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    Right can anyone help i am stuck on this question (part b, but have included part a just in case you need it):

    A curve is defined by y=ln(x+4)

    Sketch the curve indicating the points where the curve crosses the co-ordinate axis

    part b

    The point A on the curve is such that the tangent to the curve at A is parallel to the line with equation 2y=x-4

    Find the equation of the tangent to the curve at the point A
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    Part B

    Okay this is quite a long winded question.

    You need to find the general gradient of the curve and the line by finding their derivatives which are .. 1/(x+4) and 0.5. As they are parallel you know that they are equal so that enables you to find x. x = -2. Feed that back into the original curve function to find y which is = 0.6931471806

    Now you know the gradient and the co-ordinates you can use the formula (y - y1) = m(x-x1) to find the equation.

    Final Answer = y = 0.5x + 1.693141781. Hope this helps
 
 
 
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Updated: April 5, 2011
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