Lova Posted April 12, 2009 Report Share Posted April 12, 2009 Is this true for Q5:D=|SL - SR - SM|where SL = x2-x1, SR = x4-x3 and SM = x6-x5i'm mixing them all up just please tell me if this is correct or not Reply Link to post Share on other sites More sharing options...
nevermore2010 Posted April 13, 2009 Report Share Posted April 13, 2009 Hi fellows,i really need a hand from you guys, for the q5, should we follow all the way we did for q1-4, then how we determined the x1,x2,x3, and so on, since the way it's determined are changed in q3. And the most important is should we keep using linier line to intersect the cubic-higher order polynomials or we can use a quadratic lines or so? Reply Link to post Share on other sites More sharing options...
Sunshine :D Posted April 26, 2009 Report Share Posted April 26, 2009 HI!I am doing this portfolio also. the parabola investigation, but i'm not sure if it's the same thing as yours, cuz i got a different conjecture. anyways.for number 3, investigate your conjecture for any real value of a and any placement of the vertex, refine your conjecture as necessary and prove it. maitina the labeling conventions used in parts 1 and 3 by having the intersections of the first line to be x2 and x3 and the intersections with the second line to be x1 and x4.i need help with this question. I was wondering, if we have to actually graph all the possible parabolas for a<0 and in the different quadrants and then firgure out if the conjecture is the same or not. i was thinking that there can be a different and easier way to figure this out instead of doing trial and error.also, can someone please tell me how viete's formula x1x2 = c/a and x1+x2 = -b/a applies to this investigation?... i'm not sure how that that works and in what way it works .any guidance will be of help! please. it's actually due tomorrow- i know you'll say omg why are you doing it now, but there's bin A LOT of stuff going on.thanks. Reply Link to post Share on other sites More sharing options...
Lova Posted April 26, 2009 Report Share Posted April 26, 2009 HI!I am doing this portfolio also. the parabola investigation, but i'm not sure if it's the same thing as yours, cuz i got a different conjecture. anyways.for number 3, investigate your conjecture for any real value of a and any placement of the vertex, refine your conjecture as necessary and prove it. maitina the labeling conventions used in parts 1 and 3 by having the intersections of the first line to be x2 and x3 and the intersections with the second line to be x1 and x4.i need help with this question. I was wondering, if we have to actually graph all the possible parabolas for a<0 and in the different quadrants and then firgure out if the conjecture is the same or not. i was thinking that there can be a different and easier way to figure this out instead of doing trial and error.also, can someone please tell me how viete's formula x1x2 = c/a and x1+x2 = -b/a applies to this investigation?... i'm not sure how that that works and in what way it works .any guidance will be of help! please. it's actually due tomorrow- i know you'll say omg why are you doing it now, but there's bin A LOT of stuff going on.thanks.yes there is an easier way than by trial and error, but it means that you have to proove it. I know two ways, and the simplest one, or at least most obvious, is by using the quadratic formula. First of all, make expressions for each one of the x-values from the quadratic formula and then plug them in in D=|sl- sr|=|(x2-x1)-(x4-x3)|. you will get the conjecture and it works for any lines and any parabolas. the other proof involves the roots of the polynomial but it is much tricker and it takes time (or it did for me anyway). you are also supposed to use the roots for 5 and 6 although i'm not exactly sure how yet. Reply Link to post Share on other sites More sharing options...
Lova Posted April 26, 2009 Report Share Posted April 26, 2009 and i also need some help... for 5 and 6, diffrence in the sum of the roots is zero, how is that supposed to work out? it works for q1-4, but then...? im completely lost here can someone plez help me Reply Link to post Share on other sites More sharing options...
SharkSpider Posted April 26, 2009 Report Share Posted April 26, 2009 yes there is an easier way than by trial and error, but it means that you have to proove it. I know two ways, and the simplest one, or at least most obvious, is by using the quadratic formula. First of all, make expressions for each one of the x-values from the quadratic formula and then plug them in in D=|sl- sr|=|(x2-x1)-(x4-x3)|. you will get the conjecture and it works for any lines and any parabolas. the other proof involves the roots of the polynomial but it is much tricker and it takes time (or it did for me anyway). you are also supposed to use the roots for 5 and 6 although i'm not exactly sure how yet.Please don't tell people to use the quadratic formula for this. It's a key sign to any examiner that you've missed the boat completely, because it's a long proof and it can be done in a much, much easier manner. The one hint I can give is that you need to look at both forms of a polynomial, and equate them. Doing so will make this proof a 3 liner, instead of half a page. Reply Link to post Share on other sites More sharing options...
