Question

如图,已知抛物线y=ax2+bx+c与x轴相交于A(-2,0),B(4,0)两点,且经过点E(2,

如图,已知抛物线y=ax2+bx+c与x轴相交于A(-2,0),B(4,0)两点,且经过点E(2,4)

Answer 1

(1)
过(-2, 0), (4, 0), 则可表达为y = a(x + 2)(x - 4)
过(2, 4): -8a = 4, a = -1/2
y = (-1/2)(x + 2)(x - 4)
C(0, 4)

(3)
对称轴：x = (-2 + 4)/2 = 1
P(1, p)
C, E的纵坐标相同，则关于对称轴对称。AE与对称轴的交点即为P(AC固定，只需P与A和C的距离之和最小即可, AE与对称性交于P，令CE与对称轴的交点为F，CPF和EPF全等，PE= PC；又两点间直线距离最短).
AE: (y - 0)/(4 - 0) = (x + 2)/(2 + 2), y = x + 2

P(1, 3)

(4)
B关于对称轴的对称点为A，AC的方程为: x/(-2) + y/4 = 1, y = 2x + 4
AC与对称轴的交点即为Q (http://zuoye.baidu.com/question/79bc1318fb4c41a67508f1a084918610.html)
x = 1, y = 6
Q(1, 6)