抛物线Y=ax2+bx-3a经过A(-1,0),C(0,-3)两点,与X轴交于另一点B,顶点为点D.
1个回答

(1)y = ax² + bx - 3a 中,x=-1时,y=0;x=0时,y=-3.

分别代入解得:a = 1 b = - 2

故所求解析式为y = x² - 2x - 3

(2)y = x² - 2x - 3中,x = m时,y = - m - 1.代入得:

m² - m - 2 = 0

(m - 2)(m+1)= 0

解得:m = 2 或 m = - 1(不合题意,舍去)

则D(2,- 3)

y = x² - 2x - 3中令y=0得x = - 1或x = 3,故B(3,0)

由B(3,0),C(0,-3)易得直线BC的解析式为:y = x - 3

由 DD‘⊥BC,D(2,- 3)易得直线DD’的解析式为:y = - x - 1,其与y轴交于(0,- 1).

(0,- 1)与D(2,- 3)关于BC对称

∴ D‘(0,- 1)

(3)存在符合条件的点P.

易得BD的解析式为:y = 3x - 9

作CE∥BD交x轴于点P,则∠PCB=∠CBD.此时,BP = CD = 2,故P(1,0)

作∠BCP =∠CBD,CP交x轴于点P,(在点B右侧)

设直线BD交y轴于点M,易得M(0,- 9) 显然有△OCP≌△OBM,则P(9,0)

故符合条件的点P有(1,0)和(9,0)