Finding two lines simultaneously tangent to both parabolas.

C

cpotts13

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Nov 21, 2008
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The problem is:
Graph the two parabolas (y=x[sup:rcbaob74]2[/sup:rcbaob74]) and (y=-x[sup:rcbaob74]2[/sup:rcbaob74]+2x-5) in the same coordinate plane. Find equations of the two lines simultaneously tangent to both parabolas.
Any help with this problem would be appreciated. Thanks
 
cpotts13 said:
... Any help with this problem would be appreciated ...
Click to expand...


Hello C Potts:

Any help with what you need to know in order to start anything at all on this problem would be appreciated.

Learning how to graph y = x^2 is taught in intermediate algebra classes. Have you taken any courses in algebra?

Intermediate algebra classes also cover how to write the equation of a line.

I'm certainly not going to type up five quarters' worth of class lectures for you. Please, read the post titled "Read Before Posting"; it outlines your responsibilities for seeking guidance at this web site. In summary, you need to show some initiative so that we do not need to guess at why you cannot do anything at all on this exercise.

Here's an equation that you can start with.

no work shown or questions asked by you = no help from me

Cheers,

~ Howard I. Noe :?

 
I'm sorry if I wasn't clear enough. I thought the subject stated my problem well enough. I can graph the equations just fine. The part I am having trouble with is finding the tangent lines. I know how to find one for each equation but not one that is tangent to both at the same time.
 
cpotts13 said:
... the subject stated my problem well enough ...
Click to expand...


The subject states the exercise, but it says nothing about your problem.

You still have not asked any questions about how to start or complete this exercise.

You still have not shown any initiative at all.

Did you read the post titled "Read Before Posting"?

HAVE YOU TRIED ANYTHING AT ALL?

~ Howard I. Noe :?

 
Here's a start:

Let \(\displaystyle P(p,p^{2})\) be a point on \(\displaystyle y=x^{2}\)

The derivative is \(\displaystyle y'=2x\)

The slope at \(\displaystyle x=p \;\ is \;\ 2p\)

Let \(\displaystyle Q(q,-q^{2}+2q-5)\) be a point on \(\displaystyle y=-x^{2}+2x-5\)

The derivative is \(\displaystyle y'=-2x+2\)

The slope at \(\displaystyle x=q \;\ is \;\ -2q+2\)

We want the slope to be the same, so we equate them and get \(\displaystyle 2p=-2q+2\)

\(\displaystyle p=1-q\)

The slope of PQ is \(\displaystyle \frac{p^{2}-(-q^{2}+2q-5)}{p-q}\)

This must equal the slope at P.

Can you continue on?. Let me know if you get hung up.
 
THANK YOU GALACTUS!
That was just what I needed to get me started.
ROCK ON!
cpotts13
 
Howard,
I actually did take the time to read the aforementioned post and I am wondering now, Have you yourself read it thoroughly? The only reason I can see for your complaint is that i posted a problem with no work shown. That is because I was not sure how to start the problem. In the post, "Read Before Posting", it states that by showing no work, it can be assumed that I am having trouble staring the problem. You also seemed to think that I was having trouble with graphing the functions in the problem and implied that I needed help with inremediate algebra. I posted this problem in the calculus section for a reason. Finally, you claim that I did not show any initiative. If I did not have any initiative then I would certainly not even have attempted to get help with the problem. Please do not assume that eveyone has the same mathematical abilities as yourself. People come here to have their questions answered, not to be scrutinized. I appreciate anyone who volunteers their time to tutor, but remember that you are only tutoring, not lecturing.
Sincerely,
Kilroy Robertson/cpotts13
 
cpotts13 said:
... Have you yourself read it thoroughly? Yes.

... The only reason I can see for your complaint ... Who told you that I complained?

... you claim that I did not show any initiative ... I still do.

... Please do not assume ... Please do not assume what I assume.

... People come here to have their questions answered ... You never asked any questions. Clearly, you just don't get it.

... remember that you are ... not lecturing. Oh, yes I am. Furthermore, I am free to do so.

I see from your excited response to Galactus that you are more than satisfied with having somebody else do your independent thinking for you via handing to you on a silver platter the major portion of this exercise.

Are you one of the students who spend the entire class time sitting in the back row sucking their lower lip? (This is a rhetorical question.)

~ Howard I. Noe :roll:

Click to expand...
 
For thre record, Howard, I have an 87 percent in Calc 2 and I achieved that through my own hard work and thinking. Clearly you do not have anything better to do than argue, over the internet about someone you do not know. So, if you were wondering, I do not know you, and I do not care about what you think about me. Please remember that this is a respectable math help/tutorig forum, not my Ethics class (a class you should consider taking). Anything you say to me on this forum will no longer be taken seriously. I wish happiness and good luck to you and your family and I hope your day will be as pleasant as mine has been.
With my sincerest regards,
Kilroy Robertson
 
Mr. Potts,

In the subject line of your problem, you said "Finding two lines...". I saw that you were spoon-fed the method of finding one line. Did you find the second line?
 
cpotts13 said:
For [the] record, Howard, I have an 87 percent in Calc 2 and I achieved that through my own hard work and thinking ...

... Clearly you do not have anything better to do than argue ...

... So, if you were wondering, I do not know you ...

... remember that this is a respectable math help/[tutoring] forum, not my [ethics] class ...

... a class you should consider taking ...

... Anything you say to me on this forum will no longer be taken seriously ...
Click to expand...


Hello Kilroy:

Thank you for your thoughtful remarks. You may consider them all acknowledged by me.

I am encouraged to learn that you do well at school; most people in the world are not educated.

I am not sure which definition of the verb "to argue" you intend in your comment, but I have posted many contributions to discussions at this web site that I believe contain no arguing of any kind. (I will interpret your use of the pronoun "anything" as an unintentional exaggeration.)

The moderators keep this public forum a respectable place; I will hear from them if I cross any boundaries.

Of course, I cannot be sure in what type of ethics class you are currently enrolled, but I often sit in as a guest at many differing lectures at the University of Washington. I will take your suggestion under advisement.

Everyone is free to take anything from this forum in which they find value and discard the rest. It's always been that way.

I would like you to consider the task of tutoring somebody whom you do not know. You do not know anything about their math education. You do not know anything about why they are stuck on any given exercise. They are not sitting next to you, so it is not possible to quiz them. How would you determine where to start assisting?

I would, at the risk of upsetting you further, like to ask that when you post requests here for guidance in the future, you try to provide us with some clues as to where you're at with the exercise and what you are thinking. These types of explicit statements make the entire tutoring-by-Internet process proceed much more smoothly. Since you have a fairly solid grasp of the material, I speculate that it would not take much effort on your part to make a start on an exercise and express questions that arise in your mind while doing so.

Cheers,

~ Mark :)

 "Spoon feeding, in the long run, teaches us nothing but the shape of the spoon." ~ E.M. Forster

 
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