Author:
QUITTNER
[ Edit | View ]
|
Date Posted: 06:21:51 05/23/03 Fri
Joanna, I assume that you know how to invert a matrix. One of he equations of a parabola is A+Bx+Cx^2=y, which can also be written as Ax^0+Bx^1+Cx^2=y, where x^0 is always 1. A Polynomial of nth degree needs n+1 points; here we have a 2nd degree polynomial and 3 points, that's OK.
..... Let matrix M1 equal the left side of the equation, and fill in all the x's as follows:
Take one of the points (in any order), (3,-1) first. Its x is 3, therefore the top line of M1 is 1, for x'0, 3 for x, and 9 for x'2. The 2nd line is for point (1,-7): 1,1,1. The 3rd line for point (-2,14) is 1,-2,4. Therefore, to repeat, M1 is
| 1 3 9 | for point 1,
| 1 1 1 } 2,
| 1 =2 4 | 3.
The right side of the equation of the parabola is only y, we then make Matrix M2 equal to
| -1 | for point 1,
| -7 | 2,
| 14 | 3.
Next invert M1 to get matrix M3:
| 1/5 1 1/5 |
| 1/10 1/6 -15/4 |
| 1/10 -1/6 1/15 |
To get the results matrix, multiply M2 by M3, giving
| -4 |
| -5 |
| 2 | Therefore the equation of the parabola is
A+Bx+Cx^2=y where A=-4, B=-5 and C=2. Make sure you check this by evaluating the equation for each of the given x values of each of the 3 points and see that you get the given y's.
[ Post a Reply to This Message ]
|
Feedback: 