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    A curve has the equation y=2(x-1)/x^2 +3 and crosses the x-axis at point A.
    a) show that the normal to the curve at A has the equation y=2-2x
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    (Original post by SpongyMesophyll)
    A curve has the equation y=2(x-1)/x^2 +3 and crosses the x-axis at point A.
    a) show that the normal to the curve at A has the equation y=2-2x
    What have you tried so far?
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    (Original post by SeanFM)
    What have you tried so far?
    I've tried using the quotient rule, where I get to (-2x^2 +4+6)/(x^2 +3)^2
    I don't know where to go from there, or even if I'm right
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    (Original post by SpongyMesophyll)
    I've tried using the quotient rule, where I get to (-2x^2 +4+6)/(x^2 +3)^2
    I don't know where to go from there, or even if I'm right
    Edit: I think you mean 4x there, unfortunate typo

    After that, think about what you've actualy found using the quotient rule and what is required/given in the question.
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    sorry i meant to put 4x instead of 4. I don't know how to get to 2x or how to simplify from there. sorry xD
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    (Original post by SpongyMesophyll)
    sorry i meant to put 4x instead of 4. I don't know how to get to 2x or how to simplify from there. sorry xD
    Apologies, you are right - it is 4x.

    So what have you actually calculated using the quotient rule, and how will it be useful to the question?

    What parts of the question can you now do stuff with?
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    I believe that I have found the gradient. apart from that I am lost
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    (Original post by SpongyMesophyll)
    I believe that I have found the gradient. apart from that I am lost
    Please quote me so I know you have replied

    Correct, you've found f'(x), a function to find the gradient at a given point x.

    The question talks about a point A, gives you some information about it and asks you to find something about it. How can you do those things? The function you've found plays a part and I think you will know how to use it when you need it. Becauae you've differentiated f (x) for some reason which is right, which leads me to suspect that you know you need to use it for something.
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    (Original post by SeanFM)
    Please quote me so I know you have replied

    Correct, you've found f'(x), a function to find the gradient at a given point x.

    The question talks about a point A, gives you some information about it and asks you to find something about it. How can you do those things? The function you've found plays a part and I think you will know how to use it when you need it. Becauae you've differentiated f (x) for some reason which is right, which leads me to suspect that you know you need to use it for something.
    I only integrated because that was the topic in which the question was in. I really am lost xD.
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    (Original post by SpongyMesophyll)
    I only integrated because that was the topic in which the question was in. I really am lost xD.
    What's the point A? As in, what are it's coordinates?
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    (Original post by SpongyMesophyll)
    I only integrated because that was the topic in which the question was in. I really am lost xD.
    Think about the questiom asked above. The natural step is to find A first and then you may see how to use it.
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    (Original post by SeanFM)
    Think about the questiom asked above. The natural step is to find A first and then you may see how to use it.
    I got that y=-2/3 when x=0. help XD
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    (Original post by SpongyMesophyll)
    I got that y=-2/3 when x=0. help XD
    Before I check it, is that good or bad.. reassuring.. puzzling?
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    (Original post by SeanFM)
    Before I check it, is that good or bad.. reassuring.. puzzling?
    I think it may be right. I'm just so lost and don't know what to do
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    (Original post by SpongyMesophyll)
    I think it may be right. I'm just so lost and don't know what to do
    Okay, let's assume it's right. The question asks you to find the normal to the curve at A.. you've found the point A. How do you find the equation of the normal to the curve at that point?


    If it helps, past this point (given your previous working) it is no longer C3 knowledge but something you have covered in AS Maths.
 
 
 
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