*oblique asymptotes of radical expressions

[karush]

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

 

[CaptainBlack]

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

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.

View attachment 430

CB

 

[aldus]

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

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 $)

 

[soroban]

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: $y \:=\:\pm(x+3)$

 

[CellCoree]

I can explain the concept of oblique asymptotes of radical expressions. An oblique asymptote is a line that the graph of a function approaches, but does not intersect, as the input values approach infinity or negative infinity. In the case of radical expressions, the oblique asymptotes can be determined by finding the limits of the function as the input values approach infinity or negative infinity.

In the given example of $\sqrt{x^2+6x}$, we can see that as the input values approach infinity, the expression inside the square root, $x^2+6x$, becomes much larger than $x$ itself. This means that the square root will be dominated by the term $x^2$ and the function will behave similarly to $\sqrt{x^2}$ which simplifies to $x$. Similarly, as the input values approach negative infinity, the expression inside the square root will be dominated by the term $x$ and the function will behave like $\sqrt{x}$.

By finding the limits of the function as the input values approach infinity and negative infinity, we can see that the function approaches the lines $y=x$ and $y=-x$ respectively. However, the given function is shifted horizontally by a constant value of 3. This means that the oblique asymptotes will be shifted by the same amount, resulting in the asymptotes $y=x+3$ and $y=-x-3$.

In conclusion, even though the given expression is not a rational expression, we can still determine the oblique asymptotes by analyzing the behavior of the function as the input values approach infinity and negative infinity. This is a common approach in mathematics and science, where we use limits to understand the behavior of a function at extreme values. I hope this explanation helps clarify the concept of oblique asymptotes of radical expressions.