Given: y=f(x)=-3x²+3x+6 ··· Eq.1 ⇒
f(x)=-3(x²-x-2) ⇒ f(x)=-3{(x²-x+¼)-¼-2} ⇒ f(x)=-3{(x-½)²-9/4}=-3(x-½)²+27/4
That is: f(x)=-3(x-½)²+27/4 From here, we try to locate the vertex of Eq.1.
Let f1(x)=-3x², the graph of f1(x) is a parabola, being convex upwards (spills water), the maximum is 0, and the point of vertex O is the origin of the coordinate plane: O(0, 0).
Let f2(x)=-3(x-½)², f2(x) is f1(x) moved off ½ units to the right, so the point of vertex R is: R(½, 0).
f(x)=-3(x-½)²+27/4, so f(x) is f2(x) moved off 27/4 units upwards, so the point of vertex S is: S(½,27/4).
Therefore, the coordinates of Eq.1's vertex is x=1/2 and y=27/4.
CK: From Eq.1, we have f'(x)=-6x+3 and f''(x)=-6<0 ⇒ if f(x)=0, x=1/2 and f(½)=27/4 Thus, the graph of Eq.1 is convex upwards, and the coordinates of vertex are x=1/2 and y=27/4. That is: Eq.1 takes its maximum y=27/4 at x=1/2. CKD.
Answer: coordinates of the vertex are x=1/2 and y=27/4