graph of function with asymtotes

**thereddevils** · March 5th 2009, 06:11 AM

Sketch the graph for each of the following functions , showing clearly the asymtotes .

(1) $y=\frac{x+1}{(x-1)(x-2)}$

(2) $y=\frac{x-1}{x(x+1)}$

For (1) I know that x=1 and x=2 are asymtotes as they make the function undefined . And that 's the only information i can gather . I'm not sure bout the rest ..

Thanks for your help.

**Soroban** · March 5th 2009, 08:00 AM

Hello, thereddevils!

Sketch the graph for each of the following functions, showing clearly the asymptotes.

$(1)\;\;y=\:\:\frac{x+1}{(x-1)(x-2)}$

I know that $x = 1, x = 2$ are asymptotes, as they make the function undefined.

You're right . . . Those are the vertical asymptotes.
We must also look for horizontal asymptotes.

They occur when $y \:=\:\lim_{x\to\infty}f(x)$ has a finite value.

We have: . $y \:=\:\lim_{x\to\infty}\frac{x+1}{x^2-3x+2}$

Divide top and bottom by $x^2\!:\;\;\frac{\dfrac{x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2} + \dfrac{2}{x^2}}$

So we have: . $y \:=\:\lim_{x\to\infty}\left(\frac{\frac{1}{x} + \frac{1}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}}\right) \:=\:\frac{0+0}{1-0+0} \:=\:0$

Therefore, $y = 0$ (the $x$ -axis) is a horizontal asymptote.

Code:

                      |   :           :*
                      |   :           :
                      |  *:           : *
                      |   :           :  *
                      |   :           :    * 
                      | * :           :           *
  --------------------+---:-----------:--------------------
    *                 *   :           :
         *           *|   :     *     :
            *      *  |   :   *   *   :
               *      |   :  *     *  :
                      |   :           :
                      |   : *       * :
                      |   :           :
                      |   :           :
                      |   :*         *:
                      |   :           :

**thereddevils** · March 6th 2009, 04:31 AM

Originally Posted by Soroban

Hello, thereddevils!

You're right . . . Those are the vertical asymptotes.
We must also look for horizontal asymptotes.

They occur when $y \:=\:\lim_{x\to\infty}f(x)$ has a finite value.

We have: . $y \:=\:\lim_{x\to\infty}\frac{x+1}{x^2-3x+2}$

Divide top and bottom by $x^2\!:\;\;\frac{\dfrac{x}{x^2} + \dfrac{1}{x^2}}{\dfrac{x^2}{x^2} - \dfrac{3x}{x^2} + \dfrac{2}{x^2}}$

So we have: . $y \:=\:\lim_{x\to\infty}\left(\frac{\frac{1}{x} + \frac{1}{x^2}}{1 - \frac{3}{x} + \frac{2}{x^2}}\right) \:=\:\frac{0+0}{1-0+0} \:=\:0$

Therefore, $y = 0$ (the $x$ -axis) is a horizontal asymptote.

Code:

                      |   :           :*
                      |   :           :
                      |  *:           : *
                      |   :           :  *
                      |   :           :    * 
                      | * :           :           *
  --------------------+---:-----------:--------------------
    *                 *   :           :
         *           *|   :     *     :
            *      *  |   :   *   *   :
               *      |   :  *     *  :
                      |   :           :
                      |   : *       * :
                      |   :           :
                      |   :           :
                      |   :*         *:
                      |   :           :

Thanks a lot Soroban . So I will have enough information to sketch by just finding the vertical and horizontal asymtotes ? I don see how you get the three curves there .

**thereddevils** · March 9th 2009, 11:02 PM

Originally Posted by thereddevils

Sketch the graph for each of the following functions , showing clearly the asymtotes .

(2) $y=\frac{x-1}{x(x+1)}$

This is my attempt for #2

The vertical asymtotes are x=0,-1 and the horizontal asymtotes is y=0 .

I tried to find the max and min points using the algebraic method . This is where i get stucked .

$yx^2+(y-1)x-x+1=0$

$(y-1)^2-4y\geq0$

$y\geq5.828$ and $y\leq0.1716$

I know that the graph doesn't exist between 0 and 0.1716 and hence it exists between 0.1716 and 5.828 .

I am not sure whether my working is correct . Can someone pls correct me ? Really Thanks .

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