I had a look through the notes but I couldn't find anything that seemed to fit what you are talking about (if you want to look at the notes yourself you need to download adobe acrobat reader

http://www.adobe.com/products/acrobat/readstep2.html - this might be why you can't access the notes). It's definately worth making the effort because the notes because they are

*very* useful.

Anyway, I'm not sure what you are asking here. I think you're saying that you are given the equation of the curve and the gradient of the tangent, and you have to find the equation of the line? I'll post how I would do this - if this isn't what you're looking for you may want to ignore this.

So you have the equation of your curve. If you differentiate this, you get a

*formula for the gradient* at any point in the curve. So in your derivative (the derivative is what you get when you differentiate) you can put in a value of x, say x=1 and this will give you the gradient of the curve at that exact point. Now you know that a tangent to the curve has the same gradient as the curve at this point.

But what happens if we don't have an x value, but we have our derivative and the gradient of the tangent. Well all we have to is set the derivative equal to the gradient (after all, we are trying to find the point at which the curve has the same gradient as the line we're trying to find, so we set them equal). You should have an equation now that is solveable for x. The x value you find from here is the the point on the curve where the gradient of the curve is the same as the gradient of the tangent.

Now you have an x-value on your curve where the gradient is the same as the line you are trying to find. The y-value will also be the same, so put your x-value you just found into the

*original equation* and solve for y. Now you will have a coordinate which is where your curve meets the line, so the point also lies on the line. You have a gradient and a coordinate, can you find the equation of the straight line from here?