Question

f(x)=ax^2+bx,f(x)=x有等根,且f(5-x)=f(x-3),求f(x)的解析式

Answer 1

f(x)=ax^2+bx=x

ax^2+(b-1)x=0有等跟

显然一个跟是x=0

所以另一个也是0

x[ax+(b-1)]=0

x=(b-1)/a=0

b=1

f(x)=ax^2+x

f(5-x)=f(x-3)

a(5-x)^2+(5-x)=a(x-3)^2+(x-3)

ax^2-10ax+25a+5-x=ax²-6ax+9a+x-3

16a+8=(4a+2)x

(4a+2)x=4(4a+2)

这是恒等式

则只有4a+2=,a=-1/2

f(x)=-1/2*x^2+x