how do you convert the standard form y = -3x^2 + 3x +6 into the vertex form

Question

could someone please give me step by step instructions on how to do this problem so I can do the rest on my own.

standard form is y = ax^2 + bx + c

vertex form is y - k = a(x-h)^2

start with:

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

Answer 1

first du y=0=-[3x^2-3x-6]

quadratik equashun giv

y=-(x-2)*(x+1)

or y=-3*(x^2-x-2)

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

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

-(y-9)/3=(x-1)^2

or (y-9)=-3*(x-1)^2

Answer 2

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