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Math Help - gradient

  1. #1
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    gradient

    hi people, my question is: find the gradient of the curve y = x^2 - 2x - 24

    a) at the point where the curve crosses the y axis
    and
    b) at each of the points wherethe curve crosses the x axis

    for a) i differentiated (i think thats what you call what i did) like this:
    \frac{dy}{dx} = 2x - 2 then made x = 0 as would be the case with the y axis and came up with - 2 which i know is right

    but for b) i made y = 0 and used the quadratic formula for x.

    \frac{2\pm\sqrt{4 + 96}}{2a} and came up with x = 6 or x = -4

    the answer for b) written in my textbook is -10, 10

    can someone show me how i'v gone wrong please?

    thankyou
    Last edited by mark; October 27th 2009 at 02:53 AM.
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  2. #2
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    ah sorry i've just realised how i need to finish that and its pretty simple haha. sorry for wasting your time.
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  3. #3
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    Quote Originally Posted by mark View Post
    ah sorry i've just realised how i need to finish that and its pretty simple haha. sorry for wasting your time.
    I assume you substituted x = 6 and x = -4 into dy/dx ....
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  4. #4
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    yep, thats what i did
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