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Tangent line at y = 9/sqrt(1+x)

  1. Oct 8, 2005 #1
    Find the y-intercept of the tangent line to the graph of y = (9)/sqrt(1 + x) at the point where x = 9.

    I used the quotient rule to find the derivative.

    I found that to be (-9/2(1+x)^-1/2)/(1+x) But then I am totally clued out...

    Thanks for helping me out.

    volc
     
  2. jcsd
  3. Oct 8, 2005 #2

    Physics Monkey

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    What is the relationship between the derivative and the tangent line?
     
  4. Oct 8, 2005 #3

    hotvette

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    You need to assess what you have:

    1. slope of the line you want to construct
    2. a point that the line goes through

    These 2 pieces of information should allow you to write the equation of the line. Once you have the equation, you can find the y intercept.
     
  5. Oct 8, 2005 #4
    Derivative of function = slope of the tgt line.
     
  6. Oct 8, 2005 #5
    I don't understand what you mean. :/
     
  7. Oct 8, 2005 #6

    Physics Monkey

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    If you have the equation of the tangent line, you can find the y-intercept, right? You need therefore to find the equation of the line. You have two pieces of information, the slope of the line and a point the line passes through. Is this suffucient to write an equation for the line?
     
  8. Oct 8, 2005 #7
    A line in the x-y plane is given by y=mx+b (I assume you know that). Taking the derivative and plugging in your x=9 will give you m. You still need to find b, which is precisely what the question asks for. If I were you, I would plug x=9 into your original equation, find your y value, then substitute both of those into y=mx+b and solve for b.

    Alex
     
  9. Oct 8, 2005 #8
    (-9/2(1+x)^-1/2)/(1+x) = slope of the tgt line right ? And the equation for the tangental line is y= mx + b, where m = (-9/2(1+x)^-1/2)/(1+x). To find y I need to find b and I can't seen to be able to find b.
     
  10. Oct 8, 2005 #9
    That is m(x). To find the slope at a point, you need to plug that x-value in!:smile: Do you understand why?

    Alex
     
    Last edited: Oct 8, 2005
  11. Oct 8, 2005 #10
    When I plug in the x value (-9/2(1+x)^-1/2)/(1+x) I get:
    (-9/2)((10)^-3/2) and this is x or m(x) ?

    volc
     
  12. Oct 8, 2005 #11
    What do you think it is? Remember that the derivative at a point will be the slope (m) of the graph at that point. Now, you have an x-value given to you. You need to find the cooresponding y-value so that you have a slope and a point. You can find an equation of a line using a point and the slope, thus find the y-intercept.

    If I were you, I'd go back and read up on what a derivative actually means geometrically.

    Alex
     
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