Finding a quadratic rule | Racquel Sanderson

Tag Archives: Finding a quadratic rule

April 23, 2025 · 9:30 am

Problem Solving

I am came across this problem and was fascinated. It’s from this book

At first I went straight to the 14-sided polygon, and tried to draw the diamonds (parallelograms), but then I thought let’s start smaller and see if there is a pattern.

Clearly a square contains 1 diamond (itself).

Pentagon

It’s not possible with a pentagon.

Hexagon

A hexagon has 6 diamonds

Septagon

I am guessing it’s not possible to fill a regular 7-sided shape with diamonds

It’s not possible with odd numbers of sides. Regular polygons with an odd number of sides have no parallel sides, so we can’t cover it with rhombi (which have opposite sides parallel).

Octagon

An octagon has 6 diamonds.

We know a decoagon has 10 diamonds (from the question)

Let’s put together what we know

$n$	$4$	$6$	$8$	$10$
Diamonds	$1$	$3$	$6$	$10$

These are the triangular numbers, so when $n=12$ the number of diamonds is $15$ , and for $n=14$ it’s $21$ .

We can work out a rule for calculating the number of diamonds given the number of sides.

Because the difference in the $n$ values is not $1$ , I am going to get $n$ and $D$ in terms of $k$ and then combine the two equations.

From the above table, $n=2k+2$

We know this rule is quadratic as the second difference is constant, hence

$D=\frac{1}{2}k^2+bk+c$

$\begin{equation*}1=\frac{1}{2}+b+c\end{equation}$

(1) $\begin{equation*}\frac{1}{2}=b+c\end{equation*}$

$\begin{equation*}3=\frac{1}{2}2^2+2b+c\end{equation}$

(2) $\begin{equation*}1=2b+c\end{equation*}$

Solve simultaneously, subtract equation $1$ from equation $2$

(3) $\begin{equation*}\frac{1}{2}=b\end{equation*}$

Substitute for $b=\frac{1}{2}$ into equation $1$

$\begin{equation*}\frac{1}{2}=\frac{1}{2}+c\end{equation}$

$c=0$ , therefore $D=\frac{1}{2}k^2+\frac{1}{2}k$

We know $n=2k+2$ hence $k=\frac{n-2}{2}$

Hence $D=\frac{1}{2}(\frac{n-2}{2})^2+\frac{1}{2}(\frac{n-2}{2})$

$D=\frac{1}{8}(n^2-4n+4)+\frac{1}{4}(n-2)=\frac{n^2}{8}-\frac{n}{2}+\frac{1}{2}+\frac{n}{4}-\frac{1}{2}=\frac{n^2}{8}-\frac{n}{4}$

$\begin{equation*}D=\frac{n^2}{8}-\frac{n}{4}\end{equation}$

Let’s test our rule for $n=14$

$\begin{equation*}D=\frac{14^2}{8}-\frac{14}{4}=\frac{49}{2}-\frac{7}{2}=\frac{42}{2}=21\end{equation}$

Quadratic Rule from a Table of Values

How do you find a quadratic rule from a table of values?

For example,

$x$	1	2	3	4
$y$	0	6	14	24

Find the first difference

First Difference

6-0=6

14-6=8

24-14=10

Find the second difference (if the second difference is a constant, then it is quadratic)

Second Difference

8-6=2

10-8=2

The general equation of a quadratic is $y=ax^2+bx+c$

The second difference is $2a$

Hence our equation is now $y=x^2+bx+c$

The $c$ value is the vertical intercept ( $x=0$ ). We can back track in the table

$x$	0	1	2	3	4
$y$	$c$	0	6	14	24

As the first differences are 6, 8, 10, the one between 0 and 1 must be 4

$0-c=4$

$c=-4$

Our equation is now $y=x^2+bx-4$ .

We can now use any other point to find the $b$ value.

Let’s use the point $(2, 6)$

$6=2^2+b(2)-4$

$6=4-2b-4$

$6=2b$

$b=3$

The function is $y=x^2+3x-4$

Let’s try another one

$x$	3	4	5	6
$y$	7	17	31	49

First differences

First difference

Second difference

Second Difference

Hence $2a=4$ , therefore $a=2$

The equation is now $y=2x^2+bx+c$

Instead of back tracking, this time I am going to use two points and simultaneous equations.

Using points $(3, 7)$ and $(4, 17)$

$7=2(3)^2+b(3)+c$

(1) $\begin{equation*}3b+c=-11\end{equation*}$

$17=2(4)^2+b(4)+c$

(2) $\begin{equation*}4b+c=-15\end{equation*}$

Equation 2 – Equation 1

$b=-4$

Substitute $b=-4$ into equation 1

$3(-4)+c=-11$

$-12+c=-11$

$c=1$

Hence the equation is $y=2x^2-4x+1$