已知抛物线y=x^2+mx+2m-m^2,根据以下条件,分别求出相应的m值

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  • 已知抛物线y=x^2+mx+2m-m^2,根据以下条件,分别求出相应的m值

    1)抛物线的最小值为-1

    y=x^2+mx+2m-m^2

    =x^2+mx+m^2/4-m^2/4+2m-m^2

    =(x+m/2)^2-5m^2/4+2m

    -5m^2/4+2m=-1

    5m^2-8m-4=0

    (5m+2)(m-2)=0

    m=-2/5 m=2

    2)抛物线与x轴两个交点间的距离为四倍根号三

    x1=(-m+(m^2-4(2m-m^2))^0.5/2

    x2=(-m-(m^2-4(2m-m^2))^0.5/2

    x1-x2=4√3

    (m^2-4(2m-m^2))^0.5=4√3

    m^2-4(2m-m^2)=48

    m^2-8m+4m^2=48

    5m^2-8m-48=0

    (5m+12)(m-4)=0

    m=-12/5 m=4

    3)抛物线的顶点在直线y=2x+1上

    y=(x+m/2)^2-5m^2/4+2m

    -5m^2/4+2m=-2*m/2+1

    5m^2-12m+4=0

    (5m-2)(m-2)=0

    m=2/5 m=2

    4)抛物线与y轴交点的纵坐标为-3

    y=x^2+mx+2m-m^2

    -3=2m-m^2

    m^2-2m-3=0

    (m-3)(m+1)=0

    m=3 m=-1