Question

已知二次函数fx满足f0=f4,且fx=0的两根平方和为10,图像过点(0.3)求fx

Answer 1

二次函数f(x)=ax^2+bx+c
f(0)=c
f(4)=16a+4b+c
f(0)=f(4)
16a+4b+c=c
4a+b=0
-b/2a=2
x1+x2=-b/2a=2
因x1^2+x2^2=10
(x1+x2)^2-2x1x2=10
2^2-2x1x2=10
x1x2=-3
x1x2=c/a=-3
图像过点(0.3)
c=3
c/a=-3
a=-1,4a+b=0
b=4
所以，a=-1,b=4,c=3
f(x)=-x^2+4x+3

Answer 2

由f(0)=f(4), 且f(0)=3
可设y=ax(x-4)+3=ax²-4ax+3
x1+x2=4, x1x2=3/a
10=x1²+x2²=(x1+x2)²-2x1x2=16-6/a, 解得：a=1
故f(x)=x²-4x+3

Answer 3

由f(0)=f(4), 且f(0)=3
可设y=ax(x-4)+3=ax²-4ax+3
x1+x2=4, x1x2=3/a
10=x1²+x2²=(x1+x2)²-2x1x2=16-6/a, 解得：a=1
故f(x)=x²-4x+3