如图,抛物线y=-x2+mx+n与x轴交于A,B两点,与y轴交于点C,抛物线的对称轴交x轴于点

1个回答

  • (1)估计原抛物线是y = -x² + mx + n

    过(0, 2), n = 2, y = -x² + mx +n

    过(-1, 0), 0 = -1 -m + 2, m = 1

    y = -x² + x + 2 = -(x + 1)(x - 2)

    B(2, 0)

    (2)

    对称轴x = (-1 + 2)/2 = 1/2

    令P(1/2, p); 看来D是对称轴与x轴的交点,D(1/2, 0)

    (i) C为顶点, CP² = CD²

    (1/2 - 0)² + (p - 2)² = 2² + (1/2)²

    p = 4, P(1/2, 4)或p = 0, P(1/2, 0)

    (ii) D为顶点,DP² = DC²

    (p - 0)² = 2² + (1/2)²

    p = ±√17/2

    P(1/2, ±√17/2)

    (iii) P为顶点,PC² = PD²

    (1/2 - 0)² + (p - 2)² = p²

    p = 17/16, P(1/2, 17/16)

    (3) BC的方程为x/2 + y/2 = 1, y = 2 - x

    E(e, 2 - e), 0 < e < 2, F(e, -e² + e + 2)

    令过EF的直线与x轴交于E'(e, 0)

    四边形CDBF的面积 = 梯形COE'F的面积 + 三角形E'BF的面积 - 三角形COD的面积

    = (1/2)*(OC + E'F)*OE' + (1/2)*E'B*E'F - (1/2)*OD*OC

    = (1/2)(2 - e² + e + 2)*e + (1/2)(2 - e)*(-e² + e + 2) - (1/2)*(1/2)*2

    = -e² + 2e + 3/2 = -(e - 1)² + 5/2

    e = 1时,面积最大,为5/2, E(1, 1)