I am not sure of the use of the term "standard form" in this context. However, this is the table of a simple quadratic function.

The table of values for the function f(x) is

x: 1 2 3 4 5 6 7

f(x): 5 0 -3 -4 -3 0 5

Note that the dependent values are symmetric. Let's try to get the independent values to be symmetric too.

If you substitute x = w+4 you get

w: -3 -2 -1 0 1 2 3

f(x): 5 0 -3 -4 -3 0 5

Now, to standardize, let's try to get the function to go through the point (0,0).

If you then substitute f(x) = g(x)-4 you get

w: -3 -2 -1 0 1 2 3

g(x): 9 4 1 0 1 4 9

So the function g(x) = w^{2}

This means that f(x) = w^{2}-4 and **f(x) = (x-4)**^{2}-4

**===============================**

Even if this table did not resolve to a quadratic function, any finite table of values like this can be resolved to a polynomial using the Difference Calculus for discrete functions.

So if you start with the original table

x: 1 2 3 4 5 6 7

f(x): 5 0 -3 -4 -3 0 5

Take the first differences of f(x) (equivalent to the derivative of a continuous function) and you get

-5, -3, -1, +1, +3, +5

Take the second differences (second derivitive) and you get a constant difference of +2.

Since the second derivitive is +2, you get the polynomial ax^{2} + bx + c. (Since the derivative is +2, we know that a = 1, but let's estimate it for now)

Substituting from the original function table, we get the equations.

a + b + c = 5

4a +2b + c = 0

9a +3b + c = -3

Solving these we get

a=1, b=-8 and c=12, or the polynomial **x**^{2}-8x+12

but note that x^{2}-8x+12 = **(x-4)**^{2}-4, the same answer as before.

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There is a shortcut to get a polynomial traversing the points given.

If you extend the differences table you get

x 0 1 2 3 4 5 6 7

function **12** 5 0 -3 -4 -3 0 5

first **-7** -5 -3 -1 1 3 5

second ** 2** 2 2 2 2 2

Note the coefficients down the diagonal on the left-hand side: 12, -7 and 2. Using a formula called Newton's Forward Differences, you can compute the polynomial which goes through these points as

**12** **-7**x + **2***(.5(x)(x-1)) = 12 - 7x +x^{2} - x

= **12 - 8x + x**^{2}

References:

## Comments

1/5, 2/0, 3/-3, 4/-4, 5/-3, 6/0, 7/5

I assume this is...

(1,5), (2,0), (3,-3), (4,-4), (5,-3), (6,0), (7,5)

The change of x is always 1.

The y goes from...

5

0

-3

-4

-3

0

5

This is a wave.