Therefore, 0= 16a-4b+c, 4=c, 0=a+b+c, while a =- 1, b =-3, c = 4, y =-x 2-3x+4 =-(x+3/2).

Vertex is (-3/2, 25/4)

(2) A right triangle does not specify what a right-angled vertex is, so it should be considered that when D is a right-angled vertex, obviously (0,0) is D that meets the conditions; When a is a right-angled vertex, d is (0,-4); When b is a right-angled vertex, d is (4,0).

(3) Let the coordinate of E be (xe, ye), and let EF be perpendicular to the X axis and intersect with F..

Then S(ABE)=S(AEF)+S(EFOB)-S(ABO)

=(x+4)*y/2+(4+y)*(-x)/2-4*4/2=2y-2x-8=-2x^2-6x+8-2x-8=-2x^2-8x=-2(x+2)^2+8

Therefore, when x=-2, the maximum value of 8 is obtained. At this time, e is (-2,6) and S = 8.

(4) P (-5,4) above AB and P (-3,4) below AB.