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

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 $y = a {x}^{2} + b x + c$, where a, b, and c are real numbers -- with $a \ne 0$.

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

We also have:
$a {1}^{2} + b \left(1\right) + c = 6$
That is, $a + b + c = 6$.

And finally:
$a {\left(2\right)}^{2} + b \left(2\right) + c = 10$
$4 a + 2 b + c = 10$

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

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

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)

$a + b + c = 6$
$- a + b - c = - 4$ (add)

$2 b = 2$
$b = 1$.

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

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

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

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

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.