Asymptote Question (1 Viewer)

[frenzal_dude]

hey i got no idea how to do this question, any help would be gr8ly appreciated thanx!

Sketch y=x + 1/x showing all stationary points and asymptotes.

Ok i got no idea how to sketch that, and id ont know what an asymptote is, pls help.

 

[klaw]

asymptotes are lines that an equation approaches but never reaches

 

[Dreamerish*~]

Here's a graph, I hope it helps.

The curves never approach 0. They come closer and closer but they never touch the x or y axis. Therefore the asymptote is 0.

 

[klaw]

y'=1-1/x²

When y'=0,
1/x²=1
x²=1
x=1 or -1

when x=1, y=2
when x=-1, y=-2

.: st points at (1,2) and (-1,-2)

 

[word.]

This is a graph with an oblique asymptote.
An asymptote is a line that the curve approaches but never reaches.

y = x + 1/x

lim x —> ±∞ [x + 1/x] = x
so there's an asymptote at y = x

then just differentiate for the stationary points as usual
and sub in a few points near 0 to see how it starts off
as it reaches the extremes of x it approaches the line y = x
well that's how I would do it anyway.

I'm fairly sure this is extension-1 and won't be asked.

 

[Antwan23q]

asymptote at y=x??? wha??

 

[frenzal_dude]

k thanx heaps guyz for your help!

also i was just wondering, in one of my exams i think it was yr 11 yearly or something, a question asked for the stationary points, so i worked them out by making dy/dx = 0 etc, and i made a table of x and dy/dx and put the 3 values in, the middle value being the stationary point, then a close value to the left of it, then to the right of it, to determine its nature, and i only got half marks because i didnt profe it using double derivative.

Would it be ok to work out stationary points using the double derivative, and skipping the table of x and dy/dx part?

 

[Antwan23q]

try it this way

y= x+1/x
then do the inverse
1/y=1/(x+1/x)
1/y=1/[(x^2+1)/x]
1/y=x/(x^2+1)

if x=0, 1/y=0, therefore y-> infinity...
so therefore, x=0,

and y=x+1/x
so y=(x^2+1)/x
x->infinity, y=0

 

[frenzal_dude]

Here's a graph, I hope it helps.

The curves never approach 0. They come closer and closer but they never touch the x or y axis. Therefore the asymptote is 0.

is your graph right? because in the answers it looks like 2 parabolas, with an assymptote of y=x, but in your graph the asymptote cant be y=x?

 

[Antwan23q]

I was more inclined to think that both ways are acceptable.
the table works, just as doing the second derivative

 

[word.]

is your graph right? because in the answers it looks like 2 parabolas, with an assymptote of y=x, but in your graph the asymptote cant be y=x?

She was just trying to explain what an asymptote was with the graph y = 1/x as an example.

 

[acmilan]

k thanx heaps guyz for your help!

also i was just wondering, in one of my exams i think it was yr 11 yearly or something, a question asked for the stationary points, so i worked them out by making dy/dx = 0 etc, and i made a table of x and dy/dx and put the 3 values in, the middle value being the stationary point, then a close value to the left of it, then to the right of it, to determine its nature, and i only got half marks because i didnt profe it using double derivative.

Would it be ok to work out stationary points using the double derivative, and skipping the table of x and dy/dx part?

If you do double derivative you dont need the table. They effectively do that same thing.

 

[insert-username]

As a general rule, whenever you have a function y = x + 1/x, you'll have an asymptote at y = x because you're always adding something onto that line, therefore the function will never cross that line. Another example: y = 3x -2 + 1/x. That function will have an asymptote of y = 3x - 2, because you're always adding 1/x to that value.


I_F

 

[acmilan]

In the example y = x + 1/x:

For large values of x, y = x is an assymptote.
For small values of x, y = 1/x is an assymptote.

 

[.ben]

what if therre were absolute values on teh graph and you couldn't use addition/subtraction of ordinates method?

e.g. (|x+4|-x^2+3)/|x|-x

 

[acmilan]

what if therre were absolute values on teh graph and you couldn't use addition/subtraction of ordinates method?

e.g. (|x+4|-x^2+3)/|x|-x

Let's not forget this is the 2 unit forum. Oblique assymptotes arent part of 2 unit.

 

[Trebla]

Sketch y=x + 1/x showing all stationary points and asymptotes.

Asymptotes:
The way I would do it is:
Put in the form (x - a)(y - b) = c
This is where x = a, y = b are asymptotes and c = coefficient or steepness of your graph.
.: y = x + 1/x
(y - x) = 1/x
x(y - x) = 1
(x - 0)(y - x) = 1
therefore the asymptotes are y = x and x = 0
Another way you could do it is by dividing all terms by the lowest power of x, which is a more common method. The method I used above only works for some functions, not all.

Stationary points:
Differentiate y = x + 1/x
dy/dx = 1 - 1/x²
Stationary points occur when dy/dx = 0
1 - 1/x² = 0
1/x² = 1
.: x² = 1
x = 1, -1
Sub x = 1, -1 into original equation
At x = 1, y = 2
At x = -1, y = -2
.: Stationary points are (1,2) and (-1,-2)

You can then work out concavity, and whether each of the stationary points are maximum or minimum points etc.
Your graph should look like a positive hyperbola (or two parabolas) stretching between the two asymptotes.

 

[MarsBarz]

So the initial graph posted on the first page was wrong? Any one brave enough to post a sketch of this thing?

 

[MarsBarz]

In ASCII. Wow, that's pretty leet. Very nice.

 

[word.]

y = x + 1/x

              |.            .
              |.          .
             4|.        .
              |.      .
              | .   .     
             2|   .
              | 
              |
<----------------------------->
  -3  -2  -1  |   1   2   3   
              |
          .   |2
        .   . |
      .      .|
    .        .|4
  .          .|
.            .|