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Thread: Find the equation of the line?

  1. #1
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    Exclamation Find the equation of the line?

    Question:
    FInd the equation of the normal lines to the curve $\displaystyle y=x^3-3x $ which is parallel to the line of $\displaystyle 2x+18y-9=0 $

    Attempt:
    $\displaystyle f'(x)=3x^2-3$
    $\displaystyle m=9 $
    $\displaystyle x=2 ;y=2 $
    $\displaystyle y-2=9(x-2) $

    Is my equation correct?
    If not please answer in steps
    Thank you
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  2. #2
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    Quote Originally Posted by mj.alawami View Post
    Question:
    FInd the equation of the normal lines to the curve $\displaystyle y=x^3-3x $ which is parallel to the line of $\displaystyle 2x+18y-9=0 $

    Attempt:
    $\displaystyle f'(x)=3x^2-3$
    $\displaystyle m=9 $
    $\displaystyle x=2 ;y=2 $
    $\displaystyle y-2=9(x-2) $

    Is my equation correct?
    If not please answer in steps
    Thank you
    HI


    $\displaystyle y=x^3-3x\Rightarrow \frac{dy}{dx}=3x^2-3$

    This is the gradient of tangent . The gradient of normal would be perpendicular to it .

    Gradient of normal , call it M2 is $\displaystyle -\frac{1}{9}$ (do u know y)

    $\displaystyle (3x^2-3)(M2)=-1$

    $\displaystyle -\frac{1}{9}=-\frac{1}{3x^2-3}$

    solving this ... $\displaystyle x=\pm 2$

    Put this into the original equation (curve) to find the y's .
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