如图,已知抛物线y=ax2+bx+c与x轴交于A(-4,0)和B(1,0),与y轴交与C(0,-2)

1个回答

  • 1. 与x轴交于A(-4,0)和B(1,0), 可以表达为 y = a(x+4)(x-1) = ax² + 3ax - 4a

    代入C点的坐标: -2 = -4a, a = 1/2

    y = x²/2 + 3x/2 -2

    2. △CDE的面积是△DBE面积的两倍, 则△DBE面积是△BCD面积的1/3.

    二者均以DB为底,所以E的纵坐标绝对值(DB上的高)为|OC|的1/3, 即的纵坐标是-2/3

    BC的解析式: x/1 + y/(-2) = 1

    取y = -2/3, x = 2/3

    E(2/3, -2/3)

    3.

    AC的解析式: x/(-4) + y/(-2) = 1

    y = -x/2 - 2

    设P(p, p²/2 + 3p/2 -2), -4 < p < 0

    F(p, -p/2 - 2)

    PF = (-p/2 - 2) - (p²/2 + 3p/2 -2) = -p²/2 - 2p

    = -(1/2)(p² + 4p + 4 - 4)

    = -(1/2)(p + 2)² + 2

    p = -2时, 线段PF的值最大

    P(-2, -3)