Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Fourth-Degree Polynomial


Date: 04/26/2001 at 09:08:06
From: Richard Miller
Subject: Fourth-degree polynomial how?

Help! Our teacher gave us a question on graphing polynomials as a 
just-for-fun problem and said it can be done. We want to come up with 
a solution.

Construct a fourth degree polynomial that contains all of the 
following points: (-1,2), (1,-2), (3,7), (2,1), (0,3).

Any help would be great! Thanks (-:


Date: 04/26/2001 at 10:20:24
From: Doctor Paul
Subject: Re: Fourth-degree polynomial how?

This is not a difficult problem. You should be aware that n points 
uniquely determine a polynomial of degree n-1. For n = 2, this fact 
reduces to the statement that two points uniquely determine a line.

You know that the general form of a 4th-degree polynomial is:

     y = ax^4 + bx^3 + cx^2 + dx + e

where a, b, c, d, and e are real numbers.

Now all you need to do is take each point (x,y) - there are five of 
them - and substitute the appropriate values for x and y into the 
general equation given above. This will give you a system of five 
equations with five unknowns (a, b, c, d, and e are the unknowns) that 
you can solve pretty easily using a TI graphing calculator.

Let's do it:

The point (-1,2) means that x = -1 and y = 2. Substituting these 
values into the equation y = ax^4 + bx^3 + cx^2 + dx + e gives:

     2 = a - b + c - d + e

Skipping the details, the other four equations are:

     (1,-2), (3,7), (2,1), (0,3)

     -2 = a + b + c + d + e
      7 = 81a + 27b + 9c + 3d + e
      1 = 16a + 8b + 4c + 2d + e
      3 = 0a + 0b + 0c + 0d + e

Now we have the necessary five equations and five unknowns. We can 
rewrite this system of equations using matrix multiplication as 
follows:

     Ax = b where 

           [ 1    -1    1    -1    1]
           [                        ]
           [ 1     1    1     1    1]
           [                        ]
      A =  [81    27    9     3    1]
           [                        ]
           [16     8    4     2    1]
           [                        ]
           [ 0     0    0     0    1]

and 

           [ 2]
           [  ]
           [-2]
           [  ]
      b =  [ 7]
           [  ]
           [ 1]
           [  ]
           [ 3]


Solving Ax = b for x gives x = A^(-1) * b. Use a TI Graphing 
calculator to compute the inverse of A, and then multiply this inverse 
by b to obtain:

           [-19]
           [---]
           [24 ]
           [   ]
           [47 ]
           [-- ]
           [12 ]
           [   ]
      x =  [-53]
           [---]
           [24 ]
           [   ]
           [-71]
           [---]
           [12 ]
           [   ]
           [ 3 ]

Interpreting these results means that

     a = -19/24
     b =  47/12
     c = -53/24
     d = -71/12
     e =  3

So the polynomial is:

     y = (-19/24)x^4 + (47/12)x^3 - (53/24)x^2 - (71/12)x + 3

- Doctor Paul, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Linear Algebra

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/