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/
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