RT tomorrow midterm exam this question is the original question, hope to have a concise answer Given that the curve C: y = x3-3x2 + 2x, the line L: y = KX, and the line L and the curve C are tangent to the point (x0, Y0) (x0 ≠ 0), the equation and tangent point coordinates of the line L are obtained

RT tomorrow midterm exam this question is the original question, hope to have a concise answer Given that the curve C: y = x3-3x2 + 2x, the line L: y = KX, and the line L and the curve C are tangent to the point (x0, Y0) (x0 ≠ 0), the equation and tangent point coordinates of the line L are obtained

The equation of simultaneous curve and straight line, that is, replace y in the curve with KX, because the abscissa of the tangent point is not 0, so both sides of the equation can be directly reduced to X; it is transformed into a quadratic equation of one variable, with only one intersection point, that is, delta = B ^ 2-4ac = 0, and the solution is k = - 1 / 4; substituting the value of K into your simultaneous equation, you can get x0 = 3 / 2, Y0 = k * x0 = - 3 / 8