Sebastian M. answered 10/10/20
Experienced HS/College Tutor for Math and Standardized Tests
So the point is that you want to find a parabola that passes through the three points. Luckily for us, when you plug in the x and y values for each point into the general quadratic equation, you get a system of three equations of just a, b and c. The point of linear algebra is to use matrices to solve such systems.
Using the three equations you gave we can create the matrix
c b a
[ 1 1 1 | -6
1 3 9 | 4
1 -2 4 | 9]
From here you can row reduce the matrix using Gaussian elimination to find the exact values of c, b and a. Let me know if you need any more help, I love tutoring Linear Algebra and I can help you with the class.
Sebastian M.
Consider the second row of the matrix (1 3 9 | 4). This came from the equation c +3b + 9a = 4. The rows of the matrix are made from the coefficients of the variables c, b and a in each equation10/10/20
Alex R.
Thank you for your response, Can you go into more detail about how you plugged them in to find the equation? That's the part I'm mostly confused about. Given each set of points how do you know that it becomes (1 1 1 |-6). Hopefully my confusion on this part isn't too troubling.10/10/20