zeros and and asymptotes

renegade05

renegade05

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Alright, I need some help understanding a couple problems im working on.

(H.A. = Horizontal Asymptote)

First one is:

\(\displaystyle h(x)=\frac{x^2-17}{(\pi x+5)(3x^2-5x-1)}\)

So the zeros are x values when y = 0, easy...

\(\displaystyle 0=\frac{x^2-17}{(\pi x+5)(3x^2-5x-1)}\)

\(\displaystyle 0=x^2-17\)

\(\displaystyle x=+-\sqrt{17}\)

Hopefully, I have not made an error so far.

Ok, now I am asked to find the horizontal asymptote of h(x).

So h(x) numerator has degree 2 and the denominator has degree 3. Therefore, H.A. = 0, right?

How can h(x) have zeros and a H.A. at y=0 ?

I tried to graph it and it looks like it has no zeros. Please explain this to me. Thanks!!

Also, i am running into another problem with this function, f(x):

\(\displaystyle f(x)=\frac{(7x-13)(11x+25)}{x^2+7x+101}\)

The problem arises when I found the H.A. = 77 but then i graphed it and it looks like f(x) does hit y=77 at 1 point but then does appear to level off at y=77 at another. Does this still make it a H.A. ?

Thanks!!!
 
renegade05 said:
Alright, I need some help understanding a couple problems im working on.

(H.A. = Horizontal Asymptote)

First one is:

\(\displaystyle h(x)=\frac{x^2-17}{(\pi x+5)(3x^2-5x-1)}\)

So the zeros are x values when y = 0, easy...

\(\displaystyle 0=\frac{x^2-17}{(\pi x+5)(3x^2-5x-1)}\)

\(\displaystyle 0=x^2-17\)

\(\displaystyle x=+-\sqrt{17}\)

Hopefully, I have not made an error so far.

Ok, now I am asked to find the horizontal asymptote of h(x).

So h(x) numerator has degree 2 and the denominator has degree 3. Therefore, H.A. = 0, right?

How can h(x) have zeros and a H.A. at y=0 ?

I tried to graph it and it looks like it has no zeros. Please explain this to me. Thanks!!

Also, i am running into another problem with this function, f(x):

\(\displaystyle f(x)=\frac{(7x-13)(11x+25)}{x^2+7x+101}\)

The problem arises when I found the H.A. = 77 but then i graphed it and it looks like f(x) does hit y=77 at 1 point but then does appear to level off at y=77 at another. Does this still make it a H.A. ?

Thanks!!!
Click to expand...

regarding the first problem - plot it with a range of x = ±6

You'll see that the given function has 4 local extrema. 2 of those are beyond |x|> ?(17). The plot will cross x-axis at ±?17 and reach maximum (or minimum) and then assymptotically approach x-axis as it is coming down (or up).

Nice way to interrogate aproblem!!!
 
So you are saying it does have zeros and a HA at y=0.

And what about the second question i had?
 


A horizontal asymptote is a horizontal line.

We don't say, "H.A. at y = 0".

We say, "the H.A. is y = 0".

The asymptotic behavior of function h occurs "at" values of x that are far away from zero.

The graphs below show:

(1) h(x) plotted over [-4, 4]

(2) Zoomed-in at the smaller function zero

(3) Asymptotic behavior as x goes from -2 toward negative infinity

[attachment=2:ulblc4cl](1).JPG[/attachment:ulblc4cl]

[attachment=1:ulblc4cl](2).JPG[/attachment:ulblc4cl]

[attachment=0:ulblc4cl](3).JPG[/attachment:ulblc4cl]

 
So to sum up. There H.A. is y =0 and the zeros are \(\displaystyle +-\sqrt{17}\) ??
 
renegade05 said:
So to sum up. There H.A. is y =0 and the zeros are \(\displaystyle +-\sqrt{17}\) ??
Click to expand...

Correct... (for problem #1)
 
renegade05 said:
i am running into another problem with … f(x)

The problem arises when I found the H.A. = 77

I see no problem there (except that it's sloppy to write H.A.=77).

The horizontal asymptote is the line given by y = 77.


but then i graphed it and it looks like f(x) does hit y = 77 at 1 point

Does this still make it a H.A.?

Oh, I see what you meant to say. The "problem" came later, when you looked at a graph and began to doubt your algebraic result.
Click to expand...

Yes, y = 77 is still the horizontal asymptote.

Locally, a graph may touch or even cross back and forth over the function's horizontal asymptote(s). The asymptotic behavior occurs globally, not locally. I mean, we see the asymptotic behavior when we examine what the graph is doing far away from the origin (as x goes toward positive and/or negative infinity).

(Some functions have graphs that actually cross a horizontal asymptote infinitely many times.)

Here is a graph of f(x), shown in green, over the restricted domain [-400, 400]. The line y = 77 is graphed in red.

Obtaining a graph that shows what you want or expect is usually a matter of choosing an appropriate view (i.e., window settings, if you're using a graphing calculator). This often takes a little experimenting.

Cheers ~ Mark

[attachment=0:2clqco8x](4).JPG[/attachment:2clqco8x]

 
mmm4444bot... thank you for your reply, as always. BUT!! i think you may be lacking in the sleep department. I posted that 2 semesters ago in OCT 2010. You posted your reply JUNE 2011.

I am way past zeros and asymptotes now :)
 
renegade05 said:
i think you may be lacking in the sleep department

True, true. Generally true, sadly.


I posted that 2 semesters ago in OCT 2010.

Yup. Sometimes I get bored, and I go looking at old posts, finding questions that never got answered.


I intended to answer your question about the graphing software, but I'm no longer able to view the images that I uploaded to this thread. ( :? ) This is why I bumped the thread -- to remind me to ask somebody what happened to my uploads.

I use MapleV Release 5, for most graphing on the computer. Galactus recently posted a link to freeware; I have not played with that software much, yet it seems to do a good job easily.

http://www.padowan.dk/graph/

Oh, sometimes I use Paint in Windows, too, to create diagrams.
Click to expand...
 
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