一道超级难的二次函数问题!如图,抛物线y=ax^2+bx+c与x轴相交于A、B两点,与y轴相交于C,其中A的坐标是(-1

8个回答

  • (1)y=(-1/2)x+(3/2)

    当x=0时,y=3/2 C:(0,3/2)

    当y=0时,x=3 B:(3,0)

    设顶点式方程为:y=m(x+1)(x-3)

    代入:(0,3/2)

    3/2=-3m m=-1/2

    抛物线的解析式:y=(-1/2)(x^2-2x-3)=-(1/2)x^2+x+(3/2)

    (2)设P(n,(-1/2)n^2+n+3/2)

    △PAB的面积=10

    (1/2)*|AB|*|(-1/2)n^2+n+3/2|=10

    (1/2)*4*|(-1/2)n^2+n+3/2|=10

    |(-1/2)n^2+n+3/2|=5

    令A=(-1/2)n^2+n-3/2=(-1/2)(n-1)^2-1≤-1

    ∴(-1/2)n^2+n+3/2=-5

    n^2-2n-13=0

    n=1±√14

    n不属于[-1,3]

    ∴不存在