Question

已知二次函数y=ax^2+bx+c的图像过点A(1,0)和点B(5,0),且它与反比例函数y=k/x的图像在第二象限内交

Answer 1

设二次函数为 y = a(x - x1)(x - x2)显然 x1 = 1,x2 = 5所以 y = a(x - 1)(x - 5)由于 二次函数经过 (-1,12)所以 12 = a(-1 - 1)(-1 - 5)所以 a = 1 y = (x - 1)(x - 5) = x^2 - 6x + 5 b = -6,c = 5因为 y = k/x 过...