16
$\begingroup$

It's known that a unique parabola of the form $y=ax^{2}+bx+c$ exists for any three distinct points, provided that the points are non-collinear and their $x$ coordinates are distinct.

Consider the following three points:

  • $A$ is a point with coordinates $(x_{1}; y_{1})$,
  • $B$ is a point with coordinates $(x_{2}; y_{2})$,
  • $C$ is a point with coordinates $(x_{3}; y_{3})$.

Let points $A$, $B$, and $C$ be non-collinear and satisfying $x_{1}\neq x_{2}$, $x_{1}\neq x_{3}$, and $x_{2}\neq x_{3}$. Therefore, through these points passes a unique parabola of the form $y=ax^{2}+bx+c$. We now proceed to express $a$, $b$, and $c$ in terms of coordinates of $A$, $B$, and $C$. As $A$, $B$, and $C$ are on the graph of the same parabola, their coordinates satisfy its equation, thus we arrive to the following system of equations:

$$ \begin{aligned} \begin{cases} y_{1} = ax_{1}^{2} + bx_{1} + c, \\ y_{2} = ax_{2}^{2} + bx_{2} + c, \\ y_{3} = ax_{3}^{2} + bx_{3} + c; \end{cases} &\Longleftrightarrow \begin{cases} y_{1} - y_{2} = (ax_{1}^{2} + bx_{1} + c) - (ax_{2}^{2} + bx_{2} + c), \\ y_{2} - y_{3} = (ax_{2}^{2} + bx_{2} + c) - (ax_{3}^{2} + bx_{3} + c); \end{cases} \\ &\Longleftrightarrow \begin{cases} y_{1} - y_{2} = a(x_{1}^{2} - x_{2}^{2}) + b(x_{1} - x_{2}), \\ y_{2} - y_{3} = a(x_{2}^{2} - x_{3}^{2}) + b(x_{2} - x_{3}); \end{cases} \\[4pt] &\Longleftrightarrow \begin{cases} y_{1} - y_{2} = a(x_{1} - x_{2})(x_{1} + x_{2}) + b(x_{1} - x_{2}), \\ y_{2} - y_{3} = a(x_{2} - x_{3})(x_{2} + x_{3}) + b(x_{2} - x_{3}); \end{cases} \\[4pt] &\Longleftrightarrow \begin{cases} y_{1} - y_{2} = (x_{1} - x_{2})(a(x_{1} + x_{2}) + b), \\ y_{2} - y_{3} = (x_{2} - x_{3})(a(x_{2} + x_{3}) + b); \end{cases} \\[4pt] &\Longleftrightarrow \begin{cases} \frac{y_{1} - y_{2}}{x_{1} - x_{2}} = a(x_{1} + x_{2}) + b, \\ \frac{y_{2} - y_{3}}{x_{2} - x_{3}} = a(x_{2} + x_{3}) + b. \end{cases} \end{aligned} $$

Now, by subtracting the second equation from the first, we get: \begin{align*} \frac{y_{1} - y_{2}}{x_{1} - x_{2}} - \frac{y_{2} - y_{3}}{x_{2} - x_{3}} &= (a(x_{1} + x_{2}) + b) - (a(x_{2} + x_{3}) + b) \\ &= a((x_{1} + x_{2}) - (x_{2} + x_{3})) + (b - b) \\ &= a(x_{1} + x_{2} - x_{2} - x_{3}) \\ &= a(x_{1} - x_{3}) \end{align*}

Thus, we express $a$ as follows:

$$ \boxed{a = \left(\frac{y_{1} - y_{2}}{x_{1} - x_{2}} - \frac{y_{2} - y_{3}}{x_{2} - x_{3}}\right) : (x_{1} - x_{3})} $$

Here, we stop.

Looking at this form of $a$, I should point out it's very easy to memorize and apply, but what I got also seems to have much in common with how one determines a slope for a line equation.

What's the connection of these 'slopes' to the leading coefficient of quadratic trinomial? Is there some sort of explanation known?

$\endgroup$
0

3 Answers 3

Reset to default
22
$\begingroup$

A nice way to think about it is $f''(x) = 2a$ for a quadratic. Your formula $$a = \frac{\frac{y_1 - y_2}{x_1 - x_2} - \frac{y_2 - y_3}{x_2 - x_3}}{x_1 - x_3}$$ is essentially the discrete second derivative. You are calculating the average rate of change of the average slope across your three points.

$\endgroup$
1
  • 1
    $\begingroup$ Then where does the factor $2$ come from (or go)? $\endgroup$ Commented Nov 7 at 5:02
8
$\begingroup$

You are discovering divided differences as they are used in Newton interpolation. These divided differences are defined recursively. For a function $y=f(x)$, the recursion starts trivially with $f[x_0]=f(x_0)$, all the higher order ones are derived from one order lower via some "slope formula", as you called it $$ f[x_0,x_1,...,x_k,x_{k+1}]=\frac{f[x_1,...x_k,x_{k+1}]-f[x_0,x_1,...,x_k]}{x_{k+1}-x_k}. $$ There is a long tradition of computing these in tables, look it up. As one can guess, the divided difference of a certain order is related to the derivative of the same order, more the value at some midpoint divided by the factorial of the order.

The Newton interpolation formula for the quadratic order looks like $$ p_2(x)=f[x_0]+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1). $$ One can see that the leading coefficient is the second order divided difference.

$\endgroup$
1
7
$\begingroup$

It's because parabolas have a special property to do with their secant slopes relating directly to their tangent slopes and that their tangent slopes change linearly.

The segment $\overline{AB}$ defines a parabolic segment, and there is a special point $P_{AB}$ on the parabola whose $x$-coordinate is the average $x$-coordinate of $A$ and $B$.

Then, the key property is that the tangent slope of the parabola at $P_{AB}$ is equal to the secant slope of $\overline{AB}$, so $$\dfrac{y_2 - y_1}{x_2 - x_1} = y'\left(\dfrac{x_1 + x_2}{2}\right)$$

For the same reason, $$\dfrac{y_2-y_3}{x_2-x_3} = y'\left(\dfrac{x_2+x_3}{2}\right)$$

So $$\dfrac{y_2 - y_1}{x_2 - x_1} - \dfrac{y_2-y_3}{x_2-x_3} = y'\left(\dfrac{x_1 + x_2}{2}\right) - y'\left(\dfrac{x_2+x_3}{2}\right)$$

But the slope of a parabola $y = ax^2 + bx + c$ is just $y' = 2ax + b$, so

So $$\dfrac{y_2 - y_1}{x_2 - x_1} - \dfrac{y_2-y_3}{x_2-x_3} = 2a\left(\dfrac{x_1+x_2}{2} - \dfrac{x_2-x_3}{2}\right) = 2a\left(\frac{x_1-x_3}{2}\right) = a(x_1 - x_3)$$

I'm not sure if this is as completely satisfying of a "why" as I'd like, but the slopes are there because they are literal slopes of the secants, and they relate to the constant $a$ because of the linearity of the derivative of a parabola (i.e. the slope of the slope is constant).

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.