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Equation of Tangent Line

  1. Jun 13, 2006 #1
    Can someone check if I did this right:

    Find the equation of the tangent line to the graph of [tex]f(x) = \frac {2x-5}{x+1}[/tex] at the point at which x = 0.

    [tex]f(x) = \frac {2(0) - 5}{(0) + 1} = -5 [/tex]

    So I got y = -5

    Then I wanted to find the derivative of the original using the quotient rule to find the slope.

    [tex] f'(x) = \frac{(2x-5)(1) - (2)(x+1)}{(x+1)^2}

    = \frac{2x-5 - 2x -2}{(x+1)^2} [/tex]

    Is the slope -7? I plugged in 0 for x.

    If so, I get :

    y -y1 = m(x-x1)

    y - -5 = m(x-0)

    y + 5 = -7x + 0

    y = -7x -5
     
    Last edited: Jun 13, 2006
  2. jcsd
  3. Jun 13, 2006 #2

    StatusX

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    I think you got the quotient rule wrong. It's low dee high minus high dee low.
     
  4. Jun 13, 2006 #3
    Not sure what you mean. I was told you can do it either of these two ways:

    [tex]F'(\frac{f}{g}) = \frac{fg' - f'g}{g^2} or \frac {f'g - fg'}{g^2} [/tex]
     
    Last edited: Jun 13, 2006
  5. Jun 13, 2006 #4

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    They can't both be right: one is the negative of the other. You want the second one.
     
  6. Jun 13, 2006 #5
    Ok so now I get: [tex] \frac{2x+1 - 2x+5}{(x+1)^2}[/tex]

    Do I plug x = 0 into that to find my slope? Which would give me m = 6
     
  7. Jun 13, 2006 #6

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    Yes, everything else you did in the first post looks right.
     
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