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

Question

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

Answer 1

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 

 

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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 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.

 

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

= 12 - 8x + x²

 

References:

http://mathworld.wolfram.com/FiniteDifference.html

http://mathworld.wolfram.com/NewtonsForwardDifferenceFormula.html

http://en.wikipedia.org/wiki/Newton_polynomial