Question

要二次函数二三问步骤谢谢大神

Answer 1

既然这样问， 应当给出(1)的答案，而且确定无误。
(1)
E(2, -1)

(2)
B(1, 0), D(-1, 0), A(3, 0), 对称轴为x = (1+3)/2 = 2, 即E为顶点
F(t, (t-1)(t-3)), t > 2
令对称性与DF的交点为G.
DF的方程为: (y - 0)/[(t - 1)(t - 3) - 0] = (x + 1)/(t + 1)
G(2, 3(t - 1)(t - 3)/(t+1))
EG = 3(t - 1)(t - 3)/(t+1) - (-1) = (3t-5)(t-2)/(t+1)
S = △DEG的面积 + △EGF的面积
= (1/2)*EG*E的横坐标 + (1/2)*EG*(F的横坐标 - E的横坐标)
= (1/2)[(3t-5)(t-2)/(t+1)]*2 + (1/2)[(3t-5)(t-2)/(t+1)]*(t - 2)
= t(t - 1)(3t - 5)/[2(t - 1)]

(3)
此时二者的底均为EF, 那么C, D与EF等距离， 即CD∥EF
EF: x/(-1) + y/3 = 1， y = 3x + 3
令EF为 y = 3x + c, 代入E(2, -1), 得c = -7
EF: y = 3x - 7
与抛物线联立: 3x - 7 = (x - 1)(x - 3)
(x - 2)(x - 5) = 0, t = 5, F(5, 8)
CF = 5√2