# Math Question

UserChevy
Posers and Puzzles 03 Oct '08 02:34
1. 03 Oct '08 02:341 edit
Actually this isn't a riddle, but a question I have concerning the graphing of an equation. If you graph a function in the y=1/x function family, something like y=(2x-7)/(3x+4), how do you determine the horizontal asymptote of the graph? The vertical asmphtote is easy enough (x=-4/3), but I am not sure how to get the horizontal one. I hope that this question makes sense...
2. 03 Oct '08 02:411 edit
Originally posted by UserChevy
Actually this isn't a riddle, but a question I have concerning the graphing of an equation. If you graph a function in the 1/x function family, something like (2x-7)/(3x+4), how do you determine the horizontal asymptote of the graph? The vertical asmphtote is easy enough (x=-4/3), but I am not sure how to get the horizontal one. I hope that this question makes sense...
If I remember correctly it would be the variable in the numerator divided by the variable in the denominator, if that makes sense. It has been three years, however, so I may be wrong. In your case:

Original Equation =

(2x-7)
-------
(3x+4)

Horizontal Asymptote =

2x
---
3x

so

2
--
3
3. 03 Oct '08 03:12
(2x-7)/(3x+4)
(6x-21)/(9x+12)
(6x+8-29)/(9x+12)
(6x+8)/(9x+12) - 29/(9x+12)
2/3 - 29/(9x+12)

The horizontal asymptote would be y=2/3.
4. 03 Oct '08 10:29
That makes sense, thanks for both answers.
5. joe shmo
Strange Egg
03 Oct '08 14:101 edit
Originally posted by UserChevy
That makes sense, thanks for both answers.
yeah..in general

if you have n(x)/d(x) = ax^m/bx^n

if m > n, then H.A does not exist

if m=n , then H.A. = a/b

if m< n, then HA is line y=0

if m > n there is a slant asymptote.. to obtain it, use long division for algebraic expressions

you will get an equation of a line + some remainder and then take the limit as x approaches + or - infinity and you get the slant (oblique asymptote)

don't know if you wanted all that, but it cant hurt!!