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    Hi, how do I do question 8iii?

    http://i.imgur.com/iomYmdR.png

    I differentiated the equation to 9x^2 - 2x^-2 - 7 = 0

    Then I used the logic that x = 0 and y = -7, yet the mark scheme said that the answer was y = -2.

    Thank you.
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    (Original post by Peppercrunch)
    Hi, how do I do question 8iii?

    I differentiated the equation to 9x^2 - 2x^-2 - 7 = 0

    Then I used the logic that x = 0 and y = -7, yet the mark scheme said that the answer was y = -2.

    Thank you.
    We can't see the question, can you post an image/link?
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    (Original post by Peppercrunch)
    Question:
    You cannot substitute x = 0 into the differential to find the y value, it is true that x = 0 at Q but you need to find the equation of the tangent.
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    (Original post by lizard54142)
    You cannot substitute x = 0 into the differential to find the y value, it is true that x = 0 at Q but you need to find the equation of the tangent.
    How would I find the equation of the tangent?
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    (Original post by Peppercrunch)
    How would I find the equation of the tangent?
    Sub x=1 into the original equation of the curve. As the tangent at the stationary point will have a gradient of 0, it will just be a line, and so the value when the tangent crosses the x axis will be whatever you get when you sub x=1 into the original equation of the curve.
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    (Original post by aoxa)
    Sub x=1 into the original equation of the curve. As the tangent at the stationary point will have a gradient of 0, it will just be a line, and so the value when the tangent crosses the x axis will be whatever you get when you sub x=1 into the original equation of the curve.
    Oh, thank you. Didn't think it would be that easy.
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    (Original post by Peppercrunch)
    Oh, thank you. Didn't think it would be that easy.
    You're welcome - if you're not sure how hard it should be, just look at the number of marks, if you're doing loads of working for a 1/2 mark question, you're probably missing something obvious
 
 
 
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