# Thread: Equation for the Quartic Polynomial Function Represented by Table

1. ## 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.

2. ## 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 $y= ax^4+ bx^3+ cx^2+ dx+ e$. 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!)

3. ## Re: Equation for the Quartic Polynomial Function Represented by Table

Originally Posted by HallsofIvy
A "quartic polynomial" is one with highest power "4". That means it can be written as $y= ax^4+ bx^3+ cx^2+ dx+ e$. 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...............................?

4. ## 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

5. ## Re: Equation for the Quartic Polynomial Function Represented by Table

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 .

6. ## 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)

7. ## Re: Equation for the Quartic Polynomial Function Represented by Table

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.

8. ## 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.

9. ## Re: Equation for the Quartic Polynomial Function Represented by Table

Still Working on it...

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?

10. ## 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.

11. ## 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 ...?