Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Find the equation of the line tangent to the curve y=x³-6x² at its point of inflection.?

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  • Philip
    Lv 6
    1 month ago

    y = x^3 -6x^2.

    y' = 3x^2 -12x.

    y'' = 6x -12.

    Point of inflection occurs at P(a,b), where y''(a) = 0 and y(a) = b.

    Putting y'' = 0 gives x = 2 and y(2) = 2^3 -6*2^2 = 8-24 = -16. Then P = (2,-16).

    Slope of tangent line at P = y'(2) = 3*4 -12*2 = -12.

    Equation of tangent line L at P is y = -12x +b.

    P is on L. Therefore -16 = -12(2) +b, ie., b = 24-16 = 8 and eqn of L is y = -12x +8.

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  • Find when y'' = 0

    y = x^3 - 6x^2

    y' = 3x^2 - 12x

    y'' = 6x - 12

    y'' = 0

    0 = 6x - 12

    0 = x - 2

    x = 2

    Now, find the value of y' when x = 2

    y' = 3 * 2^2 - 12 * 2 = 12 - 24 = -12

    Now find the value of y when x = 2

    y = 2^3 - 6 * 2^2 = 8 - 24 = -16

    We need a line with a slope of -12 that passes through (2 , -16)

    y - (-16) = -12 * (x - 2)

    y + 16 = -12x + 24

    y = -12x + 8

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