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    Although there is no coordinate geometry in the c3 book this was the title i though appropiate for this qu, id appreciate any help:

    f(x)=e^(2x-1), x greater than or equal to 0
    The curve C with equation y=f(x) meets the y axis at P. The tangent to C at P crosses the x axis at Q.
    Find too 3 dp the area of triangle POQ where O is the origin.
    Id appreciate any help
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    P = (0, e^-1) by substituting x=0 into f(x)

    Now to find the tangent at P we need to differentiate to find the gradient at x = 0 which I make it to be 2e^-1 since f'(x) = 2e^(2x-1)

    The tangent is of the form y=mx+c

    => y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1

    Then to find Q,

    (2e^-1)x + e^-1 = 0 since y=0 when the tangent crosses the x-axis

    Solve to find x then you have all of the coordinates and should to be able to work out the area from there
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    ok thanks Ill try it from there
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    why here => y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1
    have you subbed in the gradient to be 2e-1 as opposed to 2e(2x-1)?
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    C is just the point where the tangent crosses the y-axis, and we want it to meet the curve f(x) when x=0 so we need c = e^-1 otherwise it wouldnt be a tangent to the curve at the point P
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    I know why we need c as e-1, but why have you put m as 2e-1 as opposed to 2e(2x-1) im confused
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    (Original post by ~NC~)
    why here => y = (2e^-1)x + e^-1 since c is where the tangent crosses the y-axis which we know already to be e^-1
    have you subbed in the gradient to be 2e-1 as opposed to 2e(2x-1)?
    The gradient of the curve at P is dy/dx at x = 0.

    dy/dx = 2e^(2x-1)

    So subbing x = 0, the gradient is 2e^-1
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    Oooo i get you
    Sorry :p: And thanks
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    thankyou I got the answer
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    No problem
 
 
 

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