曲线 $C$ 的方程为 $x^2 - 4y^2 = 4$。

1. Rewrite the equation into the standard form of a hyperbola to identify its shape.

$x^2 - 4y^2 = 4$
Dividing by $4$:
$\frac{x^2}{4} - \frac{y^2}{1} = 1$
This represents a hyperbola centred at the origin, opening horizontally.

2. Find the points of intersection with the axes.

To find $x$-intercepts, let $y = 0$:
$\frac{x^2}{4} = 1 \implies x^2 = 4 \implies x = \pm 2$
The points of intersection with the $x$-axis are $(2, 0)$ and $(-2, 0)$.

To find $y$-intercepts, let $x = 0$:
$-\frac{y^2}{1} = 1 \implies y^2 = -1$
Since there are no real solutions, the curve does not intersect the $y$-axis.

3. Determine the equations of the asymptotes.

For a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the asymptotes are given by $y = \pm \frac{b}{a}x$.
Here $a = 2$ and $b = 1$, so the asymptotes are:
$y = \pm \frac{1}{2}x$

4. Sketch the hyperbola $C$.

1. Draw the coordinate axes.
2. Plot the $x$-intercepts at $(-2, 0)$ and $(2, 0)$.
3. Draw and label the dashed lines for the asymptotes $y = \frac{1}{2}x$ and $y = -\frac{1}{2}x$.
4. Sketch the two branches of the hyperbola opening to the left and right, passing through the intercepts and becoming asymptotic to the dashed lines as $x \to \pm\infty$.

1. Identify the algebraic substitutions that link the original equation to the new equation.

The new equation is $(x - 2)^2 - y^2 = 4$.
This can be rewritten to match the form of the original equation $C$:
$(x - 2)^2 - 4\left(\frac{y}{2}\right)^2 = 4$
Comparing this with $x^2 - 4y^2 = 4$, we observe that $x$ is replaced by $(x - 2)$ and $y$ is replaced by $\frac{y}{2}$.

2. Describe the geometric transformations corresponding to these substitutions.

Replacing $x$ with $(x - 2)$ corresponds to a translation of $2$ units in the positive $x$-direction.
Replacing $y$ with $\frac{y}{2}$ corresponds to a scaling (or stretch) parallel to the $y$-axis with a scale factor of $2$.
Since these transformations act on different, orthogonal axes, they can be applied in either order.