Ongfufu Posted April 27, 2009 Report Share Posted April 27, 2009 and i also need some help... for 5 and 6, diffrence in the sum of the roots is zero, how is that supposed to work out? it works for q1-4, but then...? im completely lost here can someone plez help meLova, as I've said before, your idea of n-1 (difference in degree of polynomials) is perfectly relevant here for #5 and 6. Reply Link to post Share on other sites More sharing options...
Lova Posted April 27, 2009 Report Share Posted April 27, 2009 (edited) Lova, as I've said before, your idea of n-1 (difference in degree of polynomials) is perfectly relevant here for #5 and 6.Yes i know, but the thing is that our teacher told us not to change the straight lines to polynomials... I think I understand though how I can make a proof and everything with your idea when looking at my calculations but he was very clear at that point so I don't know if I'm going to lose any marks or so... And thanks, I've liked all your posts Edited April 27, 2009 by Lova Reply Link to post Share on other sites More sharing options...
Sunshine :D Posted April 27, 2009 Report Share Posted April 27, 2009 does anyone know how to get exact values for the intersection point on the Ti-84 caluclator? please reply asap!thanks! Reply Link to post Share on other sites More sharing options...
Lova Posted April 27, 2009 Report Share Posted April 27, 2009 does anyone know how to get exact values for the intersection point on the Ti-84 caluclator? please reply asap!thanks!sure, when it shows the graph, press "2nd" and then "trace".choose "intersection"then you are supposed to choose line 1 and line 2, then calculator will ask you "guess?" and then press "enter". you will see: x= ....., y= ...... Reply Link to post Share on other sites More sharing options...
Sunshine :D Posted April 27, 2009 Report Share Posted April 27, 2009 sure, when it shows the graph, press "2nd" and then "trace".choose "intersection"then you are supposed to choose line 1 and line 2, then calculator will ask you "guess?" and then press "enter". you will see: x= ....., y= ......thanks i knew that, but i wanted exact values.. so like in fraction not in decimal. is that possible? Reply Link to post Share on other sites More sharing options...
Lova Posted April 27, 2009 Report Share Posted April 27, 2009 thanks i knew that, but i wanted exact values.. so like in fraction not in decimal. is that possible?I have no idea, i don't think so though... Reply Link to post Share on other sites More sharing options...
Sunshine :D Posted April 27, 2009 Report Share Posted April 27, 2009 I have no idea, i don't think so though... haha thanks!i ws wondering, for the portfolio, for #5, i've read that many people got D=0, but i got that value for maybe some of my examples and the rest were some thing else like 0.1355 and stuff like that.could you be able to give any guidance on #5? Reply Link to post Share on other sites More sharing options...
edward cullen Posted April 27, 2009 Report Share Posted April 27, 2009 Im doing 5 and 6 now and i have som trouble... i have read many forums and threads and concluded there are more than one answer to this question depending on how you define D. however, my techer told us only to use straight lines and not exchange them for polynomials, so im not going to use that solution, and that D=0 seems just wrong as there is possible to get a conjecture which is similar to the one for q1-4. i know how to define D in order to get that result, but im stuck here... so anyone who has done/is doing this question in the same way as i am, could you plese help me to proove it. because i do label the intersection points in a certain way to get the conjecture to work, but when im trying to use the sum of the roots-thing it does not work as i want because the roots on each line are not the same as if I extend the definition for the D-value. i know that this might be confusing but please if anyone has any idea what error i make/what i have to change, tell me or just give me some hints. everything is much appreciated!thanks! Reply Link to post Share on other sites More sharing options...
Ongfufu Posted April 27, 2009 Report Share Posted April 27, 2009 haha thanks!i ws wondering, for the portfolio, for #5, i've read that many people got D=0, but i got that value for maybe some of my examples and the rest were some thing else like 0.1355 and stuff like that.could you be able to give any guidance on #5?If you did #5 with a cubic and 2 linear lines, then D must be 0. Reply Link to post Share on other sites More sharing options...
