As shown in the figure, the functional relations of lines OC and BC are Y1 = x and y2 = - 2x + 6, respectively. The moving point P (x, 0) moves on ob (0 < x < 3), and the line m passing through point P is perpendicular to the X axis. (1) find the coordinates of point C, and answer when x takes what value, Y1 > Y2? (2) Let the area of the left part of the line m in △ cob be s, and the functional relationship between S and X is obtained. (3) when x is what, the line m bisects the area of △ cob?

As shown in the figure, the functional relations of lines OC and BC are Y1 = x and y2 = - 2x + 6, respectively. The moving point P (x, 0) moves on ob (0 < x < 3), and the line m passing through point P is perpendicular to the X axis. (1) find the coordinates of point C, and answer when x takes what value, Y1 > Y2? (2) Let the area of the left part of the line m in △ cob be s, and the functional relationship between S and X is obtained. (3) when x is what, the line m bisects the area of △ cob?

(1) According to the meaning of the problem, we can solve the equations y = xy = - 2x + 6, x = 2Y = 2, C point coordinates are (2, 2); according to the diagram, when x ＞ 2, Y1 ＞ Y2; (2) as shown in the figure, make CD ⊥ X axis at point d through C, then d (2, 0), ∫ line y2 = - 2x + 6 intersects with X axis at point B, ∫ B (3, 0), when 0 ＜ x ≤ 2, then the left part of the line m is △ P ′ Q ′ o, ∫ P ′ (x, 0), ∫ op ′ = X, When 2 ＜ x ＜ 3, the left part of the line m is a quadrilateral opqc, ∵ P (x, 0), ∵ OP = x, ∵ Pb = 3-x, while q is on the line y2 = - 2x + 6, ∵ PQ = - 2x + 6, ∵ s = s △ boc-s △ PBQ = 12 × CD × ob − 12 × BP × PQ = - x2 + 6x-6 (2 ＜ x ＜ 3); (3) if the line m bisects the area of △ BOC, then point P can only be in OD, That is, 0 ＜ x ＜ 2. And the area of ∵ cob is equal to 3, so 12x2 = 3 × 12. The solution is x = 3. When x = 3, the line m bisects the area of ∵ cob

It is known that the image of the function Y1 = ax ^ 2 + BX + C passes through the intersection of the image of the function y2 = - 3 / 2x + 3 and the X and Y axes, and passes through the point (1,1)
1. Find the analytic expression of quadratic function
2. The analytic formula is reduced to y = a (X-H) ^ 2 + K by the collocation method

Y2 = - 3 / 2x + 3 let X be equal to 0, find out the focus of function and Y axis, then let y be equal to 0, find out the focus of function and X axis, and then bring (1,1) and the focus obtained above into equation Y1 to get the analytical formula Y1 = 1 / 2x ^ 2-5 / 2x + 3
From y = 1 / 2 (X-5 / 2) ^ 2-13 / 4