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）

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