Given three points (-1,4) (1,6) (2,10) how do you write a quadratic function in standard form with the points?

1 Answer
May 2, 2017

Solve a system of three equations in three unknowns.

Explanation:

Let us assume that the axis of symmetry for the parabola is vertical.

The standard equation of a parabola with a vertical axis is y = ax^2 + bx + c, where a, b, and c are real numbers -- with a ne 0.

Using the coordinates of the above points, we know that...
a(-1)^2 + b(-1) + c = 4
This simplifies to a - b + c = 4.

We also have:
a1^2 + b(1) + c = 6
That is, a + b + c = 6.

And finally:
a(2)^2 + b(2) + c = 10
4a + 2b + c = 10

The three equations give us the following system, which has a unique solution (a, b, c):

a - b + c = 4
a + b + c = 6
4a + 2b + c = 10.

While we may solve these using any method, it will be convenient to subtract equation (1) from equation (2).

a + b + c = 6
a - b + c = 4 (subtract)

Change the signs and add:
a + b + c = 6
-a + b - c = -4 (add)

Adding gives us:
2b = 2
b = 1.

Our system is now reduced to just two unknowns. Putting the known value of b into the three equations gives:

a - 1 + c = 4
a + 1 + c = 6
4a + 2 + c = 10.

Simplify to:
a + c = 5
4a + c = 8.

Subtract the first from the second to obtain the equation:
3a = 3
a = 1

Substitute a = 1 into the equation a + c = 5 and obtain c = 4.

The equation of our parabola has a = 1, b = 1, and c = 4.

Therefore, it is
y = x^2 + x + 4
Check, and you will find that all three points fit the equation.