# Fit a quadratic function to 3 points

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

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