You can complete the square of the equation to get
y+(4/2)^2 = x^2+4x+(4/2)^2
y+4 = x^2 + 4x + 4
y+4 = (x+2)^2
y = (x+2)^2 - 4
This gives the form y = a(x-h)^2 + k where (h, k) is the vertex of the equation. You can also arrive at the same conclusion by making some observations of the equation. (x+2)^2 minimum value is going to be 0 since and negative values resulting from x+2 is going to become positive because of the square. So the minimum value is when x+2 is 0 or when x is equal to -2 and when it's at that minimum value of 0 it's going to have 4 subtracted from it which gives the vertex of (-2, -4)
