Finding the Tangent Line for f'(x): Solving for General Form

[Chocolaty]

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

So let's say they ask for the tangent line to the point on the graph where x=2.
I replace x by 2 in my derived equation and isolate Y. Let's say i found that Y=5
Now I need to find the equation 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?

 

[Chocolaty]

ohh please, someone has to know this!

 

[siddharth]

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

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

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

Did that help?

 

[Chocolaty]

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

 

[siddharth]

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

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

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