Let us fit a quadratic function to 3 points using the values in the table below.
| x | 1 | 2 | 3 |
| y | 0 | -3 | -10 |
Substitute the values of x and y into y = ax2 + bx + c
Using x = 1 and y = 0, 0 = a(12) + b(1) + c
We get 0 = a + b + c
Using x = 2 and y = -3, -3 = a(22) + b(2) + c
We get -3 = 4a + 2b + c
Using x = 3 and y = -10, -10 = a(32) + b(3) + c
We get -10 = 9a + 3b + c
You end up with the following system of equations with three variables.
1. a + b + c = 0
2. 4a + 2b + c = -3
3. 9a + 3b + c = -10
You can solve this system of equations with three variables using either substitution or elimination.
Let us solve by elimination. Multiply 1. by -4.
1. -4(a + b + c) = -4(0)
1. -4a + -4b + -4c = 0
Add the left sides of 1. and 2. and add the right sides of 1. and 2.
1. -4a + -4b + -4c = 0
2. 4a + 2b + c = -3
__________________
4. -2b + -3c = -3
Multiply 1. by -9.
1. -9(a + b + c) = -9(0)
1. -9a + -9b + -9c = 0
Add the left sides of 1. and 3. and add the right sides of 1. and 3.
1. -9a + -9b + -9c = 0
3. 9a + 3b + c = -10
___________________
5. -6b + -8c = -10
Write the two new equations 4. and 5. as a system.
4. -2b + -3c = -3
5. -6b + -8c = -10
Multiply 4. by -3.
4. -3(-2b + -3c) = -3(-3)
4. 6b + 9c = 9
Add the left sides of 4. and 5. and add the right sides of 4. and 5.
4. 6b + 9c = 9
5. -6b + -8c = -10
________________
c = -1
Use either equation 4. or 5. to find b
4. -2b + -3c = -3
-2b + -3(-1) = -3
-2b + 3 = -3
-2b = -6
b = 3
Use either equation 1. 2. or 3. to find a.
1. a + b + c = 0
a + 3 + -1 = 0
a + 2 = 0
a = -2
The solution is a = -2, b = 3 and c = -1
Substitute these values into the standard form.
y = -2x2 + 3x + -1