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Homework Help: 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


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    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


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    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
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