- 03 Oct '08 02:34 / 1 editActually 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...
- 03 Oct '08 02:41 / 1 edit

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:*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...**

Original Equation =

(2x-7)

-------

(3x+4)

Horizontal Asymptote =

2x

---

3x

so

2

--

3 - 03 Oct '08 14:10 / 1 edit

yeah..in general*Originally posted by UserChevy***That makes sense, thanks for both answers.**

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!!