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Based on this question, how does one analytically compute the asymptotes for the equation defined by:

$$(x+y+2)(x+y)^2 = x+9y-2 ?$$

I can easily plot the curve...

enter image description here

and can manipulate equation for large $x$, but is there a more direct function that provides the asymptotes?

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    $\begingroup$ Do you want something like eq = (x + y + 2) (x + y)^2 == x + 9 y - 2; AsymptoticSolve[eq, y, x -> Infinity, Reals] ? $\endgroup$ Commented Nov 25, 2025 at 18:32
  • $\begingroup$ @ydd: Yep... I do! (If you post this, I'll accept.). Thanks! $\endgroup$ Commented Nov 25, 2025 at 18:35
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    $\begingroup$ The link to the question you cited is dead. $\endgroup$ Commented Nov 26, 2025 at 3:25
  • $\begingroup$ @A.Kato it looks like David is well reputed enough on the math SE that the “voluntarily removed” question might still be visible to him. Perhaps, if it’s within the rules, he can post a screenshot of the question in, erm, question? $\endgroup$ Commented Dec 2, 2025 at 4:26

2 Answers 2

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I think this is a good use case for AsymptoticSolve:

eq = (x + y + 2) (x + y)^2 == x + 9 y - 2;

AsymptoticSolve[eq, y, x -> Infinity, Reals]
(*{{y -> -(2/(1/x)^(1/3)) - x}}*)
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    $\begingroup$ If you need a higher-order approximation to the asymptote: AsymptoticSolve[eq, y, {x, ∞, 5}, Reals] $\endgroup$ Commented Nov 26, 2025 at 10:37
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Linear asymptotes are determined by the top degree terms, which can be obtained by homogenizing with a new variable and then setting it to zero.

poly = (x + y + 2)*(x + y)^2 - (x + 9 y \[Minus] 2);
homogpoly = ResourceFunction["PolynomialHomogenize"][poly, {x, y}, z]
topform = homogpoly /. z -> 0

(* Out[94]= x^3 + 3 x^2 y + 3 x y^2 + y^3 + 2 x^2 z + 4 x y z +
2 y^2 z - x z^2 - 9 y z^2 + 2 z^3

Out[95]= x^3 + 3 x^2 y + 3 x y^2 + y^3 *)

Now solve for one variable in terms of the other.

Union[Solve[topform == 0, x]]

(* Out[96]= {{x -> -y}} *)
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  • $\begingroup$ Wow... an even better solution! ($+1$) $\endgroup$ Commented Nov 27, 2025 at 0:06

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