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Math Help - equation of tangent #2

  1. #1
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    equation of tangent #2

    Find the equation of the straight line that is tangent to the graph of f(x)=(x+1)^0.5 and parallel to x-6y+4=0


    my attempt:
    y=x^0.5+1^0.5
    y'=0.5x^-0.5

    M_t=0.5/sqrt(x)

    not sure if I am doing it right and where to head from here..
    please state the steps needed to finish this question. Thank you
    Last edited by mr fantastic; August 12th 2009 at 04:04 PM. Reason: Changed post title
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  2. #2
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    You need to learn how to do derivatives.
    y = \sqrt {x + 1} \, \Rightarrow \,y' = \frac{1}<br />
{{2\sqrt {x + 1} }}
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  3. #3
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    Quote Originally Posted by Plato View Post
    You need to learn how to do derivatives.
    y = \sqrt {x + 1} \, \Rightarrow \,y' = \frac{1}<br />
{{2\sqrt {x + 1} }}
    so with this derivative, how do i find the equation parallel to x-6y+4=0 now?

    so far i have:

    y=1/6x +b
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  4. #4
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    Quote Originally Posted by skeske1234 View Post
    so with this derivative, how do i find the equation parallel to x-6y+4=0 now?
    The slope of that line is \frac{1}{6}
    So you need to find the value of x giving \frac{1}{6}.
    Solve \frac{1}{{2\sqrt {x + 1} }} = \frac{1}{6}\, \Rightarrow \,x = 8.
    Thus the point of tangency is (8,3).
    Now write the tangent line.
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