Question

如果f(x+1)=x^2-5x+4,那么函数f(x)的解析式是？

Answer 1

令t=x+1,则x=t-1，f(t)=f(x+1)=x^2-5x+4=(t-1)^2-5(t-1)+4=t^2-2t+1-5t+5+4=t^2-7t+10,所以函数f(x)的解析式是f(x)=x^2-7x+10

Answer 2

f(x + 1)
= x² -5x + 4
= x² + 2x + 1 - 7x - 7 + 10
= (x + 1)² - 7(x + 1) + 10

所以 f(x) = x² - 7x + 10

Answer 3

令 t=x+1, 则 x=t-1
f(x+1)=x^2-5x+4
f(t)=(t-1)^2-5(t-1)+4
=t^2-2t+1-5t+5+4
=t^2-7t+10
即 函数f(x)的解析式是 f(x)=x^2-7x+10