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
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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Core 3/4 question watch
- Thread Starter
- 05-04-2011 16:48
- 05-04-2011 16:59
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