Math help: How do I find a 3rd degree polynomial that goes through given points

[aeroguy]

Can anyone give me some direction in finding a 3rd degree polynomial (a+bt+ct^2+dt^3) whose greaph goes through the points (0,1), (1,0), (-1,0), and (2,-15)?

I'm in a Theory of Matrices class so I need to solve it using matrices probably. We are doing a lot with Guass-Jordan elimination right now. I don't necessarily want an answer, just help on where to start.

It's probably really easy and I'm gonna kick myself later.

Thanks in advance,
Nick

 

[aeroguy]

no math love on Sunday?

Come on, I just need a hint!

 

[aux]

you have an equation y = a+bt+ct^2+dt^3 and fours sets of (t,y) points; meaning that if you substitute each pair (t,y) in the equation it should hold i.e.

1 = a+b*0+c*0^2+d*0^3
0 = a+b*1+c*1^2+d*1^3
0 = a+b*(-1)+c*(-1)^2+d*(-1)^3
-15 = a+b*2+c*2^2+d*2^3

(check again if my substitution above is correct)

this is a system of 4 linear equations with 4 unknowns (a,b,c,d,) that you should know how to solve

 

[PrincessGuard]

Evaulate (a+bt+ct^2+dt^3) at the 4 points to get 4 equations in terms of a,b,c,d.

a + 0b + 0c + 0d = 1
a + 1b + 1c + 1d = 0
a - 1b + 1c - 1d = 0
a + 2b + 4c + 8d = -15

Solve the equations using whatever method to get the 4 coefficients.

Edit: Doh, beat me to it

 

[aeroguy]

Thanks, you guys rock! I appreciate it.

That was way too easy. I knew I'd kick myself when I found out.