# General form

1. Dec 18, 2005

### Chocolaty

Okay so i have f(x)
I want to find f'(x) it's derivative
F'(x) is the slope

So lets say they ask for the tangent line to the point on the graph where x=2.
I replace x by 2 in my derived equasion and isolate Y. Lets say i found that Y=5
Now I need to find the equasion of the Tangent line. So I replace the coordonates in this equasion: (y-y1)=m(x-x1)
I isolate Y and I have my Tangent line.

Here's what I don't remember... when they ask me for the general form of the equasion, how do I find that? and also, what's the point of the general form?

2. Dec 19, 2005

### Chocolaty

ohh please, someone has to know this!

3. Dec 19, 2005

### siddharth

chocolaty, i can't figure out what you are trying to ask here.

If you have a differentiable function $y=f(x)$ then the slope of the tangent at a point $(x_0,f(x_0))$ is $f'(x_0)$ (ie, $f'(x)$ evaluated at $x_0$).

So, if you want to find the equation of the tangent line at a general point $(x_0,f(x_0))$ on a curve $y=f(x)$, you know the slope of the tangent is $f'(x_0)$ and that the line passes through $(x_0,f(x_0))$. From that you can get the general equation of the tangent.

Did that help?

Last edited: Dec 19, 2005
4. Dec 19, 2005

### Chocolaty

Maybe i wasn't clear. What i'm actually asking is how do you transform the equasion of the tangent line, once you have it, to the general form: ax+by+c=0
And, whats the point?

5. Dec 19, 2005

### siddharth

Ok, I think I see what you are trying to ask.

You will get the equation of the tangent line as
$$y-f(x_0) = f'(x_0)[x - x_0]$$

So, just expand this and collect the coefficients of x and y together. Then the equation will be of the form ax + by + c = 0