Verified answer

Vertex form => y = a(x-h)^2 + k, where (h,k) is center of vertex.

You have,

y = -3x^2 + 6x - 1

#Factor -3

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

#Pull out the -3 * 1/3, and add the 1 to make a perfect square trio-nominal, however you cannot randomly add -3 * 1, so you must also add 3 * 1, so in a sense you're adding 0.

y = -3(x^2 - 2x + 1) + (-3 * 1/3) + (3 * 1)

#Simply and factor the perfect square trio-nomial

y = -3(x - 1)^2 + (-1 + 3)

y = -3(x-1)^2 + 2

Alternative way:

You know y = ax^2 + bx^2 + c, so you know a = -3

To find the center the x coordinate is given by x = -b / (2a) = -6 / (-3 * 2) = 1

Then find the y by plugging that value of x:

y(1) = -1 + 6(1) + -3(1)^2 = -1 + 6 - 3 = 2

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

a = -3

h = 1

k = 2

Gives you the same answer.