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    Name:  20161025_215827.jpg
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Size:  526.9 KBHow would I find the equation for l. I have part a but I need b and c.
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    So you're finding the gradient of the curve at the origin, (0, 0)
    You have a point, you have the gradient of the line, that's all you need to find the equation of the line  \ell .
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    (Original post by B_9710)
    So you're finding the gradient of the curve at the origin, (0, 0)
    You have a point, you have the gradient of the line, that's all you need to find the equation of the line  \ell .
    How do I find the gradient with only one point?
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    If I remember correctly you differentiate twice then sub in the x co-ordiante to get the gradient of the curve at that point.
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    (Original post by TheStudent19)
    If I remember correctly you differentiate twice then sub in the x co-ordiante to get the gradient of the curve at that point.
    I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.
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    (Original post by Youngey4)
    I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.
    Hmm good point mate. What year paper is this? Is it edexcel?
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    (Original post by TheStudent19)
    Hmm good point mate. What year paper is this? Is it edexcel?
    Yeah its edexcel but i'm not sure what year as its just a booklet our teacher put together to do over half term
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    The only way I would go about doing b is to do c first. To do c you make the two equations equal and then find the co-ordinates I think.

    EDIT: Just realised that you can't do that. Give me another minute Do you have the answers?
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    (Original post by Youngey4)
    I know how to differentiate but this is only c1 chapters 1-5 so we havent done differentiation in core, just in FP1. Thats what confused me.
    There is another way to do it without differentiation, although I do think that this question is designed so that you use differentiation.
    You have  y=x^3+3x^2-4x and suppose we have a line through the origin  y=ax . Then solving these simultaneously you get  x^3+3x^2-(4+a)x=0 right.
    Now we know that the line is tangent to the curve at the origin.So you can write  x^3+3x^2-(4+a)x in the form  x^2(x-c) since you know that there is a repeated root at the origin - which is where the  x^2 comes from.
    So this is an identity,  x^3+3x^2-(4+a)x\equiv x^2(x-c) .
    Should be straight forward to solve for a and c and so both parts of the question are done.
    This way is similar to when you use the discriminant to find points where a line is tangent to a parabola.
    In fact for this question you could use the discriminant of a cubic to get the answer but it's quite long and most people don't know it off the top of their heads I'd imagine.
 
 
 
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