Finding a quadratic function from a table of values can seem daunting at first, but with a structured approach, it becomes manageable. Quadratic functions take the form of ( f(x) = ax^2 + bx + c ), where ( a ), ( b ), and ( c ) are constants. This guide will walk you through a stepbystep method to derive a quadratic function from a set of data.
Understanding Quadratic Functions π
Quadratic functions are polynomial functions of degree 2, characterized by their parabolic shape. The graph of a quadratic function can open upwards or downwards depending on the value of ( a ):
 If ( a > 0 ), the parabola opens upwards.
 If ( a < 0 ), it opens downwards.
Example Table of Values π
( x )  ( f(x) ) 

1  2 
2  3 
3  6 
4  11 
5  18 
Step 1: Identify the Differences
To find a quadratic function, start by calculating the first and second differences of ( f(x) ).
 First Differences: Subtract consecutive ( f(x) ) values.
 Second Differences: Subtract consecutive first differences.
First Differences Calculation
( x )  ( f(x) )  First Differences 

1  2  
2  3  ( 3  2 = 1 ) 
3  6  ( 6  3 = 3 ) 
4  11  ( 11  6 = 5 ) 
5  18  ( 18  11 = 7 ) 
First Differences: 1, 3, 5, 7
Second Differences Calculation
First Differences  Second Differences 

1  
3  ( 3  1 = 2 ) 
5  ( 5  3 = 2 ) 
7  ( 7  5 = 2 ) 
Second Differences: 2, 2, 2
Important Note: If the second differences are constant, the data can be represented by a quadratic function.
Step 2: Setting Up the System of Equations
Next, letβs represent the quadratic function in the standard form:
[ f(x) = ax^2 + bx + c ]
To find ( a ), ( b ), and ( c ), we can use three points from our table (for instance, ( (1, 2) ), ( (2, 3) ), and ( (3, 6) )) to set up a system of equations.
Creating the Equations

For ( (1, 2) ): [ a(1)^2 + b(1) + c = 2 \implies a + b + c = 2 ]

For ( (2, 3) ): [ a(2)^2 + b(2) + c = 3 \implies 4a + 2b + c = 3 ]

For ( (3, 6) ): [ a(3)^2 + b(3) + c = 6 \implies 9a + 3b + c = 6 ]
System of Equations
Now we have:
 ( a + b + c = 2 )
 ( 4a + 2b + c = 3 )
 ( 9a + 3b + c = 6 )
Step 3: Solving the System
To find ( a ), ( b ), and ( c ), we can eliminate ( c ) by subtracting the first equation from the others.

From Equation 2  Equation 1: [ (4a + 2b + c)  (a + b + c) = 3  2 \implies 3a + b = 1 \tag{4} ]

From Equation 3  Equation 1: [ (9a + 3b + c)  (a + b + c) = 6  2 \implies 8a + 2b = 4 \tag{5} ]
Now we have a simpler system:
 ( 3a + b = 1 )
 ( 8a + 2b = 4 )
Solving for ( b )
From Equation (4), we can express ( b ) in terms of ( a ):
[ b = 1  3a ]
Substituting ( b ) into Equation (5):
[ 8a + 2(1  3a) = 4 \ 8a + 2  6a = 4 \ 2a + 2 = 4 \ 2a = 2 \implies a = 1 ]
Finding ( b ) and ( c )
Now, substitute ( a = 1 ) back to find ( b ):
[ b = 1  3(1) = 2 ]
To find ( c ), use Equation (1):
[ 1  2 + c = 2 \implies c = 3 ]
Conclusion: The Quadratic Function
Thus, the quadratic function is:
[ f(x) = 1x^2  2x + 3 ]
Or simply,
[ f(x) = x^2  2x + 3 ]
Verifying the Function π
We can verify this function using the original table of values. By plugging values of ( x ) back into our derived function, we should retrieve the corresponding ( f(x) ) values.
( x )  ( f(x) )  Calculated ( f(x) = x^2  2x + 3 ) 

1  2  ( 1^2  2(1) + 3 = 2 ) 
2  3  ( 2^2  2(2) + 3 = 3 ) 
3  6  ( 3^2  2(3) + 3 = 6 ) 
4  11  ( 4^2  2(4) + 3 = 11 ) 
5  18  ( 5^2  2(5) + 3 = 18 ) 
This confirms that our quadratic function accurately represents the data from the table.
By following these steps, you can confidently find a quadratic function from any set of values represented in a table. Remember, practice makes perfect, so keep experimenting with different datasets! π