Question

如图已知抛物线y=ax²+bx+c与X轴交于A（-1,0）B（3,0），与y轴交于点C（3，0）

急！要详细过程！

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Answer 1

(1)过(-1, 0), (3, 0), 可表达为y = a(x + 1)(x - 3)
过(0, 3): x = 0, y = -3a = 3, a = -1
y = -(x + 1)(x - 3) = -x² + 2x + 3
对称轴x = (-1 + 3)/2 = 1
顶点(1, 4)

(2)
C关于对称轴的对称点为C'(2, 3)
AC'与对称轴的交点即为点P(不清楚再问)
AC'的方程：(y - 0)/(3 - 0) = (x + 1)/(2 + 1)
取x = 1, y = 2, P(1, 2)

(3)
CP的斜率为(3 - 2)/(0 - 1) = -1
DC = m, D(0, 3 - m)

DE的方程为y = -x + 3 - m
取y = 0, x = 3 - m
E(3 - m, 0)
令对称轴与x轴的交点为M'

四边形ABMC的面积 = ∆AOC的面积 + 梯形OCMM'的面积+ ∆MM'B的面积
= (1/2)*1*3 + (1/2)(3 + 4)*1 + (1/2)(3 - 1)*4
= 9
∆PDE的面积 = 1

(i)当E在OM'上, 2 < m < 3

∆PDE的面积 = 梯形ODPM'的面积 - ∆ODE的面积 - ∆PM'E的面积
= (1/2)(3 - m + 2)*1 - (1/2)(3 - m)*(3 - m) - (1/2)(1 - 3 + m)*2
= (-m² + 3m)/2 = 1
m = 1或m = 2

与前提不符合，舍去

(ii) E与M'重合，m = 2
∆PDE的面积 = (1/2)EP*P的横坐标
= (1/2)*2*1 = 1
符合要求

(iii) E在M'B上, 0 < m < 2
∆PDE的面积 = 梯形ODPM'的面积＋ ∆PM'E的面积　-∆ODE的面积

＝(1/2)(3 - m + 2)*1 + (1/2)*2*(3 - m - 1) - (1/2)(3 - m)*(3 - m)
= (-m² + 3m)/2 = 1
m =1或m = 2
与前提不符，舍去

三者结合, m = 2