Write f(x)=-2x^2-6x+1 in vertex form by completing the square. Please show work?

y=2x^2-6x+1
y-1= 2x^2-6x
y-1=2(x^2-3x)
y-1+2(3/2)^2= 2(x^2-3x+(3/2)
y-1+9/2=2(x-3/2)2
y+7/2=2(x-3/2)^2
(x-3/2)^2=1/2(y+7/2) vertex(3/2. -7/2) tocheck
axis of symetry x=-b/2a x=-(-6)/2.2=6/4=3/2
substitute 3/2 to get the same result for y

Vertex form is:

f(x) = a(x - h)² + k where (h, k) is the vertex of the parabola.

So starting with:

f(x) = -2x² - 6x + 1

In order to complete the square, we need to get the right side of this equation to:

x² + bx

form. so to do that, let's start with subtracting 1 from both sides, then dividing both sides by -2. That gives us:

f(x) - 1 = -2x² - 6x
-(f(x) - 1) / 2 = x² + 3x

Now that we're here, to complete the square, follow the following steps:
Start with the coefficient of x (3),
half it (3/2)
square it (9/4)

add it to both sides:

-(f(x) - 1) / 2 + 9/4 = x² + 3x + 9/4

Now the right side is a perfect square trinomial so we can factor, then solve for f(x):

-(f(x) - 1) / 2 + 9/4 = (x + 3/2)²

Subtract 9/4 from both sides:

-(f(x) - 1) / 2 = (x + 3/2)² - 9/4

Multiply both sides by -2:

f(x) - 1 = -2(x + 3/2)² + 9/2

Then finally, add 1 to both sides:

f(x) = -2(x + 3/2)² + 11/2

y = -2x² - 6x + 1

factor out the leading coefficient
y = -2(x² + 3x) + 1

complete the square
 coefficient of the x term: 3
 divide it in half: 3/2
 square the result: (3/2)²
 use (3/2)² to complete the square:
y = -2(x² + 3x + (3/2)²) + 2(3/2)² + 1
 = -2(x + 3/2)² + 9/2+1
 = -2(x + 3/2)² + 11/2