如图，抛物线y=ax눀+bx-4a经过……(请看图)求2.3两题的解答 快

(1) C(0, 4), x = 0, y = c = 4 = -4a, a = -a
y = -x² + bx + 4
x = 4, y = -16 + 4b + 4 = 0, b = 3
y = -x² + 3x + 4 = -(x + 1)(x - 4)

(2) D(m, m + 1), m > 0

m + 1 = -(m + 1)(m - 4)
m = 3
D(3, 4)
BC: x/4 + y/4 = 1, y = 4 - x, 斜率-1, DE的斜率为1，方程为y - 4 = x - 3, y = x + 1
二直线交于F(3/2, 5/2)
E(u, v), F为DE的中点
3/2 = (u + 3)/2, u = 0
5/2 = (v + 4)/2, v = 1
E(0, 1)

(3) 令BD的倾斜角为θ，tanθ = (4 - 0)/(3 - 4) = -4
BP的倾斜角为θ + π/4, 其斜率为tan(θ + π/4) = [tanθ + tan(π/4)]/[1 - tanθtan(π/4)]
= (-4 + 1)/[1 - (-4)] = -3/5
BP的方程: y - 0 = (-3/5)(x - 4)
与抛物线联立，得x = -2/5, P(-2/5, 66/25)