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Thread: Finding the equation

  1. #1
    Member Awsom Guy's Avatar
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    Finding the equation

    How do I find the equation of the tangent to the graph of f(x)=Sqrt x at the point x=9.
    I know the gradient of the tangent which is 1/6.
    But I have no idea how to do this.
    Any help will be appreciated.
    THANKS
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  2. #2
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    Quote Originally Posted by Awsom Guy View Post
    How do I find the equation of the tangent to the graph of f(x)=Sqrt x at the point x=9.
    I know the gradient of the tangent which is 1/6.
    But I have no idea how to do this.
    Any help will be appreciated.
    THANKS
    $\displaystyle f(x)=\sqrt{x}$ .. this is the equation of the curve .

    $\displaystyle f(9)=3$

    so y=3
    Last edited by mathaddict; Oct 21st 2009 at 07:01 AM.
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  3. #3
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    Hello Awsom Guy
    Quote Originally Posted by Awsom Guy View Post
    How do I find the equation of the tangent to the graph of f(x)=Sqrt x at the point x=9.
    I know the gradient of the tangent which is 1/6.
    But I have no idea how to do this.
    Any help will be appreciated.
    THANKS
    As you say, $\displaystyle f'(9) = \frac{1}{2\sqrt9}=\frac16$.

    You now need $\displaystyle f(9) = \sqrt9 = 3$

    So the tangent is at the point $\displaystyle (9,3)$.

    Now use $\displaystyle y - y_1 = m(x-x_1)$, the equation of the line with gradient $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$.

    When simplified, this comes to $\displaystyle 6y = x+9$.

    Can you fill in the details?

    Grandad
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    i dont understand how you got 1/6 for m
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  5. #5
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    Hello sjara
    Quote Originally Posted by sjara View Post
    i dont understand how you got 1/6 for m
    The value of $\displaystyle m$ is the gradient of the curve at the point where $\displaystyle x = 9$. We get this by differentiating $\displaystyle f(x)$, and then plugging in that value, $\displaystyle x = 6$.

    $\displaystyle f(x) = \sqrt{x}=x^{\frac12}$

    $\displaystyle \Rightarrow f'(x)=\tfrac12x^{-\frac12}=\frac{1}{2\sqrt{x}}$

    So when $\displaystyle x = 9, f'(9)= \frac{1}{2\sqrt{9}}=\frac16$

    And this is the value of $\displaystyle m$, the gradient of the tangent when $\displaystyle x = 9$.

    Grandad
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  6. #6
    Member Awsom Guy's Avatar
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    Thanks

    Thanks I managed to get that myself anyway. Thanks.
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