Finding asymptotes of f(x) = arccos ((262.1874-65.0884x)/80sqrt(1.06667x^2-2.1332x+17.0652))

[ParateSai]

Hey! I need to find the asymptote of the following eqn. for my IB math project. I need to find the asymptotes to be able to find a certain value for an angle for the project that I am working on:
f(x) = arccos ((262.1874-65.0884x)/80sqrt(1.06667x^2-2.1332x+17.0652))

I know that there are 2 asymptotes as I graphed it, but I need the asymptote as x approaches negative infinity. We have learnt about limits, but we have not learnt to use limits as something approaches positive or negative infinity.
I think that limits are needed to solve this, but I'm not so sure. I mainly need the working to show how I got the answer (I am not allowed to directly get answer from graphing calc).

It would be very helpful if the working is also shown.
Thank you very much!!

 

[stapel]

Hey! I need to find the asymptote of the following eqn. for my IB math project. I need to find the asymptotes to be able to find a certain value for an angle for the project that I am working on:
f(x) = arccos ((262.1874-65.0884x)/80sqrt(1.06667x^2-2.1332x+17.0652))

I know that there are 2 asymptotes as I graphed it, but I need the asymptote as x approaches negative infinity. We have learnt about limits, but we have not learnt to use limits as something approaches positive or negative infinity.
I think that limits are needed to solve this, but I'm not so sure. I mainly need the working to show how I got the answer (I am not allowed to directly get answer from graphing calc).
View attachment 36327

Isn't the range of the inverse cosine restricted?

It would be very helpful if the working is also shown.

Actually, our doing this for you would likely *not* be "very helpful", since you already have worked examples in your textbook and in your class notes. Instead, it would be helpful (for your own learning) for *you* to work on this.

Please re-read the "Read Before Posting" message, and reply with the requested information. Thank you!

 

[Dr.Peterson]

Hey! I need to find the asymptote of the following eqn. for my IB math project. I need to find the asymptotes to be able to find a certain value for an angle for the project that I am working on:
$$f(x) =\arccos \left(\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}}\right)$$

I know that there are 2 asymptotes as I graphed it, but I need the asymptote as x approaches negative infinity. We have learnt about limits, but we have not learnt to use limits as something approaches positive or negative infinity.
I think that limits are needed to solve this, but I'm not so sure. I mainly need the working to show how I got the answer (I am not allowed to directly get answer from graphing calc).
View attachment 36327
It would be very helpful if the working is also shown.
Thank you very much!!

Have you tried searching (in your textbook's index or online) for information about limits at infinity?

You can start by finding the limits of the argument of the arccos, [imath]\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}}[/imath]. The last example here is similar.

Then show us your work, and we can help you refine it.

 

[BigBeachBanana]

[math]\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}} = \dfrac{1}{80} \times \sqrt{\dfrac{(262.1874-65.0884x)^2}{1.06667x^2-2.1332x+17.0652}}[/math]
Expand the numerator under the radical and evaluate the limit.

 

[Dr.Peterson]

[math]\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}} = \dfrac{1}{80} \times \sqrt{\dfrac{(262.1874-65.0884x)^2}{1.06667x^2-2.1332x+17.0652}}[/math]
Expand the numerator under the radical and evaluate the limit.

That's not quite exactly true; you have to think carefully about signs. That's why the limits to the left and right are different. My approach is a little different, but has the same issue with signs.

 

[Steven G]

[math]\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}} = \dfrac{1}{80} \times \sqrt{\dfrac{(262.1874-65.0884x)^2}{1.06667x^2-2.1332x+17.0652}}[/math]
Expand the numerator under the radical and evaluate the limit.

That is only true if the numerator is positive. Of course, it would be positive if x is approaching neg infinity.

Ooops, Dr Peterson beat me to it.

 

[khansaheb]

Hey! I need to find the asymptote of the following eqn. for my IB math project. I need to find the asymptotes to be able to find a certain value for an angle for the project that I am working on:
f(x) = arccos ((262.1874-65.0884x)/80sqrt(1.06667x^2-2.1332x+17.0652))

I know that there are 2 asymptotes as I graphed it, but I need the asymptote as x approaches negative infinity. We have learnt about limits, but we have not learnt to use limits as something approaches positive or negative infinity.
I think that limits are needed to solve this, but I'm not so sure. I mainly need the working to show how I got the answer (I am not allowed to directly get answer from graphing calc).
View attachment 36327
It would be very helpful if the working is also shown.
Thank you very much!!

Split the problem into three parts - first part x < 0, second part when x >0 and finally x =0.

What do you get when you factor out 'x' from the numerator and the denominator (for the terms inside the arguments of the 'inverse cosine'), simplify and then take the limit of the argument.

 

[ParateSai]

Isn't the range of the inverse cosine restricted?



Actually, our doing this for you would likely *not* be "very helpful", since you already have worked examples in your textbook and in your class notes. Instead, it would be helpful (for your own learning) for *you* to work on this.

Please re-read the "Read Before Posting" message, and reply with the requested information. Thank you!

The range of inverse cosine is restricted but the internal function has a range which is within the appropriate range of inverse cosine.

Also this is not from a textbook, it is for a project. We only learn limits that do not involve infinity, but the project is a project that we have to complete ourselves and sometimes it can come up with equations which are not part of our syllabus

 

[ParateSai]

Have you tried searching (in your textbook's index or online) for information about limits at infinity?

You can start by finding the limits of the argument of the arccos, [imath]\dfrac{262.1874-65.0884x}{80\sqrt{1.06667x^2-2.1332x+17.0652}}[/imath]. The last example here is similar.

Then show us your work, and we can help you refine it.

Thank you, let me try this and see how that helps

 

[ParateSai]

I am trying he methods that have been discussed. When I get an answer, I shall update.
Thanks!

 

[ParateSai]

Hey!
I used the methods provided and I got the answer. I believe it should be correct. The asymptote for the negative values of x was:
y=0.6636, which when converted to degrees is around 38.02.

The calculations that I did give me this answer which I compared from the graphing calculator, and this answer is correct!

Thank you guys very much