曲线y=ax平方+bx-2在点（-1.3处与直线y=4x+7相切，求a，b

一、用导数计算曲线在切点的导数等于切线的斜率；即 y' = 2ax + b，y'(-1) = -2a + b = 4；①切点坐标代入曲线方程，a - b - 2 = 3，a - b = 5；②① + ②，-a = 9，a = -9；b =a - 5 = -9 - 5 = -14；曲线方程y = -9x^2 - 14x - 2 。二、不用导数计算切点坐标代入曲线方程，a - b - 2 = 3， b = a - 5；在切点，ax^2 + bx - 2 = 4x + 7将 b = a - 5代入，ax^2 + ( a - 5 - 4 )x - 9 = 0( ax - 9 )( x + 1 ) = 0切线与曲线只交于一点，所以方程在x = -1 有2个相等的根，即ax - 9 = x + 1 = 0；故 a * (-1) - 9 = 0，a = -9；b = -9 - 5 = -14；曲线方程 y = -9x^2 - 14x - 2 。