Question 40873
Since one variable is squared and the other isn't, you know right away that this is a parabola...
(y-2)^2 = 16(x+4)  now let's continue and solve for x...
(1/16)(y-2)^2 = x+4
x = (1/16)(y - 2)^2 - 4
vertex at (-4, 2)
It is noticeably wider than usual and concave right...