已知f(x)为二次函数，且f(0)=3,f(x+2)-f(x)=4x+2，求f(x)

已知f(x)为二次函数，且f(0)=3,f(x+2)-f(x)=4x+2，求f(x)

不妨设f(x)=ax²+bx+c，由f(0)=3知c=3；
f(x+2)=a(x+2)²+b(x+2)+3=ax²+(2a+b)x+4a+2b+3
f(x+2)-f(x)=2ax+4a+2b=4x+2,
解出a=2，b=-3，c=3；

设f(x) = ax² + bx + c f(0) = 3，代入 c = 3 f(x + 2) - f(x) = 4x + 2 f(- 2 + 2) - f(- 2) = 4(- 2) + 2 3 - f(- 2) = - 6 f(- 2) = 9，代入 9 = 4a - 2b + 3 ==> 4a - 2b = 6 f(- 4 + 2) - f(- 4) = 4(- 4) + 2 9 - f(- 4) = - 14 f(- 4) = 23，代入 23 = 16a - 4b + 3 ==> 4a - b = 5 (4a - 2b) - 2(4a - b) = 6 - 2(5) - 4a = - 4 ==> a = 1 b = 4(1) - 5 = - 1 所以f(x) = x² - x + 3

由题可设f(x)=ax平方+bx+3 又由f(x+2)-f(x)=4x+2可知f(2)-f(0)=4*0+2=2 (在此用到赋值法) 所以f(2)=5 同理有f(0)-f(-2)=4*(-2)+2=-6 所以f(-2)=9 则有5=4a+2b+3 9=4a-2b+3 解得a=1,b=-1 所以f(x)=x平方-x+3 ===========================================================================设二次函数f(x)=ax²+bx+c 代入f(0)=3,c=3 因此f(x)=ax²+bx+3 f(x+2)-f(x) =a(x+2)²+b(x+2)+3-(ax²+bx+3) =ax²+4ax+4a+bx+2b+3-ax²-bx-3 =4ax+4a+2b =4x+2 4a=4,a=1 4a+2b=2,b=-1 f(x)=x²-x+3