f(x)=ax²+bx+c ,a>0(1)若方程f(x)+2x=0有两个实根,X1=1,x2=3且方程f(x)+6a=0有两个相等根,求f(x)解析式.(2)ax²+bx+c两根分别是x1,x2,那么x1+x2=-b/a,x1x2=c/a.(韦达定理)若f(x)图像与X轴交于A(-3,0)B(m,0),且当-1
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![f(x)=ax²+bx+c ,a>0(1)若方程f(x)+2x=0有两个实根,X1=1,x2=3且方程f(x)+6a=0有两个相等根,求f(x)解析式.(2)ax²+bx+c两根分别是x1,x2,那么x1+x2=-b/a,x1x2=c/a.(韦达定理)若f(x)图像与X轴交于A(-3,0)B(m,0),且当-1](/uploads/image/z/8981171-35-1.jpg?t=f%28x%29%3Dax%26%23178%3B%2Bbx%2Bc+%2Ca%3E0%281%29%E8%8B%A5%E6%96%B9%E7%A8%8Bf%28x%29%2B2x%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E5%AE%9E%E6%A0%B9%2CX1%3D1%2Cx2%3D3%E4%B8%94%E6%96%B9%E7%A8%8Bf%28x%29%2B6a%3D0%E6%9C%89%E4%B8%A4%E4%B8%AA%E7%9B%B8%E7%AD%89%E6%A0%B9%2C%E6%B1%82f%28x%29%E8%A7%A3%E6%9E%90%E5%BC%8F.%282%29ax%26%23178%3B%2Bbx%2Bc%E4%B8%A4%E6%A0%B9%E5%88%86%E5%88%AB%E6%98%AFx1%2Cx2%2C%E9%82%A3%E4%B9%88x1%2Bx2%3D-b%2Fa%2Cx1x2%3Dc%2Fa.%28%E9%9F%A6%E8%BE%BE%E5%AE%9A%E7%90%86%29%E8%8B%A5f%28x%29%E5%9B%BE%E5%83%8F%E4%B8%8EX%E8%BD%B4%E4%BA%A4%E4%BA%8EA%28-3%2C0%29B%28m%2C0%29%2C%E4%B8%94%E5%BD%93-1)
f(x)=ax²+bx+c ,a>0(1)若方程f(x)+2x=0有两个实根,X1=1,x2=3且方程f(x)+6a=0有两个相等根,求f(x)解析式.(2)ax²+bx+c两根分别是x1,x2,那么x1+x2=-b/a,x1x2=c/a.(韦达定理)若f(x)图像与X轴交于A(-3,0)B(m,0),且当-1
f(x)=ax²+bx+c ,a>0
(1)若方程f(x)+2x=0有两个实根,X1=1,x2=3且方程f(x)+6a=0有两个相等根,求f(x)解析式.
(2)ax²+bx+c两根分别是x1,x2,那么x1+x2=-b/a,x1x2=c/a.(韦达定理)
若f(x)图像与X轴交于A(-3,0)B(m,0),且当-1小于等于x小于等于0时,f(x)小于等于0恒成立,求实数m取值范围.
f(x)=ax²+bx+c ,a>0(1)若方程f(x)+2x=0有两个实根,X1=1,x2=3且方程f(x)+6a=0有两个相等根,求f(x)解析式.(2)ax²+bx+c两根分别是x1,x2,那么x1+x2=-b/a,x1x2=c/a.(韦达定理)若f(x)图像与X轴交于A(-3,0)B(m,0),且当-1
1).f(x)+2x=ax^2+(b+2)x+c=0
x1+x2=1+3=4=-(b+2)/a--> b=-4a-2
x1x2=3=c/a--> c=3a
f(x)+6a=ax^2+bx+c+6a=0有两个等根,delta=b^2-4a(c+6a)=0
(4a+2)^2-4a(3a+6a)=0
(5a+1)(-a+1)=0
所以a=1,(-1/5舍去)
b=-4a-2=-6,c=3a=3
f(x)=x^2-6x+3
2)f(x)=0根为-3,m
-1=