已知f（x）=ax²+bx+c,若f（0）=1并且f（x+1）-f（x）=2x,则f（1）=____ f（-1）____a=____b=____.
问题描述：
已知f（x）=ax²+bx+c,若f（0）=1并且f（x+1）-f（x）=2x,则f（1）=____ f（-1）____a=____b=____.
答
f(x) =ax^2+bx+cf(0) =c =1f(x+1) -f(x) = 2xa(x+1)^2 +b(x+1) +1 - (ax^2+bx+1) = 2x(2a-2)x + a+b =0coef.of x => a=1coef of constanta+b=0b = -1f(x) = x^2-x+1f(1)= 1f(-1) = 3a=1b=-1coef.of x => a=1coef of constant 看不懂啊x的系数 => a=1常数的系数coef.of x => a=1 这部怎么求出来的2a-2=0怎么出来的a(x+1)^2 +b(x+1) +1 - (ax^2+bx+1) = 2x2ax+a+b = 2x2a=2a=1