The tutor shows an example to find the equation of a quadratic function from three points. The method is to construct and solve a three-variable linear system.

Example: Find the quadratic function that passes through the points (2,17), (6,5), and (8,-10).

Solution: We don’t know the vertex, nor two points of the same height, so we imagine our equation is of the form

y=ax²+bx+c

and plug in our known points for x and y.

First, based on (2,17):

a(2)² + b(2) + c = 17 ⇒ 4a+2b+c=17

Next, (6,5):

a(6)² + b(6) + c = 5 ⇒ 36a+6b+c=5

Finally, (8,-10):

a(8)² + b(8) + c = -10 ⇒ 64a+8b+c=-10

So we arrive at the three-variable system

1. 4a+2b+c=17
2. 36a+6b+c=5
3. 64a+8b+c=-10

This system can be solved manually, with a spread sheet, or with a calculator. Presently I’m using the Sharp el-520w. I press MODE 2 1 to get into three-variable system mode, then enter the parameters row by row. (See my post here for insight.)

The solution: a=-.75, b=3, c=14

Therefore, the quadratic function that contains the points (2,17), (6,5), and (8,-10) is

y=-.75x² + 3x + 14.

HTH:)

Source:

Johnson, Lee W. et al. Introduction to Linear Algebra, second ed. Don Mills: Addison-Wesley, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.