KNOWN POINTS ON AN UNKNOWN POLYNOMIAL FUNCTION

Two Points       Three Points       Four Points       Five Points       Six Points   .   .   .   n Points

Linear           Quadratic           Cubic           Quartic           Quintic     .   .   .     Polynonial

The set of points given in coordinate form must be a function for the ideas covered in the following methods. This means that no two points in the set have the same first coordinate and the given number of points are distinct. All sets of points used in this Web page will be a function.

In general:

two distinct points will determine a straight line or linear function,

three distinct points will determine a quadratic function
(assume the three points do not lie on a straight line),

four distinct points will determine a cubic function
(assume the four points do not lie on a straight line or on a quadratic),

and so on.

The methods used in the following examples assumes a knowledge of solving systems of equations using matrix solutions. Review this method in your algebra course if you need more information.

In general, a matrix equation consisting of a constant matrix A times a variable matrix X equal to a constant matrix B may be solved for variable matrix X by taking the inverse of matrix A times matrix B.

If         A X = B,

then         X = A^(-1) B.         Equation 1.

In practice, A^(-1), the inverse of constant matrix A, may not exist. Note that matrix multiplication is not commutative.

EXAMPLES:

Given Two Points Find A Linear Function

If points (-1, 1) and (0, 3) are given as points on a linear function:

y = a x + b                 FUNction 1.

then we can find a and b.

Replace x and y with the coordinate values given to get the system:

3 = a ( 0 ) + b ( 1 )
1 = a ( -1 ) + b ( 1 )

Note that b = b ( 1 ) in both equations.

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then                     a = 2     and     b = 3.

If points (-1, 1) and (0, 3) are given as points on a linear function then:

y = 2 x + 3

is the linear function.

Given Three Points Find A Quadratic Function

If points (1, 1), (2, 5) and (-1, -1) are given as points on a quadratic function:

y = a x ² + b x + c                FUNction 2.

then we can find a, b and c.

Note that FUNction 2. can be written:

y = a x ^2 + b x + c                FUNction 2a.

Replace x and y with the coordinate values given to get the system:

1 = a ( 1 ) + b ( 1 ) + c ( 1 )

5 = a ( 4 ) + b ( 2 ) + c ( 1 )

-1 = a ( 1 ) - b ( 1 ) + c ( 1 )

Note that c = c ( 1 ) in each equation.

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then                     a = 1     b = 1 and     c = -1.

If points (1, 1), (2, 5) and (-1, -1) are given as points on a quadratic function then:

y = x ² + x - 1

Given Four Points Find A Cubic Function

If points (1, -3), (-1, -1), (2, -13) and (-2, -21) are given as points on a quadratic function:

y = a x ³ + b x ² + c x + d                FUNction 3.

then we can find a, b and c.

Note that FUNction 3. can be written:

y = a x ^3 + b x ^2 + c x + d          FUNction 3a.

Replace x and y with the coordinate values given to get the system:

-3 = a ( 1 ) + b ( 1 ) + c ( 1 ) + d ( 1 )

-1 = a ( -1 ) + b ( 1 ) + c ( -1 ) + d ( 1 )

-13 = a ( 8 ) + b ( 4 ) + c ( 2 ) + d ( 1 )

-21 = a ( -8 ) + b ( 4 ) + c ( -2 ) + d ( 1 )

Note that d = d ( 1 ) in each equation.

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then                     a = ____ ,     b = ____ ,     c = ____ and     d = ____.

If points (1, -3), (-1, -1), (2, -13) and (-2, -21) are given as points on a cubic function then:

y = _____ x ³ - _____ x ² - _____ x + _____

is the cubic function.

Given Five Points Find A Quartic Function

If points (-1, 10), (1, 4), (2, 10), (3, 22) and (4, 10) are given as points on a quartic function:

y = a x^4 + b x ³ + c x ² + d x + e                FUNction 4.

then we can find a, b, c, d and e.

Replace x and y with the coordinate values given to get the system:

10 = a ( 1 ) + b ( -1 ) + c ( 1 ) + d ( -1 ) + e ( 1 )

4 = a ( 1 ) + b ( 1 ) + c ( 1 ) + d ( 1 ) + e ( 1 )

10 = a ( 16 ) + b ( 8 ) + c ( 4 ) + d ( 2 ) + e ( 1 )

22 = a ( 81 ) + b ( 27 ) + c ( 9 ) + d ( 3 ) + e ( 1 )

10 = a ( 256 ) + b ( 64 ) + c ( 16 ) + d ( 4 ) + e ( 1 )

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then               a = ____ ,     b = ____ ,     c = ____ and     d = ____ and     e = ____ .

