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Math Help - How do I find two lines that pass through points (x, y) and are tangent to the curve?

  1. #1
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    How do I find two lines that pass through points (x, y) and are tangent to the curve?

    For example, lets say the function of the curve is y = x^3

    How would you find two lines that pass through points (2, 8) and are tangent to the curve?

    Many thanks.
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    It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...

    Your tangent is a line, so is of the form y = mx + c.

    The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.

    So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.

    Can you get to there? Can you see what to do next?
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    Quote Originally Posted by Prove It View Post
    It's possible that there's only one tangent that goes through that line and is tangent to the curve, but anyway...

    Your tangent is a line, so is of the form y = mx + c.

    The gradient of this tangent will be the same as the gradient of the curve at a point. The gradient of the curve at any point is given by the derivative.

    So differentiate y = x^3 to get the derivative, and substitute this derivative (as a function of x) as m.

    Can you get to there? Can you see what to do next?
    y = x^3
    y' = 3x^2

    I think this is where I am stuck. I'm not sure how to proceed next
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  4. #4
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    OK, you don't know the point where the tangent will touch your curve, so you are just going to use your derivative as your m value for the moment.

    Substituting gives y = (3x^2)x + c, or y = 3x^3 + c

    Now use your point to find c...
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  5. #5
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    Am I supposed to substitute this way?

    8 = 3(2)^3 + c
    8 = 24 + c
    c = 8 - 24 or c = -16
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    Correct.

    That means your tangent line is of the form y = mx - 16.

    You are almost there, now you need to use your point to find m.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Correct.

    That means your tangent line is of the form y = mx - 16.

    You are almost there, now you need to use your point to find m.
    8 = m(2) - 16
    8 = 2m - 16
    8 +16 = 2m
    24 = 2m
    24/2 = m
    m = 12

    Therefore

    y = 12x - 16

    Is this correct?
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    Looks good to me...
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    Thanks a lot!
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