# Math Help - slant asymptote

1. ## slant asymptote

show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.

2. Originally Posted by johntuan
show that the curve sqrt(x^2+4x) has two asymptotes: y=x+2 and y+-x-2.
A 'nasty' way (mathematicians turn away):

$y = \sqrt{(x + 2)^2 - 4} \rightarrow \sqrt{(x+2)^2} = \pm(x + 2)$.

A much better way is to expand $\sqrt{x^2+4x}$ using the generalised binomial theorem and then take the limit.

3. haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function

4. Originally Posted by johntuan
haha..but i dont understand how +-(x+2) becomes asymptotes?...the way i would get S.A is long division or the limit way...but this one is weird because its not a rational function
Slant asymptotes are lines that the curve approaches as x --> +oo or x --> -oo. As x --> +oo or -oo, $(x+2)^2 - 4 \rightarrow (x+2)^2$ and the square root of this is $\pm (x+2)$. This is not a rigorous way of doing it. The mathematicians of MHF will pounce on this and make mince meat of it.

But I'm not sure you're at a level where a rigorous approach would make much sense, so you'll just have to make do with the one above.

5. ## Re: slant asymptote

Hi all. I was confused by this limit too. I am not quite sure what is wrong with this reasoning. I have a suspicion but I would like to see other thoughts.

For x > 0, sqrt(x^2 + 4x) is the same as (x^2 + 4x)/sqrt(x^2 + 4x) = (x + 4)/sqrt(1 + 4/x).
At this point, why can we just let 4/x ->0 as x -> inf and we get the limit tending to x + 4?

6. ## Re: slant asymptote

the graph of $y = \sqrt{x^2+4x}$ is the upper branch of a hyperbola with a horizontal transverse axis ...

$y^2 = x^2+4x$

$y^2+4 = (x+2)^2$

$(x+2)^2 - y^2 = 4$

note the standard form for a hyperbola is $\dfrac{(x-h)^2}{a^2} - \dfrac{(y-k)^2}{b^2} = 1$

$\dfrac{(x+2)^2}{2^2} - \dfrac{y^2}{2^2} = 1$ , a hyperbola centered at $(h,k) = (-2,0)$

Asymptotes have slope $m = \pm \dfrac{b}{a} = \pm \dfrac{2}{2} = \pm 1$

since the asymptotes pass through the hyperbola center ...

$y - 0 = -1(x + 2) \implies y = -x - 2$

and

$y-0 = 1(x+2) \implies y = x+2$

7. ## Re: slant asymptote

Originally Posted by cjfogel
I am not quite sure what is wrong with this reasoning.
The first thing I notice, is that it is not rigorous to state $(x+2)^2-4 \to x+2$. Instead we should be saying something like
$(x+2)^2-4 = (x+2)^2{(x+2)^2-4 \over (x+2)^2} = (x+2)^2\left(1 - {4 \over (x+2)^2}\right) \to (x+2)^2$
Then we get that
$y = \sqrt{x^2+4x} \to \sqrt{(x+2)^2} = \left|x+2\right|$

And the modulus signs are critical, because when $x < -2$ as it is when $x \to -\infty$, this gives us $y \to -(x+2) = -x-2$.

8. ## Re: slant asymptote

OK. I realise you meant that you have some different reasoning that you know to be wrong.
Originally Posted by cjfogel
For $x > 0$, $\sqrt{x^2 + 4x}= {x^2 + 4x \over \sqrt{x^2 + 4x}} = {x + 4 \over \sqrt{1 + \frac4x}}$.
At this point, why can we just let $\tfrac4x \to 0$ as $x \to \infty$ and we get the limit tending to $x + 4$?
The problem comes in how $\sqrt{1+\frac4x} \to 1$. The (Maclaurin) series expansion of $\sqrt{1+y}$ about $y=0$ is $\sqrt{1+y} = 1 + \tfrac12 y + \ldots$ so
\begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &= {(x+4)\sqrt{1-\frac4x} \over \sqrt{1+\frac4x}\sqrt{1-\frac4x}} \\ & \approx {(x+4)(1-\frac2x) \over 1 - \frac4x} = {x + 4 - 2 - \frac8x \over 1 - \frac4x} = {x+2 - \frac8x \over 1 -\frac4x} \\ &\approx x+2 \end{aligned}
for large $x$.

Alternatively using the Maclaurin series above and ${1 \over 1+y} = 1 - y + \ldots$, we have
\begin{aligned}{x + 4 \over \sqrt{1 + \frac4x}} &\approx {x+4 \over 1 + \frac2x} \\ &\approx (x+4) (1 - \frac2x) = x + 4 - 2 - \frac8x \\ &\approx x+2\end{aligned}