We begin with y=x^2+4x-2. In order to find the vetex form of this equation we need to factor it. If you try it, y=x^2+4x-2 is not dactorable, so now we can either complete the square or use the quadratic formula. I'm going to use the quadratic formula because it is fool-proof, but learning how to complete the square is valuable too.
The quadratic formula is x=(-b+-sqrt(b^2-4*a*c))/(2*a), where a, b, c come from ax^2 +bx+c. In our case, a=1, b =4, and c=-2.
That gives us x=(-4+-sqrt(4^2-4*1*-2))/(2*1), or (-4+-sqrt(16-(-8)))/2, which simplifies further to (-4+-sqrt(24))/2.
From here we expand sqrt(24) to 2sqrt(6), which makes the equation (-4+-2sqrt(6))/2, or -2+-sqrt(6).
So we went from x=(-4+-sqrt(4^2-4*1*-2))/(2*1) to x=-2+-sqrt(6). Now we add 2 on both sides, leaving us with +-sqrt6=x+2. From here, we need to get rid of the square root, so we'll square both sides, which will give us 6=(x+2)^2. Subtarct 6, and have 0=(x+2)^2-6. Since we're looking for the eqaution when y=0 (the x-axis), we can use 0 and y interchanagbly.
Thus, 0=(x+2)^2-6 is the same thing as y=(x+2)^2-6. Nice work, we ow have the equation in Vertex form!