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    This has preceding questions so I'll just fill you in:

    Part I)

    Describe fully the transformation that transforms the curve y = \frac{1}{x} to the curve y = \frac{1}{x + 2}.

    I know that's a translation of 2 units in the negative x direction parallel to the x axis.

    Then it asks to differentiate the first curve which gives -x^{-2}

    Now this last part I can't understand how I'm doing it wrong:

    Part iii)

    Use the preceding parts of the question to find the gradient of the curve y = \frac{1}{x + 2} at the point where it crosses the y axis.

    Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?
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    Hmm I understand but you get the same result dont you?

    Call me an idiot but can it not be written as:

    y = \frac{1}{2} + x^{-1}
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    Differentiating that gives exactly the same as the first curve - anyone help?
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    (Original post by Dededex)
    This has preceding questions so I'll just fill you in:

    -x^{-2}

    find the gradient of the curve y = \frac{1}{x + 2} at the point where it crosses the y axis.

    Well surely when it crosses the y axis, x = 0 - if I differentiate the curve and put in x as zero you get infinity..?
    I think it's asking for the gradient value of the (x+2)^-1 not the x^-1 graph,

    for y = (x+2)^-1
    when x = 0, y = 1/2

    The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
    So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

    Sketch the graphs.
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    (Original post by GabsyWabsyWoo)
    The y = (x+2)^-1 graph has only been translated to the left by 2 relative to y=x^-1
    So the gradient values of the new curve could be thought of as translated to the left by 2 as well.

    Sketch the graphs.
    Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0
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    (Original post by Dededex)
    Kind of - I've drawn them but still don't understand why you cant get the correct answer with x = 0
    on y=(x+2)^-1 the gradient at x=0 is what we want

    on y=x^-1, the gradient at x=2 is -1/4 as found by the gradient function y'=-(x^-2)

    if tangents are equal but translated we can assume what the y=(x+2)^-1 gradient is at x=0
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    Sorry for the bad wording. :/ in C1 you "shouldn't know" how to differentiate (x+2)^-1
    You learn that in C3; you use what's called the Chain Rule.
 
 
 
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