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Math Help - Slope of all Tangent Lines through the origin. Method 1.

  1. #1
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    Slope of all Tangent Lines through the origin. Method 1.

    https://skydrive.live.com/redir?resid=DB0189AA1CD22777!139676&v=3
    https://skydrive.live.com/redir?resid=DB0189AA1CD22777!139675&v=3

    To start with I graphed the equation 1+(x-1)^2. I see that the origin of the parabola is shifted one unit up and one unit to the right and the parabola faces upward. So method one applies the formula on a line in point slope form. So an arbitrary point a is applied and x and y are set to zero. The resulting equation is solved for a. I think I understand this far.

    What I am still figuring is y=2( (sqrt 2) - 1 )x and y=-2( (sqrt2) + 1 )x

    I don't see yet what that really means. Maybe it's just the notation but I could really use some explanation.

    Slope of all Tangent Lines through the origin. Method 1.-tangentoriginquestion.png
    Slope of all Tangent Lines through the origin. Method 1.-tangentoriginsolution.png

    Thanks in advance...
    Last edited by sepoto; October 28th 2013 at 11:19 PM.
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  2. #2
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    Re: Slope of all Tangent Lines through the origin. Method 1.

    Hey sepoto.

    All they are doing is getting an equation for the tangent line to the function you started off with that goes through the origin.

    They solve for the value of a and then plug that back into the equation for the tangent line and the value of a is such that it goes through the point (0,0).

    That is all that equation means: its an equation for the tangent line (of form y = mx + b) that is both tangent to the function and also goes through (0,0).
    Thanks from sepoto
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