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    Need help on Question please. jamielovesmusic123

    This Question is on CALCULUS:

    Find the point on the curve f(x) = x2 (X Squared) - 6x + 11 where the tangent is parallel to the x-axis?

    Thank you :)

    1. avatar image

      lalaloopsy15

      is it h.l or o.l

    2. avatar image

      Dazzla16

      Here's the solution:

      The tangent of the curve parallel to the x-axis will have a slope of 0.

      Then we find the derivative of f(x) = x^2 - 6x + 11 to find the slope of any tangent.

      dy/dx = 2x - 6

      If slope = 0,

      2x - 6 = 0

      2x = 6

      x = 3

      Then we find the corresponding y-value for x = 3

      y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2

      Therefore the point is (3,2)

    3. avatar image

      jamielovesmusic123

      Thak you Dazzla16 :)

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