(1)设二次函数解析式为 y = ax^2 + bx + c,(a≠0)因为图像经过A(1,0)B(3,0)C(0,-3)三点,有0 = a + b + c0 = 9a + 3b +c-3 = c得 a = -1 , b = 4, c = -3这个二次函数的解析式 y = -x^2 + 4x - 3(2) y = -(x^2 - 4x +3) = -(x - 2)^2 + 1所以顶点为D(2,1)设直线为y = kx +b,直线过D(2,1),A(1,0)有 1 = 2k + b0 = k + b得 k = 1,b = -1直线的表达式为 y = x - 1...
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