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anonymous

  • 5 years ago

I am trying to sketch the curve y=x+1/x Following are the steps that are generally followed. But I don’t understand one thing First I note that while as x tends to infinity, 1/x tends to 0. Then we conclude that y tends x, since 1/x tends to 0. The thing that I don’t understand is if x tends to infinity, so why don't we say y tends to infinity too.

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  1. amistre64
    • 5 years ago
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    y = (x+1)/x ? or y = x + (1/x)

  2. anonymous
    • 5 years ago
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    No as it is

  3. anonymous
    • 5 years ago
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    I mean using the math rule

  4. anonymous
    • 5 years ago
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    First division then addition...................

  5. amistre64
    • 5 years ago
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    id say its y = inf + 0 as well...... is there an answer that says differently?

  6. anonymous
    • 5 years ago
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    Its curve sketching. If I say that way, the curve turns out to be wrong

  7. anonymous
    • 5 years ago
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    iamignorant, Is the following your expression? \[y=x+\frac{1}{x} \]

  8. anonymous
    • 5 years ago
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    Yes

  9. anonymous
    • 5 years ago
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    \[y==1+\frac{1}{x} \] The 2 attached plots should help visualize the function. The first plot is from x=-10 to +10. The scale of the plot is too large and should be reduced in your pdf viewer. The second is from x= -4000 to x= +4000. To the the casual observer the plot looks like a straight line drawn through the origin and making an angle of Pi/4 with the x axis.

  10. anonymous
    • 5 years ago
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    Just reviewed what was put on the net. The plots seem to be OK.

  11. anonymous
    • 5 years ago
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    Thanks robtobey

  12. anonymous
    • 5 years ago
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    You see the problem that is disturbing me is a little different

  13. anonymous
    • 5 years ago
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    I am trying to sketch the curve using the asymptote location method

  14. anonymous
    • 5 years ago
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    And I have presented the problem that I face while reasoning

  15. anonymous
    • 5 years ago
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    I understand that while 1/x tends to infinity y must tend to x, or else y=x will not be an asymptote

  16. anonymous
    • 5 years ago
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    But I don't understand where exactly I mis-reasoned to make it otherwise

  17. amistre64
    • 5 years ago
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    y = (x^2 +1)/x is an equivalent equation right?

  18. anonymous
    • 5 years ago
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    Yes sure

  19. amistre64
    • 5 years ago
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    then the asymptote is a slant/oblique right?

  20. anonymous
    • 5 years ago
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    Sorry, but how did you conclude that?

  21. amistre64
    • 5 years ago
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    the ends approach y=x

  22. anonymous
    • 5 years ago
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    Hold on a few minutes. I want to set up some Limits.

  23. amistre64
    • 5 years ago
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    the top is one degree highr than the bottom

  24. anonymous
    • 5 years ago
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    Actually I am not aware of that rule

  25. amistre64
    • 5 years ago
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    when we divide it all by x we get: x +1/x ------ 1

  26. anonymous
    • 5 years ago
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    Yes

  27. amistre64
    • 5 years ago
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    anyting with an x on the bottom goes to zero; which leaves us with y = x as the slant asymptote

  28. anonymous
    • 5 years ago
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    You mean when 1/x tends to zero anyting with an x on the bottom goes to zero

  29. anonymous
    • 5 years ago
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    Is that what you mean?

  30. amistre64
    • 5 years ago
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    so you have 2 curves that trail in the ends to the line y=x; and in the middle skirt along the x=0 axis: like this

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  31. anonymous
    • 5 years ago
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    Good job amistre64. I have to break off. Thank you.

  32. amistre64
    • 5 years ago
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    when the bottom of a fraction gets large; the value gets small: 1 -------------- = .000000...00001 1000000...0000

  33. amistre64
    • 5 years ago
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    so anything with an x on the bottom; as the x gets large; the value goes to zero.

  34. amistre64
    • 5 years ago
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    whats left over is what your asymptote is :)

  35. anonymous
    • 5 years ago
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    Actually, when x is tending towards infinity 1/x tends to zero (this is what you say), but x is tending to infinity so y=x+0 means y is tending to infinity. This is what I don't understand

  36. amistre64
    • 5 years ago
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    y is keeps going up and skirting next to the y=x line; instead of a y=3 line, its the y=x line. that line has a range of (-inf,inf)

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