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    Got a question that I'm stuck on :

    Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5

    ========

    What I do is first get the gradient of the perpendicular, which is 7 . The normal will then be -1/7 right ?

    Then differentiate the curve for gradient of the curve = 4x+3 .

    I make -1/7 = 4x+3 , to find x, but I'm not sure how to do it, I bring the +3 to the left, to give -22/7 = 4x . But after finding x , it appears wrong, as its a decimal :confused:
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    You were on the right lines. You want the normal which is perpendicular to the line y=7x-5, so yes, the normal will have gradient, -1/7. However, when the normal has gradient -1/7, what is the gradient of the tangent?

    This gradient (of the tangent) is what you need to set 4x+3 equal to.
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    Oh Okay, why do I set it for the tangent though ?

    So I done : 31 (tangent gradient is 7*4+3 right , because normal is -1/7 ? ) = 4x+3 . Bring 3 to left, 28=4x , x = 7 . y = 31 ?

    I'm way off the mark though in terms of the answer, I end up with:
    7y= -x +216 ?

    EDIT: All wrong
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    (Original post by Gelato)
    Got a question that I'm stuck on :

    Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5

    ========

    What I do is first get the gradient of the perpendicular, which is 7 . The normal will then be -1/7 right ?

    Then differentiate the curve for gradient of the curve = 4x+3 .

    I make -1/7 = 4x+3 , to find x, but I'm not sure how to do it, I bring the +3 to the left, to give -22/7 = 4x . But after finding x , it appears wrong, as its a decimal :confused:
    

y = 2x^2 + 3x +4
    y' = 4x + 3

    first, if slope here is 7,
    4x + 3 = 7
    x = 1
    so, y = 9

    now, normal =  - \frac{1}{7}
    y = - \frac{1}{7} (x-1) +9
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    I tried that method and got it right at the end.

    Could you explain the steps to me ?

    I always use the tangent to find x and Y right ?

    Then I use the gradient from the normal to find the equation ?

    Thanks
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    (Original post by Gelato)
    I always use the tangent to find x and Y right ?

    Then I use the gradient from the normal to find the equation ?
    That's correct, yes.
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    Thanks for the help =)

    I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
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    (Original post by Gelato)
    I tried that method and got it right at the end.

    Could you explain the steps to me ?

    I always use the tangent to find x and Y right ?

    Then I use the gradient from the normal to find the equation ?

    Thanks
    Find the equation of the normal to the curve y=2x^2 + 3x +4 which is perpendicular to the line y=7x-5
    1. "perpendicular to the line y=7x-5" so the gradient of the tangent would be 7
    2. find x and y using the fact that gradient of the tangent is 7
    3. So the gradient of the normal would be -1/7
    4. Find the equation using -1/7 and (x,y)

    hope this helps
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    (Original post by Gelato)
    Thanks for the help =)

    I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
    The gradient of the normal is -1/ \dfrac{dy}{dx} so the normal is -1/7 when, and ONLY at the point when the gradient (\dfrac{dy}{dx}) is 7.
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    (Original post by Gelato)
    Thanks for the help =)

    I'm not entirely sure why I need to make the DY/DX = 7 though , which was why I was confused at the beginning , as I've only been working with points so far, till this one.
    look at this picture (sorry I'm terrible at drawing.. on computers especially)
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