tan∠ACO= 1/3，AO= 1
So the coordinate of the intersection of the function image and the X axis is: a (- 1, 0) b (3,0).
Let the resolution function be the intersection point: y=a(x+ 1)(x-3).
Replace C(0, -3)
-3a=-3，a= 1
The resolution function is y=(x+ 1)(x-3), that is, y=x? -2x-3
(2)D is the vertex of parabola, and the symmetry axis is X=2/2= 1.
Substitute X= 1 and Y=-4. D( 1，-4)
Let the straight CD expression be y = kx+b.
Replace (0, -3), (1, -4)
k=- 1，b=-3
So the CD expression is y=-x-3.
Substitute y = 0, -x-3 = 0, and x =-3.
E point coordinates (-3,0)
If the quadrilateral with vertices A, E, C and F is a parallelogram, there are two situations.
1.AE is on one side of the quadrilateral. If f is to the left of c, it is obviously impossible to be on a parabola.
If f is to the right of c, the moving method from e to a is the same as that from c to f, because AE is parallel and equal to cf.
So f should be two units to the right of C, and the coordinate is (2, -3).
Substitute X=2 into the parabolic analytical formula, and Y=-3. Obviously, f is on a parabola at this time. So the coordinate of point F is (2, -3).
2.AE is diagonal. At this time, AC is parallel and equal to EF, and the moving method from C to A is the same as that from E to F.
Because C to A should be moved to the left by 1 unit and moved up by 3 units, E to F should also be moved, and the coordinate of point F is (-4,3).
Substitute X=-4 into parabolic analytical formula, Y=2 1. Obviously f is not on the parabola.
To sum up, the existence of F(2, -3) makes the quadrilateral AECF a parallelogram.
(3) Because the point m (the intersection of a straight line parallel to the X axis and the left side of the parabola) is in the parabola Y=X? On -2X-3, the coordinate of point M is (x, x? -2X-3).
Because the straight line is parallel to the X axis and the ordinate of M and N is equal, it is symmetrical about the symmetry axis. The symmetry axis is X= 1.
So the distance from m to x = 1, 1-X is the radius of the circle.
When the straight line parallel to the X axis is above the X axis, because the circle is tangent to the X axis, the distance from the center of the circle to the X axis is x? -2X-3, also the radius.
So x? -2X-3= 1-X .x？ -X-4=0
X 1 = (1+√17)/2, X2=( 1-√ 17)/2, because the radius is 1-X cannot be negative, so x/is omitted.
X=( 1-√ 17)/2, and the radius is:1-(1-√17)/2 = (1+√1)
When the straight line parallel to the X axis is below the X axis, the distance from the center of the circle to the X axis is -X? +2X+3
So -X？ +2X+3= 1-X .x？ -3X-2=0
X 1=(3+√ 17)/2，X2=(3-√ 17)/2。 Because the radius of 1-X cannot be negative, X 1 is omitted.
X=(3-√ 17)/2, and the radius is:1-(3-√17)/2 = (√17-)/2.
After writing this topic, I really want to go to bed at once. I'm too tired to type.