Q.What parting advice does Mrs Fitzgerald give after the change-back, and how does the play end? (Answer in 30–40 words.)

Character Transformation — From Intuition to Precision

Imagine you're looking at a function $f(x)$ that describes some shape — say, a smooth curve. Now suppose you want to shift that curve to the right by 3 units. You might think: "I'll add 3 to every $x$." But if you try $f(x+3)$, the curve actually moves left. That feels backwards. Why?

The reason is subtle but crucial: when you replace $x$ with $x+3$, you're asking "what value of the original function sits 3 units ahead of the current $x$?" So the graph gets pulled backward. This is the seed of character transformation: you transform the input, not the output, to move the graph.

The Core Idea

A character transformation is a change you make to the variable (the "character" $x$) inside a function, rather than to the function's output. The effect on the graph is the opposite of what you might naively expect.

Why? Because to get the same $y$-value as the original $f(x)$, the new $x$ must be larger by $h$ when the shift is $x-h$. So the whole graph moves right.

The Precise Statement

Let $f$ be a function. For any real number $c > 0$:

• Horizontal shift: $y = f(x - c)$ shifts the graph of $f$ right by $c$ units. $y = f(x + c)$ shifts it left by $c$ units.
• Horizontal stretch/compression: $y = f(kx)$ with $k > 1$ compresses the graph horizontally by a factor of $1/k$ (the graph gets narrower). $y = f(x/k)$ stretches it horizontally by a factor of $k$ (wider).
• Reflection across the y-axis: $y = f(-x)$ flips the graph horizontally.

The pattern is always: what you do to $x$ inside the function, the graph does the opposite to the $x$-coordinate.

Why "Character Transformation"? …