It is known that there are three points a (x1, Y1) and B (X2, Y2) on the image of quadratic function y = - 3x ^ 2 + 6x + K. if | x1-1 | < | x2-1 |, then the corresponding function values Y1 and Y2 have the size relationship

It is known that there are three points a (x1, Y1) and B (X2, Y2) on the image of quadratic function y = - 3x ^ 2 + 6x + K. if | x1-1 | < | x2-1 |, then the corresponding function values Y1 and Y2 have the size relationship

Given that the coordinates of point a are x1, Y1, and the coordinates of point B are X2, Y2, on the image of quadratic function y = (X-2) square + 3, if X1 > x2 > 2, calculate Y1_____ y2
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Because X1 > x2 > 2, Y1 > Y2

We know that the image of quadratic function y = ax ^ 2 + BX + C (a ≠ 0) passes through point a (2,4) and intersects with X axis at points B (x1,0), C (x2,0), X1 ^ 2 + x2 ^ 2 = 13, and the vertex is horizontal
It is known that the image of quadratic function y = ax ^ 2 + BX + C (a ≠ 0) passes through point a (2,4) and intersects with X axis at points B (x1,0), C (x2,0), X1 ^ 2 + x2 ^ 2 = 13, and the abscissa of vertex is 1 / 2
(1) Find the analytic expression of this function
(2) Is there d on the parabola above the x-axis such that s △ ABC = 2S △ DBC? If it exists, find the point d satisfying the condition. If it does not exist, explain the reason

From the title:
x1+x2=-b/a,x1*x2=c/a;
x1^2+x2^2=(x1+x2)^2-2x1*x2=(b/a)^2-2c/a=13 (1)
Since the function is over a (2,4), the following formula is obtained:
4=4a+2b+c (2)
The abscissa of the fixed point is 1 / 2
-b/(2a)=1/2 (3)
The three formulas work together
a=-1,b=1,c=6
y=-x^2+x+6
2. The triangle ABC and triangle DBC have the same base, so the ratio of triangle area is equal to the ratio of height. The height of triangle ABC is 4, so the height of triangle DBC is 2, and 2 is the corresponding point on the quadratic function image
So the equations 2 = - x ^ 2 + X + 6 and - 2 = - x ^ 2 + X + 6 can be formulated
The solution of the first equation is X1 = [(17 under the root) + 1] / 2, X2 = negative [(17 under the root) + 1] / 2
The solution of the second equation is X3 = [(33) + 1] / 2, X4 = negative [(33) + 1] / 2