If points (-1, 10), (1, 4), (2, 10), (3, 22), (4, 10) and (9, 3372) are given as points on a quartic function then:

y = - _____ x^4 + _____ x ³ - _____ x ² - _____ x + _____

is the quartic function.

Given Six Points Find A Quintic Function

If points (0, -288), (1, 12), (3, 0), (7, 192), (8, 1020) and (9, 3372) are given as points on a quintic function:

y = a x^5 + b x^4 + c x ³ + d x ² + e x + f                FUNction 5.

then we can find a, b, c, d, e and f.

Replace x and y with the coordinate values given to get the system:

-228 = a ( 0 ) + b ( 0 ) + c ( 0 ) + d ( 0 ) + e ( 0 ) + f ( 1 )

12 = a ( 1 ) + b ( 1 ) + c ( 1 ) + d ( 1 ) + e ( 1 ) + f ( 1 )

0 = a ( 243 ) + b ( 81 ) + c ( 27 ) + d ( 9 ) + e ( 3 ) + f ( 1 )

192 = a ( 16807 ) + b ( 2401 ) + c ( 343 ) + d ( 49 ) + e ( 7 ) + f ( 1 )

1020 = a ( 32768 ) + b ( 4096 ) + c ( 512 ) + d ( 64 ) + e ( 8 ) + f ( 1 )

3372 = a ( 59049 ) + b ( 6561 ) + c ( 729 ) + d ( 81 ) + e ( 9 ) + f ( 1 )

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then               a = ____ ,     b = ____ ,     c = ____ and     d = ____ ,     e = ____ and     f = ____ .

If points (0, -288), (1, 12), (3, 0), (7, 192), (8, 1020) and (9, 3372) are given as points on a quintic function then:

y = _____ x^5 - _____ x^4 + _____ x ³ - _____ x ² + _____ x - _____

is the quintic function.

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A comment about notation for this n - points section. an is one symbol with the subscript n in line with a just because it is easer to type that way in Web pages. a(n-1) is one symbol with subscript (n-1) on a. an is read "a sub n" and if n = 2 then a2 is read "a sub 2" and a(n-1) is read "a sub 1" when you substitute the 2 for n in (n-1) to get (2-1) = 1. If this is your first or second time using subscript notation then it is best if you copy the following section in your handwriting and move the superscrips, ^, up as exponents and the subscripts down in the normal subscript line.

If you do this writing you may find it easer to see and understand if your superscripts and subscripts are a bit smaller font size than the variable they are associated with.

Given n - Points Find An (n-1) Degree Polynomial Function

If points (x1, y1), (x2, y2), (x3, y3) . . . and (xn, yn) are given as points on an (n-1) degree polynomial function:

y = an x ^(n-1) + a(n-1) x ^(n-2) + . . . a4 x ³ + a3 x ² + a2 x + a1           FUNction (n-1).

then we can find an, a(n-1), . . ., a4, a3, a2 and a1.

Replace x and y with the coordinate values given to get the system:

y1 = an ( x1^(n-1) ) + a(n-1) ( x1^(n-2) ) + . . . + a3 ( x1 ² ) + a2 ( x1 ) + a1 ( 1 )

y2 = an ( x2^(n-1) ) + a(n-1) ( x2^(n-2) ) + . . . + a3 ( x2 ² ) + a2 ( x2 ) + a1 ( 1 )

y3 = an ( x3^(n-1) ) + a(n-1) ( x3^(n-2) ) + . . . + a3 ( x3 ² ) + a2 ( x3 ) + a1 ( 1 )

yn = an ( xn^(n-1) ) + a(n-1) ( xn^(n-2) ) + . . . + a3 ( xn ² ) + a2 ( xn ) + a1 ( 1 )

Note that a1 = a1 ( 1 ) in each equation.

The matrix equation for this system is: Then finding the inverse of the coefficient matrix gives: Therefore the solution of the matrix equation is found by: The solution of the matrix equation is: then           an = ____,   a(n-1) = ____ , . . ., a3 = ____,   a2 = ____ and a1 = ____ .

If points (x1, y1), (x2, y2), (x3, y3) . . . and (xn, yn) are
given as points on an (n-1) degree polynomial function:

y = ____ x ^(n-1) + ____ x ^(n-2) + . . . + ____ x ³ + ____ x ² + ____ x + ____

is the (n-1) degree polynomial function.