*oblique asymptotes of radical expressions

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Discussion Overview

The discussion centers on the oblique asymptotes of the radical expression \(y = \sqrt{x^2 + 6x}\). Participants explore how to derive these asymptotes, particularly noting that this expression is not a rational function, which is typically associated with oblique asymptotes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants note that oblique asymptotes are usually associated with rational expressions, yet \(y = \sqrt{x^2 + 6x}\) has asymptotes \(y = x + 3\) and \(y = -x - 3\).
  • One participant suggests that for large \(x\), the asymptotic behavior can be approximated as \(\sqrt{x^2 + 6x} \sim x + 3\) for \(x \to +\infty\) and \(\sqrt{x^2 + 6x} \sim -(x + 3)\) for \(x \to -\infty\), but acknowledges the need for careful handling of the square root.
  • Another participant proposes analyzing the difference \(f(x) - (ax + b)\) to derive the asymptote, providing a limit calculation to show that the function approaches the asymptote as \(x\) increases.
  • A different approach is presented where the expression is squared and manipulated into a hyperbolic form, leading to the conclusion that the asymptotes are derived from the hyperbola's properties.

Areas of Agreement / Disagreement

Participants express varying methods to derive the asymptotes, and while they agree on the existence of the asymptotes, the methods and interpretations of the radical expression's behavior lead to different perspectives. No consensus is reached on a single method of derivation.

Contextual Notes

Some limitations include the dependence on the interpretation of asymptotic behavior and the handling of the square root function, which may introduce ambiguity in the analysis.

karush
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it seems most oblique asymptotes are mostly with rational expressions but
$\sqrt{x^2+6x}$ has the asymptote of $y=x+3$ and $y=-x-3$
I don't know how this is derived since it is not a rational expression

thanks ahead:cool:
 
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karush said:
it seems most oblique asymptotes are mostly with rational expressions but
$\sqrt{x^2+6x}$ has the asymptote of $y=x+3$ and $y=-x-3$
I don't know how this is derived since it is not a rational expression

thanks ahead:cool:

For large \(x\) we have the asymtotic behavior: \[\sqrt{x^2+6x} \sim \sqrt{x^2+6x+9}=\pm (x+3)\]
Note deliberate error of use of the \(\pm\) sign, a square root is by definition positive, so as \(x \to +\infty,\ \sqrt{x^2+6x} \sim x+3\), and \(x \to -\infty,\ \sqrt{x^2+6x} \sim -(x+3)\). See plot below.

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CB
 

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karush said:
it seems most oblique asymptotes are mostly with rational expressions but
$\sqrt{x^2+6x}$ has the asymptote of $y=x+3$ and $y=-x-3$
I don't know how this is derived since it is not a rational expression

thanks ahead:cool:

As usual, you can proceed by studying the difference $f(x)-(ax+b)$ since you've got that information. For calculus, you would then use the conjuguated expression.

For $x>0$:
$\sqrt{x^2+6x}-x-3=\frac{(\sqrt{x^2+6x}-x-3)(\sqrt{x^2+6x}+x+3)}{\sqrt{x^2+6x}+x+3}=\frac{x^2+6x-(x+3)^2}{\sqrt{x^2+6x}+x+3)}=\frac{-9}{\sqrt{x^2+6x}+x+3)}$
therefore $\lim_{x->+\infty}{f(x)-x-3}=0$
you can even deduce from above that the curve is under the asymptote ($f(x)-(ax+b) < 0 $)
 
Hello, karush!

It seems most oblique asymptotes are mostly with rational expressions,
but $y \:=\:\sqrt{x^2+6x}$ has the asymptotes $y=x+3$ and $y=-x-3$
I don't know how this is derived since it is not a rational expression.
We have: .$y \:=\:\sqrt{x^2+6x} $

Square both sides: .$y^2 \:=\:x^2+6x \quad\Rightarrow\quad x^2 + 6x - y^2 \:=\:0 $

Complete the square: .$x^2 + 6x \color{red}{+ 9} - y^2 \:=\:0 \color{red}{+9} \quad\Rightarrow\quad (x+3)^2 - y^2 \:=\:9 $

Divide by 9: .$\dfrac{(x+3)^2}{9} - \dfrac{y^2}{9} \:=\:1$The graph is the upper half of this hyperbola,

. . whose asymptotes are: $y \:=\:\pm(x+3)$
 

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