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    this question has just totally confused me - pls help !

    the points P, Q and R on the curve with equation y=1-x^3 have the x- coordinates 2, -1, and -2. the curve crosses the x axis at S. Find the gradient of the curve at P, Q, R and S.
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    you need to find the gradient function and then put in suitable x values.
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    1. Substitute the x valuse for each component of the coordinate in your function, to find the y values of the points.
    2. Use m=y2-y1/x2-x1

    An alternative way to solve this, is to differentiate and plug in values...
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    Yeah I'd differentiate and substitute in the x-values to find the answer. Using m=y2-y1/x2-x1 normally only works for straight lines.

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    (Original post by nikhiltikare)
    Yeah I'd differentiate and substitute in the x-values to find the answer. Using m=y2-y1/x2-x1 normally only works for straight lines.

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    THANK YOU GUYS SO SO SO SO MUCH !!!!! really appreciate it, but would that mean that pqr and s all have different gradients and cannot have one ???
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    Yes they have different gradients, it's a x^3 curve so therefore it has 2 turning points. If it was a straight line (y = 1-x) then it'd only have 1 gradient.

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    (Original post by nikhiltikare)
    Yes they have different gradients, it's a x^3 curve so therefore it has 2 turning points. If it was a straight line (y = 1-x) then it'd only have 1 gradient.


    thank you so much!!
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    Yeah no worries
 
 
 
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Updated: October 17, 2015
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