1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General form

  1. Dec 18, 2005 #1
    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. jcsd
  3. Dec 19, 2005 #2
    ohh please, someone has to know this!
     
  4. Dec 19, 2005 #3

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    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?
     
    Last edited: Dec 19, 2005
  5. Dec 19, 2005 #4
    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?
     
  6. Dec 19, 2005 #5

    siddharth

    User Avatar
    Homework Helper
    Gold Member

    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?