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Math Help - Parallel and Tangent Lines, finding an equation

  1. #1
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    Parallel and Tangent Lines, finding an equation

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

    I've found so far that the slope of sqroot(x+1)=1/(sqroot(x+1) + sqroot(x+1))

    as well as the equation of the parallel line to be y=x/6 + 2/3.

    How do I solve for the other equation now?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    "Find an equation of the straight line that is tangent to the graph of f(x)=sqroot(x+1) and parallel to x-6y+4=0."

    I've found so far that the slope of sqroot(x+1)=1/(sqroot(x+1) + sqroot(x+1))

    as well as the equation of the parallel line to be y=x/6 + 2/3.

    How do I solve for the other equation now?
    you have an awkward way of writing the derivative. notice that \frac 1{\sqrt{x + 1} + \sqrt{x + 1}} = \frac 1{2\sqrt{x + 1}}

    anyway, remember that the derivative gives the slope. since the slope of the line is 1/6, you want the tangent line passing through the point x such that the derivative is 1/6, that is: \frac 1{2 \sqrt{x + 1}} = \frac 16

    solve for x, and you can find the coordinates of the point. then use that point in the point-slope form to find the equation of the line.

    remember the point slope form says the equation of a line is given by: y - y_1 = m(x - x_1)

    where m is the slope of the line and (x_1,y_1) is a point the line passes through
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  3. #3
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    Hello, Jeavus!

    Your derivative is incorrect . . .


    Find an equation of the straight line that is tangent to the graph of f(x)\:=\:\sqrt{x+1}
    and parallel to x-6y+4\:=\:0

    The line is: . y \:=\:\frac{1}{6}x + \frac{2}{3} . . . Hence, its slope is \frac{1}{6}

    The derivative of the function is: . f'(x)\;=\;\frac{1}{2}(x+1)^{-\frac{1}{2}} \;=\;\frac{1}{2\sqrt{x+1}}

    So we have: . \frac{1}{2\sqrt{x+1}} \:=\:\frac{1}{6}


    Can you finish it now?

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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, Jeavus!

    Your derivative is incorrect . . .
    not incorrect, just in an ugly format. it seems he used the limit definition to compute the derivative and didn't simplify afterwards
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