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Math Help - Tangent line equations

  1. #1
    Junior Member
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    Tangent line equations

    Hey guys, I need some help with a question. I've tried to do it but I keep getting it wrong.

    Here's the original question:

    Find an equation of the tangent line to the curve at the given point.



    And here's my attempt:

    Tried using the quotient rule to find the slope since the derivative of the equation is the slope:

    (gx)(f'x)-(fx)(g'x) / (gx)^2

    (x+1)^2(3) - (3x)(2x+2) / (x+1)^4
    (x^2+2x+1)(3) - (6x^2+6x) / (x+1)^4
    (3x^2+6x+3) - (6x^2+6x) / (x+1)^4
    (-3x^2+12x+3) / (x+1)^4 >> from this point on I wasn't sure what to do and got stuck....any help is appreciated!

    Thanks,
    Grace
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  2. #2
    Newbie driegert's Avatar
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    Kingston, Ontario
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    So first off, the tangent line is just going to be a straight line:

    y=mx+b

    so, in order to find this equation we can use the point slope formula:

    y-y_{1}=m(x-x_{1})

    Now that we have that in mind, we have a point (0, 0). Now all we need is the slope (i.e., the derivative at that point).

    I get (without simplification):

    \frac {dy}{dx} = \frac {3(x+1)^{2}-2(x+1)(3x)}{(x+1)^{4}}

    Why not simplify? because I can just plug in (0, 0) and most of that mess will just disappear.

    Plug your slope back into the point slope formula and voila!

    Hope that helps!
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  3. #3
    MHF Contributor
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    Quote Originally Posted by dark-ryder341 View Post
    Hey guys, I need some help with a question. I've tried to do it but I keep getting it wrong.

    Here's the original question:

    Find an equation of the tangent line to the curve at the given point.



    And here's my attempt:

    Tried using the quotient rule to find the slope since the derivative of the equation is the slope:

    (gx)(f'x)-(fx)(g'x) / (gx)^2

    (x+1)^2(3) - (3x)(2x+2) / (x+1)^4
    (x^2+2x+1)(3) - (6x^2+6x) / (x+1)^4
    (3x^2+6x+3) - (6x^2+6x) / (x+1)^4
    (-3x^2+12x+3) / (x+1)^4 >> from this point on I wasn't sure what to do and got stuck....any help is appreciated!

    Thanks,
    Grace
    Hi Grace,

    That was good,
    now the point (0,0) is on both the tangent and the curve,
    therefore placing x=0 into the slope equation gives you the slope when x=0.

    Then you can write the tangent (line) equation (y=mx, since c=0) using the tangent's actual slope and using the fact that the point (0,0) is on the tangent.
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