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## how do I write the function f(x) in standard form

using these points how would I solve to write the function for f(x) in standard form?

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

Hi L;
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.
Are you studying waves?  I do not think so.  This is not consistent with your previous questions.
Is this a misunderstanding?
You should use a graphing calculator for this problem. Press Stat Edit and put the x points in L1 and the y points in L2. Then press Stat Calc Lin Reg to get the equation.

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) = w2

This means that f(x) = w2-4  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.

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 ax2 + 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 x2-8x+12

but note that x2-8x+12 = (x-4)2-4, the same answer as before.

==================================

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 -7x + 2*(.5(x)(x-1))  = 12 - 7x +x2 - x
= 12 - 8x + x2

References: