Equation for the Quartic Polynomial Function Represented by Table

Determine an Equation for the Quartic Polynomial Function Represented by the Table of values.

x / y

-3 / 91

-2 / 21

-1 / 3

0 / 1

1 / 3

2 / 21

3 / 91

Can someone please not only help me but explain how I would do a question like this? I'm pretty sure I will see this on my test but need to understand the steps involved in order to complete this question.

Re: Equation for the Quartic Polynomial Function Represented by Table

A "quartic polynomial" is one with highest power "4". That means it can be written as . To determine that polynomial you need to find the values of the 5 coefficients, a, b, c, d, and e. Replace x and y with any 5 of the (x, y) pairs given (I recommend (-2, 21), (-1, 3), (0, 1), (1, 3), and (2, 21) just because they are simplest).

(Noticing that this is an [b]even[b] function simplifies things a lot!)

Re: Equation for the Quartic Polynomial Function Represented by Table

Quote:

Originally Posted by

**HallsofIvy** A "quartic polynomial" is one with highest power "4". That means it can be written as

. To determine that polynomial you need to find the values of the 5 coefficients, a, b, c, d, and e. Replace x and y with any 5 of the (x, y) pairs given (I recommend (-2, 21), (-1, 3), (0, 1), (1, 3), and (2, 21) just because they are simplest).

(Noticing that this is an [b]even[b] function simplifies things a lot!)

I'm not quite sure I completely get this..

Let's say I take the base: y= ax^4+ bx^3+ cx^2+ dx+ e

................Sub in (1,3) : 3= a(1)^4 + b(1)^3 + c(1)^2 + d(1) + e........ from here, I still can not solve anything? What do I do from here?

......................................3=a1 + b1 + c1 + d1 + e...............................?

Re: Equation for the Quartic Polynomial Function Represented by Table

3=a1 + b1 + c1 + d1 + e is just one equation, you have 5 unknowns so you will need 5 equations so that you can solve it like a simultaneous equation. As Hallsofivy said you wont have to do so much work if you make use of the fact that it is an even function

Re: Equation for the Quartic Polynomial Function Represented by Table

Quote:

Originally Posted by

**Shakarri** 3=a1 + b1 + c1 + d1 + e is just one equation, you have 5 unknowns so you will need 5 equations so that you can solve it like a simultaneous equation. As Hallsofivy said you wont have to do so much work if you make use of the fact that it is an

even function

Ok, So I did that, now this is what I have:

#> (x,y) -> equation that goes with that point

1> (-2,21) -> 21 = a(16) + b(-8) + c(4) +d(-2) + e

2> (-1,3) -> 3 = a(1) + b(-1) + c(1) + d(-1) + e

3> (0,1) -> 1 = a + b + c + d + e

4> (1,3) -> 3 = a(1) + b(1) + c(1) + d(1) + e

5> (2,21) -> 21 = a(16) + b(8) + c(4) + d(2) + e

How do I solve this?

*I need to get this done tonight. I just wanted to mention that I really appreciate the help :).

Re: Equation for the Quartic Polynomial Function Represented by Table

You must have been taught how to solve a system like this:

x+2y+5z=9

2x-3y+z=1

-x+y+2z=0

You need to use the same method to find a,b,c,d,e

Also note the similarity between (1) and (5), and (2) and (4)

Re: Equation for the Quartic Polynomial Function Represented by Table

Quote:

Originally Posted by

**Shakarri** You must have been taught how to solve a system like this:

x+2y+5z=9

2x-3y+z=1

-x+y+2z=0

You need to use the same method to find a,b,c,d,e

Also note the similarity between (1) and (5), and (2) and (4)

I don't recall ever seeing that before. Am I adding/subtracting downwards? Even If that's it, I still have 5 unknown variables; a,b,c,d,e.

Re: Equation for the Quartic Polynomial Function Represented by Table

Ok so I completed the table of values. Found it is the 4th difference (Quartic) and the difference is 24.

The Value of the Leading Coefficient is 1. -> (The Difference) / n!

That's all I have figured out so far. Don't know where to go from here.

Re: Equation for the Quartic Polynomial Function Represented by Table

Still Working on it...

Before I Had:

1> (-2,21) -> 21 = a(16) + b(-8) + c(4) +d(-2) + e

2> (-1,3) -> 3 = a(1) + b(-1) + c(1) + d(-1) + e

3> (0,1) -> 1 = a + b + c + d + e

4> (1,3) -> 3 = a(1) + b(1) + c(1) + d(1) + e

5> (2,21) -> 21 = a(16) + b(8) + c(4) + d(2) + e

Now...

(0,1) -> 1 = 0a + 0b + 0c + 0d + e

............1=e

3=a-b+c-d+e

3=a+b+c+d+e

21=16a-8b+4c-2d+e

21=16a+8b+4c+2d+e

So I took out e, which I know is 1.

3-1=a-b+c-d

3-1=a+b+c+d

21-1=16a-8b+4c-2d

21-1=16a+8b+4c+2d

..Am I getting closer?

Re: Equation for the Quartic Polynomial Function Represented by Table

Yes you are making good progress, notice that these equations are almost the same.

3-1=a-b+c-d

3-1=a+b+c+d

(just rewriting them with only a constant on the left hand side)

2=a-b+c-d

2=a+b+c+d

If you subtract the second from the first you get

2-2=(a-b+c-d)-(a+b+c+d)

Which simplifies a lot.

You can do something similar with these two

21-1=16a-8b+4c-2d

21-1=16a+8b+4c+2d

You get two equations for a and c which you can solve like normal simultaneous equations. Then after that you should know a, c and e; when you put the values for those into any two of your original equations you will get two simultaneous equations with only b and d in them.

Re: Equation for the Quartic Polynomial Function Represented by Table

2=a-b+c-d

__2=a+b+c+d__

0= -2b -2d

20=16a-8b+4c-2d

__20=16a+8b+4c+2d__

0= -16b -4d

Did the last two points also,

90=81a-27b+9c-3d

__90=81a+27b+9c+3d__

0= -54b -6d

All the results only give the same two variables, b and d. Since "0=" how would I even 'sub in' anything to try to solve for either? Sorry, I'm confused, how would I get an equation for 'a' and 'c' if they cancel out ie: 81a-81a ...?