CocoPop Posted May 11, 2009 Report Share Posted May 11, 2009 Hi, I just started this investigation today, but I'm a bit confused. I've been experimenting with numbers 5 and 6, using Vieta's Theorem.As many have, I find that D=0 for powers greater than 2 when you intersect the polynomial with a straight line, and I get a different answer with quadratics (which is correct from what I've gathered).Is it sufficient for me to prove that D=0 using Vieta's Theorem, or do I need to go further than that. Am I going in the right direction? Can someone guide me into the right direction?Thanks Reply Link to post Share on other sites More sharing options...
Ongfufu Posted May 14, 2009 Report Share Posted May 14, 2009 Hi, I just started this investigation today, but I'm a bit confused. I've been experimenting with numbers 5 and 6, using Vieta's Theorem.As many have, I find that D=0 for powers greater than 2 when you intersect the polynomial with a straight line, and I get a different answer with quadratics (which is correct from what I've gathered).Is it sufficient for me to prove that D=0 using Vieta's Theorem, or do I need to go further than that. Am I going in the right direction? Can someone guide me into the right direction?ThanksYes, you can use Vieta's Theorem to prove that D=0 when a cubic intersects 2 linears. You can even prove the value of D for a quadratic intersecting 2 linears. But you need one more proof. Think about it, you have linear, quadratic and cubic, what other combination are you missing? 1 Reply Link to post Share on other sites More sharing options...
CocoPop Posted May 15, 2009 Report Share Posted May 15, 2009 Hi,Thanks for the response. I was able to do the proofs using Vieta's theorem, and I also used another way of proving them (a slightly less sophisticated and less holistic way for the cubics). Will the examiner think it is redundant if I show two different proofs for the same thing? Basically I talk about Vieta's theorem and show those proofs right at the end of my coursework to show that my own proofs are in fact in concordance with it, but I feel reluctant to include it in my main body because I feel that the examiner will think I'm just taking an existing formula that's already been proven, which isn't as impressive.I also wanted to know whether it would help my mark in any way if I mentioned something about polynomials of any order intersecting with other polynomials of any order, rather than just with linears. Do I only do what's mentioned on the sheet or can extending it further gain me extra marks? Reply Link to post Share on other sites More sharing options...
Ongfufu Posted May 15, 2009 Report Share Posted May 15, 2009 Hi,Thanks for the response. I was able to do the proofs using Vieta's theorem, and I also used another way of proving them (a slightly less sophisticated and less holistic way for the cubics). Will the examiner think it is redundant if I show two different proofs for the same thing? Basically I talk about Vieta's theorem and show those proofs right at the end of my coursework to show that my own proofs are in fact in concordance with it, but I feel reluctant to include it in my main body because I feel that the examiner will think I'm just taking an existing formula that's already been proven, which isn't as impressive.I also wanted to know whether it would help my mark in any way if I mentioned something about polynomials of any order intersecting with other polynomials of any order, rather than just with linears. Do I only do what's mentioned on the sheet or can extending it further gain me extra marks?You could do polynomials with any order, rather than just linears. Especially with cubic, which was asked in the question. You've done cubic with 2 linears, then its cubics and quadratics. You'll find something very interesting. Reply Link to post Share on other sites More sharing options...
CocoPop Posted May 15, 2009 Report Share Posted May 15, 2009 Hi,Thanks for the response. I was able to do the proofs using Vieta's theorem, and I also used another way of proving them (a slightly less sophisticated and less holistic way for the cubics). Will the examiner think it is redundant if I show two different proofs for the same thing? Basically I talk about Vieta's theorem and show those proofs right at the end of my coursework to show that my own proofs are in fact in concordance with it, but I feel reluctant to include it in my main body because I feel that the examiner will think I'm just taking an existing formula that's already been proven, which isn't as impressive.I also wanted to know whether it would help my mark in any way if I mentioned something about polynomials of any order intersecting with other polynomials of any order, rather than just with linears. Do I only do what's mentioned on the sheet or can extending it further gain me extra marks?You could do polynomials with any order, rather than just linears. Especially with cubic, which was asked in the question. You've done cubic with 2 linears, then its cubics and quadratics. You'll find something very interesting.Yeah I noticed that when I was looking at Vieta's Theorem. Neat stuff! Reply Link to post Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.