Question

已知二次函数f(x)=ax^2+bx+c,若f(x)+f(x+1）=2x^2-2x+13 求函数f(x)的解析式

Answer 1

f(x)+f(x+1)=ax²+bx+c+a(x+1)²+b(x+1)+c=2ax²+(2a+2b)x+a+2c+b=2x²-2x+13;

2a=2;a=1;

2a+2b=-2;

b=-2;

a+b+2c=13;

2c=14;

c=7;

解析式为f(x)=x²-2x+7;