∫ point a (3,4) is on the straight line y=x+m,

∴ 4=3+m

∴ m= 1

Find the analytical formula of straight line y=x+ 1.

Let the relationship of quadratic function be y = a (x- 1) 2.

∵ point a (3,4) is on the image of quadratic function y = a (x- 1) 2.

∴ 4=a (3- 1)^2

∴ a= 1。

∴ The relationship of quadratic function is y = (x- 1) 2.

That is y = x 2-2x+ 1.

(2) Let the ordinate of p and e be yp and ye respectively.

∴PE = h = yp-ye =(x+ 1)-(x^2-2x+ 1)=-x^2+3x.

That is, h =-x 2+3x (0 < x < 3).

(3) existence.

PE=DC is necessary to make the quadrangle DCEP a parallelogram.

Point d is on a straight line, y=x+ 1,

The coordinate of point ∴d is (1, 2),

∴ -x^2+3x=2。

X 2-3x+2 = 0。

If you solve it, you will get x 1=2, x2= 1 (irrelevant, so it is omitted).

When the coordinate of point P is (2,3), the quadrangle DCEP is a parallelogram.