Verified answer

x = - b/2a = - 6/(2 * -3) = -6/-6 = 1 <--- line of symmetry

f(1) = -3 * 1^2 + 6 * 1 - 4 = -3 + 6 - 4 = -1 vertex at (1 , -1)

interior the kind y = ax^2 + bx + c vertex will be at x = -b / 2a - 3 x^2 - 6x + 4 b = - 6 ; a = - 3 - (-6 / (2 * (-3))) = - a million <=== x coorcoordinatevertex y = - 3 *(-a million)^2 - 6 * (-a million) + 4 = 7 <=== y coorcoordinatevertex (- a million , 7) <===== vertex

Complete the squares to get..

y = -3x² + 6x - 3 - 4 + 3

= -3(x² - 2x + 1) - 1

= -3(x - 1)² - 1

Hence, by y = a(x - h)² + k form, where the vertex is (h,k), the vertex is (1, −1)

I hope this helps!

(1, -1) take the derivative and set equal to zero. this will give you the x value you need. then plug this in for x to get y.