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Thread: Turning point of the curve

  1. #1
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    Turning point of the curve

    The curve $\displaystyle y=hx^2+\frac{k}{x}$ has a turning point at $\displaystyle P(\frac{1}{2},6)$.Find

    (i)the value of $\displaystyle h$ and $\displaystyle k$

    (ii)the equation of tangent to the curve at $\displaystyle x=1$
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  2. #2
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    Quote Originally Posted by mastermin346 View Post
    The curve $\displaystyle y=hx^2+\frac{k}{x}$ has a turning point at $\displaystyle P(\frac{1}{2},6)$.Find

    (i)the value of $\displaystyle h$ and $\displaystyle k$

    (ii)the equation of tangent to the curve at $\displaystyle x=1$
    the point $\displaystyle P\left(\frac{1}{2} , 6 \right)$ is on the curve ...

    $\displaystyle 6 = \frac{h}{4} + 2k$


    since $\displaystyle P$ is a turning point ...

    $\displaystyle y'\left(\frac{1}{2}\right) = 0$

    use the above equation to get a second equation in terms of $\displaystyle h$ and $\displaystyle k$ ... then solve the system of equations.
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    still dont understand..please help
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    Quote Originally Posted by mastermin346 View Post
    still dont understand..please help
    show me the derivative of y , then I'll provide more direction.
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