How do you write the quadratic in vertex form given #f(x) = -3x^2 + 6x -2#?

Question

Abigail Allen · Accepted Answer

The vertex form of a quadratic function is given by
#y = a(x - h)^2 + k#, where #(h, k)# is the vertex of the parabola.  To enter this into the Vertex Form, we can utilize the Completing the Square procedure. #y=-3x^2+6x-2# #-> y + 2 = -3x^2 + 6x# (Transposed -2 to the Left Hand Side) #-> y + 2 = -3(x^2 - 2x)# (Made the coefficient of #x^2# as 1) Now we subtract #3# from each side to complete the square #-> y + 2 - 3 = -3(x^2 - 2x + 1^2)# #->  y - 1 = -3(x-1)^2 # # -> color(green)(y  = -3{x - 1}^2 + 1# is the Vertex Form The vertex of the Parabola is# {1 , 1}#

Julia Adams · Answer

undefined To write the quadratic function $ f(x) = -3x^2 + 6x - 2 $ in vertex form, follow these steps:

1. Complete the square for the quadratic term:
   $$ f(x) = -3(x^2 - 2x) - 2 $$

2. To complete the square inside the parentheses, halve the coefficient of the linear term (6x) and square it:
   $$ (-3(x^2 - 2x + 1 - 1)) - 2 $$

3. Rewrite the expression inside the parentheses:
   $$ -3((x - 1)^2 - 1) - 2 $$

4. Distribute the -3:
   $$ -3(x - 1)^2 + 3 - 2 $$

5. Combine like terms:
   $$ -3(x - 1)^2 + 1 $$

So, the quadratic function $ f(x) = -3x^2 + 6x - 2 $ in vertex form is:
$$ f(x) = -3(x - 1)^2 + 1 $$