03 Oct 08
1 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...
03 Oct 08
1 edit
Originally posted by UserChevyIf 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:
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
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03 Oct 08
1 edit
Originally posted by UserChevyyeah..in general
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!!