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 11-26-2012 #1 (permalink) ggFTW Addict Games Awards   Join Date: Apr 2011 Posts: 222 Blog Entries: 15 iTrader: (5) IGN: shenni?? Class: HL 3rd degree polynomial (math) I don't know whether it's a good idea asking you guys, since I probably don't know the right translations for the following math problem: (am really a math noob, so idk what to do) A 3rd degree polynomial has the same zeroing as function g(x)=x²-x-2. It subtends the y - axis (ordinate) with P(0|2) and has the slope of -3. What's the equation of the 3rd degree polynomial? So far I know that the "base frame" for the polynomial is: f(x)= ax³+ bx² + cx +d and the first derivation is f′(x)= 3ax² + 2bx +c Also the zeroings of g(x) are Z1(-1|0) and Z2(2|0) Idk what to do next cos i dont really understand a thing and what I just did there lol. Also dunno whether you guys understand me (me and my translator lol). But I would appreciate any helps! __________________ follow me on ▲ instagram: shannus
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 11-26-2012 #2 (permalink) Nonsense Games Awards   Join Date: Dec 2008 Location: The lower troposphere. Posts: 999 Blog Entries: 8 iTrader: (12) IGN: Enthalpy/ Reminder Class: Duelist Guild: Guardians You are on the right track. There was be an easier way that I have forgotten however this will solve it. Use the base frame and the derivative of the base frame. First work out f(0)=2 and f'(0)=-3. (Ie substitute x for 0) You will get two equations. f(0)=a0^3+b0^2+c0+d=2 This gives d=2 And c=-3 from the derviative. Now you do it for the two zeros. Ie f(-1)=0 and f(2)=0 This will give two equations with a,b, c and d. So you can the sustitue what c and d are into them (since you already know them) and then solve two simulataneous equations. That is the method. You just need to do it now Let me know if you are still stuck. And your notation is interesting. __________________
 11-26-2012 #3 (permalink) ggFTW Addict Games Awards   Join Date: Apr 2011 Posts: 222 Blog Entries: 15 iTrader: (5) IGN: shenni?? Class: HL Amg thanks for your fast reply! And thanks, I also kind of understood what to do : so I did this like you said: f(2) = 0 8a+4b-3(2)+2 = 0 0=8a+4b-4 f(-1)=0 -a+b-3+2=0 -a+b-1 = 0 | +1 -a+b = 1 |+a b= 1+a Now I substitued b into 0=8a+4b-4 which will be: 8a+4(1+a)-4 0=8a+4+4a-4 0=12a|:12 a=0 Substitued a into 1+a: b=1+0 =1 The equation is: f(x)=x²+3x+2 Is this right? Because it's only 2. Degree polynomial thanks in advance! __________________ follow me on ▲ instagram: shannus
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 Originally Posted by Degoutant Amg thanks for your fast reply! And thanks, I also kind of understood what to do : so I did this like you said: f(2) = 0 8a+4b-3(2)+2 = 0 0=8a+4b-4 f(-1)=0 f(-1)=-a+(-1)^2*b-(-1)3+2 -a+b+3+2=0 Etc -a+b-1 = 0 | +1 -a+b = 1 |+a b= 1+a Now I substitued b into 0=8a+4b-4 which will be: 8a+4(1+a)-4 0=8a+4+4a-4 0=12a|:12 a=0 Substitued a into 1+a: b=1+0 =1 The equation is: f(x)=x²+3x+2 Is this right? Because it's only 2. Degree polynomial thanks in advance!
It is good that you are checking and seeing that the answer does not make sense.

You just made an small in substitution. I have highlighted it out. Try again and see if that fixes it.

When you work out a and b again, see if f(0)=2 just to check.
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 11-28-2012 #5 (permalink) ggFTW Addict Games Awards   Join Date: Apr 2011 Posts: 222 Blog Entries: 15 iTrader: (5) IGN: shenni?? Class: HL ah right! I forgot that the double negative becomes + f(-1)=-a+(-1)^2*b-(-1)3+2 so that'd be: 0=-a+b+5|+a a=b+5 This time a into b 0=8a+4b-4|+4 4=8(b+5)+4b 4=12b+40|-40 -36=12b|:12 b=-3 and then b back into a a=-3+5 a=2 the polynomial is: f(x)=2x³-3x²-3x+2 right? __________________ follow me on ▲ instagram: shannus
 11-28-2012 #6 (permalink) Nonsense Games Awards   Join Date: Dec 2008 Location: The lower troposphere. Posts: 999 Blog Entries: 8 iTrader: (12) IGN: Enthalpy/ Reminder Class: Duelist Guild: Guardians I checked that with each of the conditions you specified it works. Just make sure that f(-1)=0, f(2)=0, f(0)=2 (if you are in a hurry, check one of the first two as f(0) is a bad check in this case). __________________
 12-01-2012 #7 (permalink) ggFTW Addict Games Awards   Join Date: Apr 2011 Posts: 222 Blog Entries: 15 iTrader: (5) IGN: shenni?? Class: HL alright, thanks for that tip! And thanks for your time and help! I appreciate it a lot! : D __________________ follow me on ▲ instagram: shannus
 12-01-2012 #8 (permalink) Nonsense Games Awards   Join Date: Dec 2008 Location: The lower troposphere. Posts: 999 Blog Entries: 8 iTrader: (12) IGN: Enthalpy/ Reminder Class: Duelist Guild: Guardians You are welcome . Post again if you have any other questions. I should be able to do most final year highschool stuff. (If I have not forgotten, lol) __________________

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