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    im probably not approaching the question properly, need a bit of support, please.
    Question: The curve C has the equation y=2e^{x}-6lnx and passes through the point P with x-coordinate 1.

    a) Find an equation for the tangent to C at P.

    The tangent to C at P meets the coordinate axes at the points Q and R.

    b) Show that the area of triangle OQR, where O is the origin, is \frac{9}{3-e}


    so, i've figured out that:

    y=2e^{x}-6lnx

    P(1, ?)

    subbing, x=1 into y

    2e-6ln1

    2e

    P(1, 2e)

    y=2e^{x}-6lnx

    \frac{dy}{dx}=2e^{x}-\frac{6}{x}

    sub x = 1

    \frac{dy}{dx}=2e-6

    \frac{dy}{dx}=2(e-3)

    here, I'm unsure if dy/dx can equal m or the gradient...???

    (y-2e)=2(e-3)(x-1)

    y-2e=2[ex-e-3x+3]

    y=2ex-2e-6x+6+2e

    y=2ex-6x+6

    y=ex-3x+3

    i really dunno if this correct or not.
    please help.
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    When you sub x = 1 into dy/dx the value you get will be the gradient, m, of the tangent. Therefore your tangent equation will be y = 2(e-3)x + c. I'm guessing you know how to find c.
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    thats fine up till where you randomly divide by 2 in the 2nd to last line of working....
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    part b) just subbing y=0 and x=0 into the eqn of the line will find the two coordinates. and then should be some simple area of triagnle = 1/2 x b x h...
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    (Original post by Dagrath)
    When you sub x = 1 into dy/dx the value you get will be the gradient, m, of the tangent. Therefore your tangent equation will be y = 2(e-3)x + c. I'm guessing you know how to find c.
    c=y-intercept so, x=0
    sub x=0 into the equation?
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    (Original post by GHOSH-5)
    part b) just subbing y=0 and x=0 into the eqn of the line will find the two coordinates. and then should be some simple area of triagnle = 1/2 x b x h...
    thank you!
 
 
 
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