Parabola through three points

Find the parabola through the three points $(0, 1)$ , $(1, 0)$ and $(2, 3)$ . Through three points with distinct $x$ -coordinates passes exactly one quadratic function¹.

Setting up the equations

Write it as $y = ax^2 + bx + c$ and substitute each point.

Point	Equation
$(0, 1)$	$c = 1$
$(1, 0)$	$a + b + c = 0$
$(2, 3)$	$4a + 2b + c = 3$

Three unknowns, three conditions: exactly enough.

Solving

Using $c = 1$ leaves two equations.

\begin{align*} a + b &= -1 \\ 2a + b &= 1 \end{align*}

Subtracting gives $a = 2$ and then $b = -3$ . Hence $y = 2x^2 - 3x + 1$ .

Checking

$x$	$2x^2 - 3x + 1$	Required
$0$	$1$	$1$
$1$	$0$	$0$
$2$	$3$	$3$

Why three points suffice

A quadratic carries three degrees of freedom, $a$ , $b$ and $c$ . Passing through one point is one equation, so three points supply three equations and pin the curve down with nothing to spare. Given only two points, infinitely many parabolas pass through them.

Curve	Degrees of freedom	Points needed
Line	$2$	$2$
Parabola	$3$	$3$
Degree $n$ polynomial	$n + 1$	$n + 1$

The structure is identical to that of finding the circle through three points.

What else the answer tells us

$y = 2x^2 - 3x + 1$ factors as $(2x - 1)(x - 1)$ , so it meets the $x$ -axis at $\left(\dfrac{1}{2}, 0\right)$ and $(1, 0)$ . One of the three given points was on the $x$ -axis to begin with.

Completing the square turns it into $2\left(x - \dfrac{3}{4}\right)^2 - \dfrac{1}{8}$ , so the vertex is $\left(\dfrac{3}{4}, -\dfrac{1}{8}\right)$ . From the bare requirement of passing through three points, the position and the width have both been settled.

Interpolation

Finding a polynomial that passes through a given set of points is called interpolation, and it is the basic tool for drawing a smooth curve through scattered measurements. The large dots mark the three points.

Polynomial interpolation, Wikipedia