LinuxBeta Posted March 3, 2009 Report Share Posted March 3, 2009 There is no strict number. Just make sure you cover the appropriate cases. In an investigation, before you reach a conjecture you need to investigate what happens in any situation, and state what happens, while stating any limitations to your conjecture. Reply Link to post Share on other sites More sharing options...
Ongfufu Posted March 4, 2009 Report Share Posted March 4, 2009 LinuxBeta, you said you can't use Vieta's formula? I wonder why. I used it in my portfolio to prove my conjectures algebraically and I got 18/20. Was my teacher wrong? Reply Link to post Share on other sites More sharing options...
LinuxBeta Posted March 4, 2009 Report Share Posted March 4, 2009 I have no clue...I got 18/20 and I didn't use it. A friend of mine who also didn't use it got 19/20. I remember when I was doing mine, everyone on the forum was talking about Vieta's formula...so I tried to use it, and was missing a parameter. I forget what...but I asked two people (with Masters in math) about this too, and they agreed you can't use it. If you found a way, and it's correct, then congrats. My point wasn't whether or not it's wrong to use it...just that you don't exactly need it. As long as you explain things and have a logical basis for your conjecture, and you prove it using direct examples, and examine the limitations of your conjecture, you will be fine. Reply Link to post Share on other sites More sharing options...
Ongfufu Posted March 5, 2009 Report Share Posted March 5, 2009 Cool. Thanks. Reply Link to post Share on other sites More sharing options...
toa Posted March 6, 2009 Report Share Posted March 6, 2009 Actually you do not need Vieta's formulas but a small part of the theory behind the formulas gives you proof for cubics. I am working on 6 now and yeah it takes a while to find the proof for 5 but you must find it before going on to 6 because you use it in both cases I think. I'll give you a hint for 5:if ax^2 + bx + c = 0 can be simplified to the form a(x-f)(x-g) = 0 where f and g are the solutions for x (a cannot = 0), then try to simplify the cubic to a similar form. Then try to play around with the value of D. If you can't find it now I recommend going over the topic on polynomials. If you are using the Haese & Harris publications book called "Mathematics HL (Core)" then look at question 4 on page 176. Reply Link to post Share on other sites More sharing options...
Shreeyash Saboo Posted March 6, 2009 Report Share Posted March 6, 2009 Actually you do not need Vieta's formulas but a small part of the theory behind the formulas gives you proof for cubics. I am working on 6 now and yeah it takes a while to find the proof for 5 but you must find it before going on to 6 because you use it in both cases I think. I'll give you a hint for 5:if ax^2 + bx + c = 0 can be simplified to the form a(x-f)(x-g) = 0 where f and g are the solutions for x (a cannot = 0), then try to simplify the cubic to a similar form. Then try to play around with the value of D. If you can't find it now I recommend going over the topic on polynomials. If you are using the Haese & Harris publications book called "Mathematics HL (Core)" then look at question 4 on page 176.Can u help me with question 5 I'm not quite sure on how to go about it Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) I just wanna give out a hint to everyone doing the parabola investigation...Numbers 5 and 6 are NOT 0.*That's enough help * Edited March 6, 2009 by Aboo Took out the answers - Aboo Reply Link to post Share on other sites More sharing options...
Shreeyash Saboo Posted March 6, 2009 Report Share Posted March 6, 2009 I just wanna give out a hint to everyone doing the parabola investigation...Numbers 5 and 6 are NOT 0, they hold the exact same conjecture as number 4 and it's provable, you just need to figure out what to change. You've gotta change something and you'll get the exact same conjecture.Does that apply for cubic polynomials and also for higher order ones I'm not understanding what to do Help would be appreciated Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) I'm sorry I can't help you more than that, it's supposed so be an investigation. I have hinted you, now you may investigate.And yes it applies to all polynomials of any higher degree. Even if you had a x^11 polynomial in there.I'll help you out by saying that by now, I'm sure you have noticed that no matter the value of 'a', the placement of the vertex, or the value of the slope of the intersecting lines, the conjecture still holds. (Up to quadratics and linears that is). That means you've gotta change something in the structure of the question. Not just change the numbers. One more thing: Make sure the intersecting lines or functions you choose don't cross each other inside your higher order polynomial, that'll just screw everything up. Edited March 6, 2009 by Anonymouser Reply Link to post Share on other sites More sharing options...
