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Tangent lines to a f(x)

  1. Sep 28, 2006 #1
    Seems like a simple problem, but I have no idea how to accomplish this!

    f(x) = x^2 +4

    I have to find a tangent line to the graph that passes through the point (0,0).

    I know the derivative is 2x, but I don't know how to mathematically figure out the which (x,f(x)) the tangent line would pass through..!
    Any help would be appreciated
     
  2. jcsd
  3. Sep 28, 2006 #2
    If the derivative is [tex] 2x [/tex], then the slope at [tex] (0,0) [/tex] will be [tex] 0 [/tex]. So its just a horizontal line through [tex] (0,0) [/tex]
     
  4. Sep 28, 2006 #3

    I know that.

    You seee (0,0) is not a point on the graph. (0,0) is a given point (outside the graph) which I need to find a line tangent to the graph with.

    [​IMG]
     
  5. Sep 28, 2006 #4
    you have to use Newtons method [tex] x_{n+1} = x_{n} - \frac{f(x_{n})}{f'(x_{n})} [/tex]
     
  6. Sep 28, 2006 #5
    You could do it in a simpler way.
    [tex]f(x) = x^2 + 4[/tex]
    [tex]f'(x) = 2x[/tex]

    Let [tex]g(x) = f'(x).x[/tex], i.e. write the equation of the tangent line in terms of x and [tex]f'(x)[/tex].

    So you need a point where [tex]f(x) = g(x)[/tex]
    So: [tex] x^2 + 4 = (2x)x[/tex]
     
    Last edited: Sep 28, 2006
  7. Sep 29, 2006 #6

    HallsofIvy

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    The problem did not say "tangent at (0,0)".

    First, distinguish between the variable, x, and the x-value of the point of tangency. Let's call the x value at the point of tangency x0. Then the derivative at that point is 2x0 and so the tangent line has equation y= 2x0x. In order to be a tangent, that line must touch the parabola at x0: at x= x0 we must have y= 2x02= x02+ 4. Because of the symmetry, there are two lines through (0,0) tangent to y= x2+ 4
     
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