二次函数代入式怎么解

设y=ax^2+bx+c三个点(xi,yi)代入得：ax1^2+bx1+c=y1ax2^2+bx2+c=y2ax3^2+bx3+c=y3这是个三元一次方程组，可以解得：a=y1/[(x1-x2)(x1-x3)]+y2/[(x2-x1)(x2-x3)]+y3/[(x3-x1)(x3-x2)]b=-y1(x2+x3)/[(x1-x2)(x1-x3)]-y2(x1+x3)/[(x2-x1)(x2-x3)]-y3(x1+x2)/[(x3-x1)(x3-x2)]c=y1x2x3/[(x1-x2)(x1-x3)]+y2x1x3/[(x2-x1)(x2-x3)]+y3x1x2/[(x3-x1)(x3-x2)]都是轮换对称的式子。将A点[-1，-1]、B[9,-9]分别带入y=ax 2次方-4x+c,得：a+4+c=-1，即a+c=-581a-36+c=-9,即81a+c=27两式相减得：80a=32得：a=0.4故c=-5-a=-5.4y=0.4x^2-4x-5.4

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