anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
y = (x+1)/x ? or y = x + (1/x)
anonymous
  • anonymous
No as it is
anonymous
  • anonymous
I mean using the math rule

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anonymous
  • anonymous
First division then addition...................
amistre64
  • amistre64
id say its y = inf + 0 as well...... is there an answer that says differently?
anonymous
  • anonymous
Its curve sketching. If I say that way, the curve turns out to be wrong
anonymous
  • anonymous
iamignorant, Is the following your expression? \[y=x+\frac{1}{x} \]
anonymous
  • anonymous
Yes
anonymous
  • anonymous
\[y==1+\frac{1}{x} \] 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.
anonymous
  • anonymous
Just reviewed what was put on the net. The plots seem to be OK.
anonymous
  • anonymous
Thanks robtobey
anonymous
  • anonymous
You see the problem that is disturbing me is a little different
anonymous
  • anonymous
I am trying to sketch the curve using the asymptote location method
anonymous
  • anonymous
And I have presented the problem that I face while reasoning
anonymous
  • anonymous
I understand that while 1/x tends to infinity y must tend to x, or else y=x will not be an asymptote
anonymous
  • anonymous
But I don't understand where exactly I mis-reasoned to make it otherwise
amistre64
  • amistre64
y = (x^2 +1)/x is an equivalent equation right?
anonymous
  • anonymous
Yes sure
amistre64
  • amistre64
then the asymptote is a slant/oblique right?
anonymous
  • anonymous
Sorry, but how did you conclude that?
amistre64
  • amistre64
the ends approach y=x
anonymous
  • anonymous
Hold on a few minutes. I want to set up some Limits.
amistre64
  • amistre64
the top is one degree highr than the bottom
anonymous
  • anonymous
Actually I am not aware of that rule
amistre64
  • amistre64
when we divide it all by x we get: x +1/x ------ 1
anonymous
  • anonymous
Yes
amistre64
  • amistre64
anyting with an x on the bottom goes to zero; which leaves us with y = x as the slant asymptote
anonymous
  • anonymous
You mean when 1/x tends to zero anyting with an x on the bottom goes to zero
anonymous
  • anonymous
Is that what you mean?
amistre64
  • amistre64
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
1 Attachment
anonymous
  • anonymous
Good job amistre64. I have to break off. Thank you.
amistre64
  • amistre64
when the bottom of a fraction gets large; the value gets small: 1 -------------- = .000000...00001 1000000...0000
amistre64
  • amistre64
so anything with an x on the bottom; as the x gets large; the value goes to zero.
amistre64
  • amistre64
whats left over is what your asymptote is :)
anonymous
  • anonymous
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
amistre64
  • amistre64
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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