x Turn on thread page Beta
 You are Here: Home >< Maths

# Core 3/4 question watch

1. 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
2. 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

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 5, 2011
Today on TSR

### University rankings 2019

Cambridge at number one

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams