Tables

Some of the other mathematicians have got much better techniques for doing these but I resorted to trial and error and just looking for patterns.

I started by adding 3 to the y value because 0 to the power of anything is 0

so .. working backwards

\(y=2x^4-3\)

I worked this out quite quickly because I have had a lot of practice but you should get a better answer off someone else  

Very nice detective work, Melody!!!

I haven't seen this done before, but it appears to work!!!

Since the polynomial is symmetric about y and appears to intercept the x axis twice, I would have guessed that it was a quadratic......but yours perfectly models the data....

Geno's method is known as the sum of differences method......this would have probably worked here, but it's a bit of work.....I like your approach better......

\(\displaystyle \begin{array}{rrrrrr} x & y\\ -3 & 159\\ & & -130 \\ -2 & 29 & & 100\\ & & -30 & & -72\\ -1 & -1 & & 28 & & 48\\ & & -2 && -24 & \\ 0 & -3 & & 4 & & 48\\ & & 2 && 24 \\ 1 & -1 & & 28 & & 48 \\ & & 30 & &72 \\ 2 & 29 & & 100\\ & & 130\\ 3 & 159\\ \end{array} \)

If the function is a polynomial tabulated exactly (no decimal approximations), its degree can be found using a difference table, (see above).

If the function is linear, the entries in the first difference column (the third column in the table), will be constant

If it's a quadratic the entries in the second difference column will be constant, and so on.

In the table above, entries in the fourth difference column are constant, so it's a quartic.

The table with its differences can then be used to calculate the coefficients of the polynomial.

What is the degree of the function that generates the data shown?

\(\displaystyle \begin{array}{|r|rrrrr|} \hline x & y\\ \hline -3 & 159 \\ & & -130 \\ -2 & 29 & & 100 \\ & & -30 & & -72 \\ -1 & -1 & & 28 & & 48 \\ & & -2 & & -24 & \\ 0 & -3 & & 4 & & 48 \\ & & 2 & & 24 & \\ 1 & \color{red}d_0=-1 & & 28 & & 48 \\ & & \color{red}d_1=30 & & 72 & \\ 2 & 29 & & \color{red}d_2=100 & & 48 \\ & & 130 & & \color{red}d_3=120 & \\ 3 & 159 & & 220 & & \color{red}d_4=48 \\ & & 350 & & 168 & \\ 4 & 509 & & 388 & & \\ & & 738 & & \\ 5 & 1247& & & & \\ \hline \end{array}\)

\(\displaystyle \begin{array}{|rcll|} \hline y(x) &=& \binom{x-1}{0}\cdot {\color{red}d_0 } + \binom{x-1}{1}\cdot {\color{red}d_1 } + \binom{x-1}{2}\cdot {\color{red}d_2 } + \binom{x-1}{3}\cdot {\color{red}d_3 } + \binom{x-1}{4}\cdot {\color{red}d_4 } \\\\ &=& \binom{x-1}{0}\cdot ( {\color{red}-1 } ) + \binom{x-1}{1}\cdot {\color{red}30 } + \binom{x-1}{2}\cdot {\color{red}100 } + \binom{x-1}{3}\cdot {\color{red}120 } + \binom{x-1}{4}\cdot {\color{red}48 } \\\\ &=& -1 \\ &&+~ \underbrace{30 \cdot(x-1)}_{\text{max degree } 1} \\ &&+~ \underbrace{100 \cdot\left(\frac{x-1}{2}\right)\cdot \left(\frac{x-2}{1}\right)}_{\text{max degree } 2} \\ &&+~ \underbrace{120 \cdot \left(\frac{x-1}{3}\right)\cdot \left(\frac{x-2}{2}\right)\cdot \left(\frac{x-3}{1}\right)}_{\text{max degree } 3} \\ &&+~ \underbrace{48 \cdot\left(\frac{x-1}{4}\right)\cdot \left(\frac{x-2}{3}\right)\cdot \left(\frac{x-3}{2}\right)\cdot \left(\frac{x-4}{1}\right)}_{\text{max degree } \color{red}4} \\ &=& \mathbf{2x^4-3} \\ \hline \end{array}\)

expand see WolframAlpha:

If the difference \(d_i\) is constant, so the degree of the polynomial to calculate the coefficients is \(i\).

\({\color{red}d_4}=48\) is constant, so the degree is \({\color{red}4}\)

If the difference \(d_i \) is constant, so the degree of the polynomial to calculate the sum of the coefficients is \(i+1\).

Thanks to Geno, Melody, Heureka and Guest  !!!

Here's an algebraic solution.......

As the Guest pointed out.....this will be a 4th degree polynomial in the form

ax^4 + bx^3 + cx^2 + dx + e

Since (0, -3)  is on the graph, then  e  =  - 3

We can first solve for a and c

So we have that

a(-2)^4 + b(-2)^3 + c(-2)^2 + d(-2)  - 3  = 29

a(2)^4  + b(2)^3  + c(2)^2  + d(2)   - 3   = 29

Adding these, we have that

32a  + 8c  -  6   =  58

32a + 8c  =  64

4a + c   =  8        (1)

And  we have that

a(1)^4 + b(1)^3 + c(1)^2 + d(1)  - 3  = -1

a(-1)^4  + b(-1)^3  + c(-1)^2  + d(-1)   - 3   = -1

Adding these we get that

2a  + 2c   =  4

a + c   =  2      (2)

Subtract   (2)  from (1)

3a  =  6   ⇒   a   =  2

And    2 + c =  2    ⇒   c  = 0

And we have that

2 (1)^4  + b(1)^3  + d(1)  - 3  =  -1

2(-1)^4  + b(-1)^3  + d(-1) - 3  = - 1

Subtract these

2b +  2d  =  0

b + d  = 0       (3)

And

2(2)^4 + b(2)^3 + d(2)  - 3  = 29

2(-2)^4 + b(-2)^3 + d(-2)  - 3  = 29

Subtract these

16b + 4d  = 0

4b + d   = 0           (4)

Subtract (3)  from (4)

3b = 0

b  = 0

And    b + d   = 0  ⇒   0 + d = 0  ⇒   d  = 0

So......the function is

2x^4  - 3