Question

如图，抛物线y=ax2+bx+c经过点A（-3，0）B（1,0）C（-2,1），交y轴于点M

Answer 1

(1)

y = a(x + 3)(x - 1)
过(-2, 1): -3a = 1, a = -1/3
y = -x²/3 -2x/3 + 1

(2)
M(0, 1)
E(e, 0), D(e, (-1/3)(e + 3)(e - 1))
AM: x/(-3) + y/1 = 1，y = x/3 + 1
F(e, e/3 + 1)
DF = (-1/3)(e + 3)(e - 1) - (e/3 + 1)
= (-1/3)e(e+3)
e = (0 - 3)/2 = -3/2时，DF最大，为3/4

(3)
图不全，无法做。但显然二者都是直角三角形，只要直角边成比例即可。