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    The curve C has equation

    y=(x+3)(x-8) ,x>0
    x
    A). Find dy in its simplest form.
    dx
    B). Find an equation of the tangent to C at the point where x=2

    Any ideas please?
    X
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    Expand the brackets then split into 3 fractions


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    y=(x+3)(x-8) : Expand
    y=x^2 -5x-24 : Now differentiate
    dy/dx= 2x-5 : Part a complete
    Alternitavely you could use the product rule.
    Find the equation of the tangent. You know dy/dx is the gradient of the curve at that point so it is also the gradient of the tangent at that point (since classically tangents were used to calculate rates of change. Its why we use Δy/Δx)
    dy/dx = 2x-5 : Plug in x coordinate
    dy/dx = 2(2) -5 : Get dy/dx by arithmetic methods
    dy/dx = -1
    Hence since dy/dx = m therefore
    y=-x+c : Since you need a y coordinate plug in x into the original equation to get your y value at that point.
    y=(2+3)(2-8)
    y=-30
    Back to the tangent equation
    y=-x+c :since we have both x and y we can find c. (2,-30)
    -30=-(2)+c therefor c= -28
    Therefor the equation y=-x-28
    I hope this helps.
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    There seems to be a divide by x in the original question
    Just FYI posting full solutions is generally discouraged in the maths forum unless after a bit of advice the person still really needs it


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    Ooo didnt see that. As for the whole full solution posting, i didnt know that so thanks. Back to the question, the advice above mine is pretty much the best way to go.
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    If you expand the brackets you get a quadratic over x
    divide that by x, eg X^2 becomes x
    for a start

    does that help put you in the right direction?
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    Hi everyone,
    Thanks for all the help it was great and I have handed in the work now just to let you know.
    So yeah cheers
 
 
 

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