Question 38909
From 
y^2 - 4y - 4x = 0
we complete the square, also add 4x to both sides
y^2 - 4y + 4 = 4x + 4
(y - 2)^2 = 4(x + 1)
now divide by four and subtract one and get
(1/4)(y - 2)^2 = x + 1
x = (1/4)(y - 2)^2 - 1
This is a wider-than-normal parabola, concave right, with vertex at (-1, 2).