# Puzzles

## 6 December

\(p(x)\) is a quadratic with real coefficients. For all real numbers \(x\),

$$x^2+4x+14\leq p(x)\leq 2x^2+8x+18$$

\(p(2)=34\). What is \(p(6)\)?

## Between quadratics

\(p(x)\) is a quadratic polynomial with real coefficients. For all real numbers \(x\),

$$x^2-2x+2\leq p(x)\leq 2x^2-4x+3$$

\(p(11)=181\). Find \(p(16)\).

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$$x^2-2x+2=(x-1)^2+1$$
$$2x^2-4x+3=2(x-1)^2+1$$

Therefore the minimum point of both of these quadratics is \((1,1)\). \(p(x)\) will only be between these if:

$$p(x)=a(x-1)^2+1\quad\text{where }1\leq a\leq 2$$

We know that \(p(11)=181\), so:

$$\begin{array}{rl}
181&=p(11)\\
&=a(11-1)^2+1\\
&=100a+1
\end{array}$$

Therefore \(a=1.8\). This means that:

$$\begin{array}{rl}
p(16)&=1.8(16-1)^2+1\\
&=1.8\times225+1\\
&=406
\end{array}$$

## Parabola

On a graph of \(y=x^2\), two lines are drawn at \(x=a\) and \(x=-b\) (for \(a,b>0\). The points where these lines intersect the parabola are connected.

What is the y-coordinate of the point where this line intersects the y-axis?

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The co-ordinates of the points where the lines intersect the parabola are \((a,a^2)\) and \((-b,b^2)\). Hence the gradient of the line between them is:

$$\frac{a^2-b^2}{a-(-b)}=\frac{(a+b)(a-b)}{a+b}=a-b$$

Therefore the y-coordinate is:

$$b^2 + b(a-b) = ba$$

Ferdinand Möbius, who discovered this property called the curve a *Multiplicationsmaschine* or 'multiplication machine' as it could be used to perform multiplication.

#### Extension

How could you use the graph of \(y=x^2\) to divide 100 by 7?