(1) According to the meaning of the question, get (2 points).

solve

∴ The function expression corresponding to parabola is y = x2-2x-3. (3 points)

(2) existence.

In y=x2-2x-3, let x=0 and get y =-3.

Let y=0, x2-2x-3=0.

∴x 1=- 1,x2=3.

∴A(- 1,0),B(3,0),C(0,-3).

y=(x- 1)2-4，

∴ Vertex M( 1, -4). (5 points)

The expression of the straight line CM is y =-x-3.

In y=-x-3, let y=0 and get x =-3.

∴N(-3,0)，

∴ Ann = 2. (6 points)

In y=x2-2x-3, let y=-3 and get x 1=0, x2 = 2.

∴CP=2，

∴AN=CP.

A CP,

∴ Quadrilateral ANCP is a parallelogram, where P(2, -3). (8 points)

(3)△AEF is an isosceles right triangle.

Reason: in y=-x+3, let x=0 and get y=3, let y=0 and get x = 3.

The intersection of the straight line y=-x+3 and the coordinate axis is d (0 0,3) and b (3 3,0).

∴OD=OB，

∴∠OBD=45 degree. (9 points)

Point C(0, -3),

∴OB=OC.

∴∠OBC=45 degree. (10)

It can be seen from the figure that ∠ AEF = ∠ ABF = 45, ∠AFE =∠ Abe = 45. ( 1 1)

∴∠ EAF = 90，AE = AF。

∴△AEF is an isosceles right triangle. (12)

(4) When point E is any point on the straight line y=-x+3, the conclusion in (3) holds. (14)