SharkSpider Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) I just wanna give out a hint to everyone doing the parabola investigation...Numbers 5 and 6 are NOT 0, they hold the exact same conjecture as number 4 and it's provable, you just need to figure out what to change. You've gotta change something and you'll get the exact same conjecture.You're dead wrong, here. The sum of the roots is NOT changed when higher order polynomials are intersected by lines, and if you're looking at getting a conjecture similar to four you aren't approaching the problem mathematically (I'd assume you're doing a visual or 2d approach to it instead) I'd argue further, but doing so would really give away the key to the portfolio, which I'm not supposed to do, considering I know my teacher's evaluation, which was an 18, since I got a 0 in the use of technology section.EDIT: And yes, I hate using the "appeal to high scores" fallacy, but I can't really post my portfolio. Edited March 6, 2009 by SharkSpider Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) If you take higher order polynomials and intersect them by lines, you know as well as I do that the difference between the sum of the roots on each intersecting line will lead to zero, right?And what I mean by exact same conjecture is that you will still get the difference between the slopes of the intersecting lines over 'a'. (I'd use the more appropriate word than 'slope' but I'll be giving away the answer, so you may understand the word 'slope' as the way we should )Oh and just for the record, it's a 20/20 for me =P Edited March 6, 2009 by Anonymouser Reply Link to post Share on other sites More sharing options...
SharkSpider Posted March 6, 2009 Report Share Posted March 6, 2009 Ah, okay, I misinterpreted your statement, and yeah, you're right in that regard, I just thought you were talking as if the lower coefficients mattered for the intersections, and I wanted to correct that.Oh, and just for the record, my teacher gives all 0s in tech use to protest IB's including the criteria =P Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 6, 2009 Report Share Posted March 6, 2009 (edited) Haha nope they don't at all. I'm just glad that someone else has figured this out the way I did. Thumbs up man! I was the only one I knew who has figured this out with the use of 'sum of the roots'.. My whole class and even the teacher! were convinced that D=0. Thumbs up again! =PHaha that's unfair tho! Edited March 6, 2009 by Anonymouser Reply Link to post Share on other sites More sharing options...
Shreeyash Saboo Posted March 7, 2009 Report Share Posted March 7, 2009 As long as you can prove the conjecture to specify 5 i'm sure you would be correct Can some one guide me as to how many graphs / equations would be required in question 6 Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 7, 2009 Report Share Posted March 7, 2009 Shreeyash, keep in mind proof means proof in variables, not numbers. (same applies for parts 3 and 4) If you were able to prove by variables that it's equal to zero then you should be able to see the correct answer right in front of you.And there's no specific number, all you need is a few examples that support your conjecture. Reply Link to post Share on other sites More sharing options...
Ongfufu Posted March 7, 2009 Report Share Posted March 7, 2009 There's no specific requirement for graphs or examples in #6, it just says "determine" if there is a conjecture, not prove. In fact, I had no examples or graphs at all for #6, only about 5 lines of conjecture. My teacher took off 2 marks for that so I got 18. So my advice is to do 3 or 4 examples with graphs. Reply Link to post Share on other sites More sharing options...
confusedito Posted March 7, 2009 Report Share Posted March 7, 2009 Can someone please help? I worked through the first couple parts really easily, but now that I am proving my conjecture for number 3, I am stuck. I am reasonably sure that I have the correct conjecture. Any help would be greatly appreciated, even what type of proof you used. Reply Link to post Share on other sites More sharing options...
Shreeyash Saboo Posted March 7, 2009 Report Share Posted March 7, 2009 Can someone please help? I worked through the first couple parts really easily, but now that I am proving my conjecture for number 3, I am stuck. I am reasonably sure that I have the correct conjecture. Any help would be greatly appreciated, even what type of proof you used.I think you should prove it algebraically Reply Link to post Share on other sites More sharing options...
confusedito Posted March 8, 2009 Report Share Posted March 8, 2009 But by induction, contradiction, etc? I am just at a complete loss. Reply Link to post Share on other sites More sharing options...
Anonymouser Posted March 8, 2009 Report Share Posted March 8, 2009 By formal proof. As in do the same thing you're doing but with variables rather than numbers. Take the general form of a quadratic and prove that when you find SL and SR and then find D, you'd get the conjecture you have.And I didn't post the answer this time Aboo! =P Reply Link to post Share on other sites More sharing options...
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