# Math Help - please how to find vertical and horizontal asymtopes

1. ## please how to find vertical and horizontal asymtopes

Hello everybody, how are you doing?

I'm having a hard time trying to figure out how to find vertical and horizontal asymtopes which if I'm not wrong are essentials to be able to graph a function correctly.
So, could you please teach me how to find the vertical and horizontal asymtopes as in this example

f(x) = 1/x+2 -1 , yeah I know is hard to read, I didn't know how to type it correctly, it is read: 1 over x -2 minus 1, (minus 1 is substrating both the denominador and the numerator).

if you didn't understand how I wrote that exercise, you could also see it here

http://www.accd.edu/sac/math/faculty.../PdfT1Rev2.pdf on question 4

thank you so much.

2. Asymptotes will occur where you have a denominator equal to 0. So the first one is obvious: when x = -2, you'll get one, so the equation of that asymptote is x = -2. Just rearrange the equation to get x in terms of f(x) and then you'll see the value of f(x) that produces the asympote. This equation will be y = f(a) where a is the x co-ordinate.

3. Rudipoo is correct so far. to summarize:

we find vertical asymptotes by finding the x's for which the function is undefined.

to find horizontal asymptotes, we find the limits $\lim_{x \to \infty} f(x)$ AND $\lim_{x \to - \infty}f(x)$, if either limit exists and is finite, then the finite number(s) is(are) the horizontal asymptotes.

4. Originally Posted by Jhevon
Rudipoo is correct so far. to summarize:

we find vertical asymptotes by finding the x's for which the function is undefined.
Is there a vertical asymptote for $f(x) = \sqrt{x}$? I note that f(-1) is not defined.

-Dan

5. Originally Posted by topsquark
Is there a vertical asymptote for $f(x) = \sqrt{x}$? I note that f(-1) is not defined.

-Dan
Ah, point taken, maybe i should choose my words more carefully. i was speaking of rational functions as we have here. we have vertical asymptotes when the denominator is zero, that is, when the rational function is undefined