Horizontal Asymtote

Question

Horizontal Asymtote

anonymous · Answer

@e.mccormick

jdoe0001 · Answer

if the "degree" of the polynomial below, is GREATER than the one atop
then the horizontal asymptote is at the x-axis, that is y = 0

anonymous · Answer

apparently that is not correct

e.mccormick · Answer

Because of the root over a square, the degree comparison is not quite that easy.

anonymous · Answer

yes it is difficult...i started and ended up getting $$\sqrt{-3/2}$$   which is incorrect.

e.mccormick · Answer

Hmmm....  Well, it is not that.  Did you look at a graph of it?

anonymous · Answer

no let me check

anonymous · Answer

well it looks like its b/w -1 and 0. I think @e.mccormick

e.mccormick · Answer

I got 2 of them... https://www.desmos.com/calculator/3xfoxgi7no

anonymous · Answer

-infinity and +infinity?

e.mccormick · Answer

Horozontal asymptotes.  So you are looking for end behavior.

anonymous · Answer

yes does'nt it look like its going to +/- infinity?

anonymous · Answer

do you know how to do this algebraically?

e.mccormick · Answer

Because of the square, $2x^2+3$ is always positive.  Therefore any value is valid in the domain for that.  The domain restriction is based on the 4x+2.  But you want the end behavior, and because this is a top over bottom power that match, that is usually a single number.  However, the top and bottom do not change signs at the same time.  When a positive is put in the top it remains positive as does the bottom.  But when a negative is in the top it is positive and the bottom is negative.  Thus the two asymptotes.

e.mccormick · Answer

Well, the degree of the top and bottom are the same, right?

anonymous · Answer

no the degree of the top is 2 and the bottom is 1

e.mccormick · Answer

No.  The degree of the top is $x^{\frac{2}{2}}$ which is 1.

anonymous · Answer

so since both are the same would'nt it be 2/4?

e.mccormick · Answer

Not quite.  That is where the not so simple comes in.  The coefficent of the first term on the top is not 2.

anonymous · Answer

4?

e.mccormick · Answer

$\sqrt{2x^2+3}$  the 2 is there.  But what else is there?

e.mccormick · Answer

As in, what is modifying the value of 2?

anonymous · Answer

Honestly, I am not sure

e.mccormick · Answer

Well, then let me show you a strange but mathematically valid thing.
$\sqrt{2x^2+3} = \sqrt{2}\sqrt{x^2+\frac{3}{2}}$

e.mccormick · Answer

Now you might think that is irrattional... but hey, it has $\sqrt{2}$ in it...  so a little irrational is normal!  =P

e.mccormick · Answer

So what is the coefficent of the top now?

anonymous · Answer

1?

e.mccormick · Answer

I factored it out to make it easier to see.....

anonymous · Answer

it is not 1?

e.mccormick · Answer

Nope.  It involves 2 but is not 2 because 2 is modified... the 2 is still under the root.

anonymous · Answer

so would it be the sqrt of 2?

e.mccormick · Answer

Yes.

Now, put those three facts together.
1) It is the same degree on top and bottom.
2) Because the top has the root of a square it is always positive, but the bottom can be negative, so there are two h-asymp.
3) You have corefficents of $\sqrt{2}$ on top and 4 on bottom.

anonymous · Answer

so it would be sqrt(2)/4?

e.mccormick · Answer

One of them is.  What is the other?

anonymous · Answer

-sqrt(2)/4

e.mccormick · Answer

Ta Da: https://www.desmos.com/calculator/bemlugqvow

anonymous · Answer

i tried both the answers and it was not accepted

anonymous · Answer

nvm it worked!

e.mccormick · Answer

As a comma separated list with symbolic notation and fractions?

anonymous · Answer

Yes that was the problem

e.mccormick · Answer

Ah, kk.  Hehe.

anonymous · Answer

Now I have a question...how did you modify the top and how did you know to do so?

e.mccormick · Answer

I didn't modify the top.  I knew that it was a square under a root and therefore the highest degree of the top was 1.

anonymous · Answer

oh because the exponent cancels because of the square root. correct?

e.mccormick · Answer

Exactly!  Because in general $\sqrt{a^2}=a$ .  The definition of absolute value.

Then, because the top and the bottom were the same degree, the asymptotes would be the ratio of their coefficents.

The tricky part was finding that and realizing the sign change could happen.  Absolute value over something that is not an absolute value.

anonymous · Answer

but why did the sqrt stay...shouldnt it have disapeared because of the 2nd power?

e.mccormick · Answer

And by tricky finding that I meant tricky finding the coeficents.  Not exactly the numbers.

e.mccormick · Answer

It dissapeared on the x.  $(2x)^2\ne 2x^2$

anonymous · Answer

ahh okay gotcha! thanks again

e.mccormick · Answer

np.  Yah, this one took a little more doing and thinking, but hopefully you will remember how this worked.