1. ## vertical asymptote help!!!!!

the problem is:

h(x)=x^2/x^2-3x+2

i need to identity the vertical and horizontal asymptote as well as the domain

so i got, horizontal asympt y=1
vertical asymptote = 2,1
and domain is all numbers except 2 and 1

to obtain the vertical asymptote i set the denominator to equal zero, but apprently i did something wrong somewhere because the answer says that the vertical asymptote CANNOT be 2. is there another way to obtain the vertical asymptote? can i use this method for all problems?

why can the vertical asymptote NOT be 2?

thanks!

thanks!

2. ## Re: vertical asymptote help!!!!!

Originally Posted by noork85
the problem is:

h(x)=x^2/x^2-3x+2

i need to identity the vertical and horizontal asymptote as well as the domain

so i got, horizontal asympt y=1
vertical asymptote = 2,1
and domain is all numbers except 2 and 1

to obtain the vertical asymptote i set the denominator to equal zero, but apprently i did something wrong somewhere because the answer says that the vertical asymptote CANNOT be 2. is there another way to obtain the vertical asymptote? can i use this method for all problems?

why can the vertical asymptote NOT be 2?

thanks!

thanks!
Since :

$\displaystyle \displaystyle \lim_{x \to \2^{-}}f(x)=-\infty ~\text{and}~ \displaystyle \lim_{x \to \2^{+}}f(x)=+\infty$

the line $\displaystyle x=2$ is a vertical asymptote .

3. ## Re: vertical asymptote help!!!!!

im confused....

4. ## Re: vertical asymptote help!!!!!

Originally Posted by noork85
im confused....
Vertical asymptotes

5. ## Re: vertical asymptote help!!!!!

(x^2)/(x^2-3x+2) Vertical asymptotes occure where the function is undefined. A function is undefined when the denominator is zero. Where is the denominator zero? Factor the denominator (x-2)(x-1) and set them to zero. x=2 and x=1 Those are your vertical. For horizontal, the degree of the numerator and the degree of the denominator are equal, so long division on the leading expressions gives us y=1. Your horizontal. I hope that helps.

6. ## Re: vertical asymptote help!!!!!

x=1 and x=2 are vertical asymptotes. Go to 'drawing graphs' on Google to see the graph.

7. ## Re: vertical asymptote help!!!!!

Are you sure you have copied the problem correctly? x= 2 certainly is a vertical asymptote for x^2/(x^2- 3x+ 2).

On the other hand, x= 2 is not a vertical asymptote for (x- 2)/(x^2- 3x+ 2) even though the denominator is 0 there. Do you see why?