You need to remember what the vertex form of a quadratic is such that:

`y = a(x-h)^2 + k`

Notice that (h,k) expresses the vertex of parabola.

The problem provides the equation `y=-2x^2+6x+1` , hence you should collect the terms `-2x^2+6x` , to factor out -2 and to complete the square`x^2 - 3x` such that:

`x^2 - 2*(1/2)*3x + (3/2)^2`

Notice that the term that completes the square is `(3/2)^2` , hence, the binomial raised to square is `(x - 3/2)^2` .

`y = -2(x - 3/2)^2 - 9/4 + 1`

Notice that the term `9/4` is added to complete the square, but you need to remember that you need to subtract `9/4` also to keep the equation balanced)

`y = -2(x - 3/2)^2 - 5/4`

You need to use the vertex form to write the equation `y=x^2+4x+1` , hence you need to complete the square `x^2 + 4x` such that:

`y = (x^2 + 4x + 4) - 4 + 1`

`y = (x+2)^2 - 3`

**Hence, converting the given standard forms of quadratics into vertex forms yields `y = -2(x - 3/2)^2 - 5/4` and `y = (x+2)^2 - 3.` **

=x^2+4x+1

= (x^2+4x)+1

divide the coefficient of x(4) by 2'then suare it

= [x^2+4x+(4/2)^2]+1 that is [(4/2)^2=2^2=4]

=(x^2+4x+4-4)+1

= (x^2+4x+4)-4+1

=(x^2+2)-3

This is another method of coverting quaratic equations to vertex form.