1. Oct 19, 2006

### Johnnycab

This is a example in my book

a rational function like f(x) = [(x+2)(x-5)]/[(x-3)(x+1)] has vertical asymptotes at x=3 and x=(-1)

Is all you do is switch the addition and subtraction signs in [(x-3)(x+1)]?

The same equation has a horizontal asymptotes at y=1, which i cant seem to understand where it came from

The is the best i could explain this, thank you for reading this

Last edited: Oct 19, 2006
2. Oct 19, 2006

Do you mean, a horizontal asymptote? Here, this rough sketch might help: http://archives.math.utk.edu/visual.calculus/1/horizontal.5/index.html" [Broken].

Last edited by a moderator: May 2, 2017
3. Oct 19, 2006

### Johnnycab

-sorry iam unable to view the flash pictures, because i cant install macromedia plugin (wont verify)
-could someone give me a written explanation please, iam still lost

4. Oct 20, 2006

### HallsofIvy

No, you don't ever just arbitrarily "switch signs"! A rational function has vertical asymptotes where the denominator is 0 (and the numerator isn't). The denominator of f(x) is (x-3)(x+1) which equals 0 when x- 3= 0 or x+ 1= 0. To solve x- 3= 0, add 3 to both sides: x- 3+ 3= x= 0+ 3= 3. To solve x+ 1= 0, subtract 1 from both sides: x+ 1- 1= x= 0- 1= -1. See? I didn't "switch signs" anywhere!

I feel I should also point out that the "+" and "-" signs in
x= +3 and x= -1 are not "addition" and "subtraction" signs- they are "positive" and "negative" signs. The difference is subtle but important. (TI calculators have separate "subtraction" and "negative" keys for that reason.)

Last edited by a moderator: Oct 20, 2006
5. Oct 21, 2006