# Thread: graph of function with asymtotes

1. ## graph of function with asymtotes

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

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

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

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 ..

2. Hello, thereddevils!

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

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

I know that $\displaystyle x = 1, x = 2$ 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 $\displaystyle y \:=\:\lim_{x\to\infty}f(x)$ has a finite value.

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

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

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

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

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3. Originally Posted by Soroban
Hello, thereddevils!

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

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

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

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

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

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

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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 .

4. Originally Posted by thereddevils
Sketch the graph for each of the following functions , showing clearly the asymtotes .

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

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 .

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

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

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

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 .