Question

已知抛物线y=-x^2+(m-4)x+2m+4与x轴交于A(x1,0),B(x2,0)两点？

Answer 1

1.x1+2x2=0x2=-(x1+x2)=4-m代入方程得：-(4-m)^2+(m-4)(4-m)+2m+4=0化简m^2-9m+14=0m1=2,m2=7将m1代入原方程得y=-x^2-2x+8=-(x+4)(x-2),x1=-4,x2=2将m2代入原方程得y=-x^2+3x+18=-(x+3)(x-6),x1=-3,x2=6 (不符合题意,舍去)解得A(-4,0),B(2,0),C(0,8)D(4,0)设过BCD的抛物线解析式为y=ax^2+bx+c4a+2b+c=0c=816a+4b+c=0a=1,b=-6,c=8y=x^2-6x+82.y=x^2-6x+8=(x-3)^2-1P(3,-1)S△CBD=2*8/2=8设H(x,y)S△HBD=2*|y|/2=8|y|=8|x^2-6x+8|=8解得x1=0,x2=6H(6,8)直线PH的解析式：(y-8)/(x-6)=(8+1)/(6-3)=33x-y-10=0,4,已知抛物线y=-x^2+(m-4)x+2m+4与x轴交于A(x1,0),B(x2,0)两点
与y轴交于C点,且x1