)^2 >= 0(y−2)2≥0 for all values of yy, but will equal 00 when y = 2y=2.
So the parabola has its vertex to the left on the yy axis at (0, 2)(0,2) and its axis is a horizontal line parallel to the xx axis with formula y = 2y=2. It is symmetric about this axis. The means that for every point (x, y)(x,y) that lies on the parabola, so does the point (x, (4-y))(x,(4−y)).
The lower 'arm' of the parabola will intersect the xx axis when y = 0y=0. So substitute y = 0y=0 in the equation of the parabola to derive xx ...
x = (y - 2)^2/8 = (0 - 2)^2 / 8 = (-2)^2/8 = 4/8 = 1/2x=(y−2)28=(0−2)28=(−2)28=48=12
In other words, the parabola intersects the xx axis at (1/2, 0)(12,0).
If you want to know any more points the parabola passes through, just choose a value of yy and substitute it into the formula for xx to get the corresponding value of xx.
