find the equation of the normal to the curve y=3x2-2x-1which

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  • find the equation of the normal to the curve y=3x2-2x-1which is perpendicular to the line y=x-3

    Solution:

    y = 3x^2 - 2x - 1 (1) Its derivative is dy/dx = 6x - 2

    Let the equation of tangent line be y = mx + c1 = x + c1 (2)

    (Tangent line is parallel to y = x - 3,so its gradient is 1)

    Let 6x - 2 = 1,we obtain x = 1/2

    Sub x = 1/2 into (1),we get y = 3/4 - 1 - 1 = - 5/4

    Coordinates of tangent point are (1/2,-5/4)

    Let the equation of normal line be y = -x + c2 (3)

    Sub x = 1/2 and y = -5/4 into (3),

    we have c2 = 1/2 - 5/4 = -1/4

    Sub c2 = -1/4 back into (3),finally,the normal equation is

    y = -x - 1/4 (Ans)