A(0，6): y=(0)a+(0)b+c=c=6

B(-3，0): y=(9)a+(-3)b+c=9a-3+c=0

C(6，0): y=(36)a+(6)b+c=36a+6b+c=0

The joint solution is: a=- 1/3, b= 1, c=6.

The parabolic equation is: y =-( 1/3) x 2+x+6.

Let P(x, 0), please draw your own picture according to the meaning of the question: P(x, 0) and PE//AB, AC to E.

|BC|=9，|AB|=45^.5=3(5^.5)，|AC|=72^.5=6(2^.5)

| PE | = | ab | |pc|/|bc|=(45^.5)(6-x)/9=(5/9)^.5(6-x)

| AE | = | AC | |bp|/|bc|=(72^.5)(x+3)/9=(8/9)^.5(x+3)

Ape area of triangle = | PE | | | AE | sin (angle AEP) = (6-x) (x+3) (40/81) .5 sin (angle AEP).

(APE area of triangle)' = (-2x+3) [(40/8 1) .5 sin (angle AEP)] = 0 >; x= 1.5

The maximum vertex area of the triangle appears at p (1.5,0). The maximum area can be calculated by the above formula, but it can be obtained by the particularity of geometric figures here. P is the midpoint of BC, and then E is the midpoint of AC, so it is given by (APC) area =(APB) area, (APE) area = (BPE) area = (ABC) area/4 = (1/2) (9) (6)/4 = 6.75.

3 Let G (x, -( 1/3) x 2+x+6), please draw your own picture according to the meaning of the question: G(x, y)[ on a parabola], connecting GA and GC.

The equation of straight line AC is y=6-x, that is, x+y-6=0. The vertical distance from g to straight line AC is:

d = |(x)+(-( 1/3)x^2+x+6)+(-6)|/( 1+ 1)^.5

= |-( 1/3)x^2+2x|/(2^.5)

Therefore, the (AGC) area is | AC | d/2 = (9/2) |-(1/3) x2+2x |/(2.5).

Let (AGC) area =(AEP) area, that is

(9/2)|-( 1/3)x^2+2x|/(2^.5)=27/4

Solve this binary linear equation and get two solutions: x=3( 1+/-0.5), that is

At g (3/2,27/4) or G(9/2, 15/4), (AGC) area =(APE)=27/4.

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