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