Question

把抛物线l1：y=-x2向右平移1个单位长度,再向上平移4个单位长度,得到抛物线l2．如图,点A、B分别是抛物线l2与x

Answer 1

(1)

y = -x²向右平移1个单位变为y = -(x - 1)²

再向上平移4个单位长度,变为y = -(x - 1)² + 4 = -x² + 2x + 3 = -(x + 1)(x - 3)

对称轴 x = 1

(2)

题中没有图,这里假设A为左交点(-1,0)

C(0,3)

A关于对称轴 x = 1的对称点为A'(3,0)

A'C的解析式:x/3 + y/3 = 1

令x = 1,y = 2

P(1,2) (不清楚可以再问)

(3) E(t,0)

①四边形DCEB为平行四边形,则CD与x轴平行且CD = EB = 3 - t

C,D关于对称轴 x = 1对称,D(2,3)

CD = 2 = t

EB = 3 - t = 1

二者矛盾,不可能

②

t = 2时,四边形CEBD为梯形,D(2,3)

另外须考虑DB与CE平行

E(t,0),D(t,-t² + 2t + 3)

CE的斜率m = (3 - 0)/(0 - t) = -3/t

DB的斜率n = (-t² + 2t + 3 - 0)/(t - 3) = -(t + 1)

m = n,-3/t = -(t + 1)

t = (√13 - 1)/2 (另一解< 0,舍去)

D((√13 - 1)/2,(10 - 3√13)/2)