Question

急！初三函数题，求解

Answer 1

(1)
抛物线开口向上, 按图, c<0
抛物线可以表达为y = (x+1)(x+c) = x² + (c+1)x + c
OB = OC = 2, 则B的横坐标为-c = 2, c = -2, b = c+1=-1

(2)
按(1), A(-1, 0), OB = 5, 则B(5, 0), C(0, -5)
条件过多，但不矛盾，可以自己验证ABC的面积为15
抛物线为y = (x + 1)(x - 5) = x² - 4x - 5 = (x - 2)² -9
顶点D(2, -9)

(3)
有两种可能:
(i) PC∥BD
BD的斜率为k = (-9 - 0)/(2 - 5) = 3
PC的方程为y = 3x - 5 (斜截式)
与抛物线联立得x = 7, P(7, 16)
(ii) 令PC与BD的交点为Q, QBC为等腰三角形，Q为顶点
BD: (y + 5/2)/(0 + 5/2) = (x - 2)/(5 - 2), y = 3x - 15
令Q(q, 3q - 15)
QB² = QC²
(q - 5)² + 9(q - 5)² = (q - 0)² + (3q - 15 + 5)²
q = 15/4, Q(15/4, -15/4)
QC: (y + 5)/(-15/4 + 5) = (x - 0)/(15/4 - 0)
y = x/3 - 5
与抛物线联立得P(13/3, -32/9)

(4)
令对角线的交点为R, 显然R为DQ的中点, BC与对称轴的交点为R(2, -3), 于是Q(2, 3)
R也是CG的中点, G(4, -1)
周长请自己算。