Quadratic Function

y = 3x^2/4 - 5x/11 + 4/9
= (3/4)(x^2 - 20x/33) + 4/9
= (3/4)(x^2 - 20x/33 + 100/121) + 4/9 - 75/121
= (3/4)(x - 10/11)^2 - 191/1089
so, the vertex is (10/11, -191/1089)

2015-03-03 09:01:24 補充：
Sorry, correction from the third line,
y = (3/4)(x^2 - 20/33 + 100/1089) + 4/9 - 25/363
= (3/4)(x - 10/33)^2 + 409/1089
so, the co-ordinates of the vertex is (10/33, 409/1089).

Vertex = (-b/2a , (4ac-b^2) /4a)
You may use it to ko mcq.
X coordinates = -(-5/11) / (2x3/4) = 10/33

Y coordinates = [4(3/4)(4/9)-(-5/11)^2] / [4(3/4)] = 409/1089