By a (-4,0), b (-2,2) in parabola.

Y=ax2+bx image，get: 16a-4b=0。

and

4a-2b=2

Solution: a=

-0.5

b=

-2

∴

The resolution function is: y=

-0.5x2-2x

(2) The intersection point B is BC perpendicular to the X axis, and the vertical foot is point C. 。

Branch 1: the lengths of line segments CO, CA and CB are all 2.

∴

△ABC and△△ OBC are congruent isosceles right triangles.

Company

And ∠ ABO =∠ ABC+∠ OBC =

90

△ OAB is an isosceles right triangle.

(3) (4)

As shown in the figure, rotate △OAB counterclockwise around point O 135 to get △OA'b'.

Point B falls right here.

On axis and B'A'

Axis.

∵ the length of ob' and A'B' is

Twice the root number 2.

The coordinate of the midpoint p of A'B' is (negative root number 2, twice root number 2.

)

, obviously does not satisfy the parabolic equation,

Point p is not on this parabola.

(4) existence

Pass through point O, so that OM∨AB and parabola intersect at point M.

It is easy to find the analytical formula of straight line OM: y = x.

At the same time y=x and y=

-0.5x2-2x

Get a solution

Point m (-6, -6)

Obviously, the point M (-6, -6) is about the axis of symmetry.

The symmetry point m ′ (2, -6) also meets the requirements.

Therefore, there are two points M*** that meet the conditions, and the coordinates are (-6, -6) and (2, -6) respectively.

S quadrilateral ABOM=S△ABO+S△AOM=0.5×4×2+

0.5×4×6= 16