4-1: Solving Systems of Equations With Quadratics

In some instances, you must find the equation of the quadratic when all you know are three points on the graph. Here's how:

Example: Find a quadratic function whose graph contains the points (-2,1) (2,7) and (1, 2).

Part 1: Substitute the above coordinates into the x and y values in the quadratic equation .

1 = 4a - 2b + c

7 = 4a + 2b + c

2 = a + b + c

Part 2: Subtract the first two equations

1 = 4a - 2b + c

-7 = 4a + 2b + c

-6 = 0a - 4b + 0

The new Equation: -6 = -4b

b = 1.5

Part 3: Substitute b into the second and third equations and simplify

7 = 4a + 2(1.5) + c

2 = a + 1.5 + c

Simplified equations:

4 = 4a + c

.5 = a + c

Part 4: Subtract the newly simplified equations:

4 = 4a + c

-.5 = a + c

3.5 = 3a + 0

a = 1.67

Part 5: Now that you have both a and b, plug the values of a and b into the original equations to get c.

7 = 4(1.67) + 2(1.5) + c

c = -0.667

Part 6: Now that you have all values, you can write the equation of the quadratic.

a = 1.67 b = 1.5 c = -.067

Equation: