How do you rewrite a function in vertex form?

Question

the function is this f(x)=-2x^2+2x-3
can you show how you do it please?

Answer 1

y= -2x^2 +2x -3
y= -2(x^2-x ) -3
y= -2(x^2 -x +1/4) -3+1/2
y = -2 (x-1/2)^2 - 2.5
So vertex is at (.5, -2.5)

Answer 2

You want to make a perfect square appear by itself, with just surrounding constants (no "x" term). Do it like that:

f(x) = -2 [x^2 + 2x(1/2)] -3

The square bracket is "almost" a peffect square. It's just missing a constant, which we add inside the bracket and subtract outside of it:

f(x) = -2 [x^2 -2x(1/2) + (1/2)^2] + 2(1/2)^2 -3

Finally, since the bracket is a square (no pun intended):

f(x) = -2 (x-1/2)^2 -5/2