# Thread: Slope of a Tangent Line.

1. ## Slope of a Tangent Line.

I see that one of the ways to find the slope of a tangent line P->Q is to is to build a table like the one I attached but the problem is that it is not the exact slope since a value for one is erroneous. I think that a better solution is provided with

y-y0=f'(x0)(x-x0)

I have attached the image of that formula which I believe was covered in algebra in one form. Here there is the use of f' I see so it's a little bit different. My question is about a) which is asking about the tangent line to 1/(2x+1) at x=1. I have also attached what I see to be f' or the derivative of 1/(2x+1) which is -2/(2x+1)^2 as I have in my notes. Hopefully I have all that right I'm just starting these exercises. So when I look at the formula for the slope of the tangent line I am trying to put together still what y-y0=f'(x0)(x-x0) actually looks like for 1/(2x+1) at x=1. I have a sample but there are some abbreviated steps and I'm finding I'm having a bit of trouble putting together exactly how to plug the right values into the slope of the tangent line formula and have things come out and how things cancel and so forth. I'm hoping maybe someone could outline the steps in a) so I might apply those steps to the other problems.

Thanks in advance...

2. ## Re: Slope of a Tangent Line.

I think it is y-1/3=2/9(x-1). This was explained in a reduced form. So if I understand correctly then it is solved for y so that when x goes in the slope of the tangent line comes out.