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 Degoutant 11-26-2012 11:34 AM

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!

 Nebula 11-26-2012 12:28 PM

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.

 Degoutant 11-26-2012 01:20 PM

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

 Nebula 11-26-2012 02:52 PM

Quote:
 Originally Posted by Degoutant (Post 1786420) 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.

 Degoutant 11-28-2012 10:51 AM

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?

 Nebula 11-28-2012 12:23 PM

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).

 Degoutant 12-01-2012 12:40 PM

alright, thanks for that tip!
And thanks for your time and help! I appreciate it a lot! : D

 Nebula 12-01-2012 01:55 PM

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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