画出 $(y+2)^2 = 4(x-1)^2 - 9$ 的图像，并标出其主要特征。 由此求 $k$ 的取值范围，使得方程 $9 + (y+2)^2 = 4(k^2 - 1)^2.$ 无实根。 下图显示了 $y = h(3 - x)$ 的图像。在不同的图上，分别画出以下函数的图像： $y = h(x+3)$， $y = \tfrac{1}{2}h(x)$。

1. Rearrange the equation into the standard form of a hyperbola to find its centre, vertices, and asymptotes.

$4(x-1)^2 - (y+2)^2 = 9$
$\frac{(x-1)^2}{\frac{9}{4}} - \frac{(y+2)^2}{9} = 1$
$\frac{(x-1)^2}{(\frac{3}{2})^2} - \frac{(y+2)^2}{3^2} = 1$

The centre is $(1, -2)$.
Since the $x$-term is positive, it is a horizontal hyperbola.
Vertices: $(1 - \frac{3}{2}, -2)$ and $(1 + \frac{3}{2}, -2) \implies (-0.5, -2)$ and $(2.5, -2)$.

Asymptotes: $y - (-2) = \pm \frac{3}{3/2}(x - 1)$
$y + 2 = \pm 2(x - 1)$
$y = 2x - 4$ \text{ and } y = -2x$

2. Determine the intersections with the coordinate axes.

To find $y$-intercepts, let $x = 0$:
$(y+2)^2 = 4(-1)^2 - 9 = -5$
Since there are no real roots, there are no $y$-intercepts.

To find $x$-intercepts, let $y = 0$:
$(0+2)^2 = 4(x-1)^2 - 9$
$4 = 4(x-1)^2 - 9$
$(x-1)^2 = \frac{13}{4}$
$x - 1 = \pm \frac{\sqrt{13}}{2}$
$x = 1 \pm \frac{\sqrt{13}}{2}$
The $x$-intercepts are $(1 - \frac{\sqrt{13}}{2}, 0)$ and $(1 + \frac{\sqrt{13}}{2}, 0)$.

3. Sketch the hyperbola with its main features clearly labelled.

Sketch two branches opening left and right, ensuring they approach the asymptotes $y=2x-4$ and $y=-2x$ as $x \to \pm\infty$.
Label the vertices at $(-0.5, -2)$ and $(2.5, -2)$.
Label the $x$-intercepts at $1 - \frac{\sqrt{13}}{2}$ and $1 + \frac{\sqrt{13}}{2}$.

1. Rearrange the given equation to match the structure of the curve from part (a)(i).

We are given the equation $9 + (y+2)^2 = 4(k^2 - 1)^2$. Rearranging to make $(y+2)^2$ the subject gives:
$(y+2)^2 = 4(k^2 - 1)^2 - 9$
Comparing this with the equation of the sketched curve $(y+2)^2 = 4(x - 1)^2 - 9$, we can see that this corresponds to the intersection of the curve with the vertical line $x = k^2$.

2. Determine the range of $x$ for which the graph yields no real values for $y$.

For the equation to have no real roots for $y$, the value of $4(x-1)^2 - 9$ must be negative, corresponding to the gap between the two branches of the hyperbola sketched in (a)(i).
Setting $(y+2)^2 = 0$ identifies the $x$-coordinates of the vertices:
$4(x-1)^2 - 9 = 0 \implies (x-1)^2 = \frac{9}{4} \implies x = \frac{5}{2} \text{ or } x = -\frac{1}{2}$
Thus, there are no real roots when the $x$-value lies strictly between the vertices:
$-\frac{1}{2} < x < \frac{5}{2}$

3. Substitute $x = k^2$ into the inequality and solve for $k$.

Substituting $x = k^2$ into the inequality yields:
$-\frac{1}{2} < k^2 < \frac{5}{2}$
Since $k^2 \ge 0$ is always true for any real $k$, the left inequality ($-\frac{1}{2} < k^2$) is automatically satisfied. We only need to solve:
$k^2 < \frac{5}{2}$
$-\sqrt{\frac{5}{2}} < k < \sqrt{\frac{5}{2}}$
Rationalising the denominators, we obtain:
$-\frac{\sqrt{10}}{2} < k < \frac{\sqrt{10}}{2}$

1. Identify the transformation required to map the given graph $y = h(3-x)$ to the target graph $y = h(x+3)$.

Let the given function be $\mathrm{f}(x) = h(3 - x)$.
We want to sketch the graph of $y = h(x + 3)$.
Expressing this in terms of $\mathrm{f}(x)$:
$h(x + 3) = h(3 - (-x)) = \mathrm{f}(-x)$.
Therefore, the graph of $y = h(x + 3)$ is obtained by reflecting the graph of $y = h(3 - x)$ in the $y$-axis.

2. Determine the coordinates of key points and equations of asymptotes after the reflection.

Under a reflection in the $y$-axis, the $x$-coordinates are negated while the $y$-coordinates remain unchanged:

• The known point $(-3, -2)$ reflects to $(3, -2)$.
• The minimum point of the dip at $(1, -4)$ reflects to $(-1, -4)$.
• The vertical asymptote $x = 2$ reflects to $x = -2$.
• The vertical asymptote $x = 0$ remains $x = 0$.
• The horizontal asymptote $y = 0$ as $x \to \infty$ becomes $y = 0$ as $x \to -\infty$.

1. Determine the transformation mapping from the given graph to the target graph.

Let the given graph be $y = f(x) = h(3-x)$.
By substituting $u = 3-x$ (which means $x = 3-u$), we get $f(3-u) = h(u)$, so $h(x) = f(3-x)$.
The target graph is $y = \frac{1}{2}h(x) = \frac{1}{2}f(3-x)$.
If a point $(x, y)$ lies on $f(x)$, its corresponding point $(x', y')$ on the target graph satisfies $x = 3 - x' \implies x' = 3 - x$ and $y' = \frac{1}{2}y$.

2. Map the given key points to their new coordinates.

Apply the mapping $(x, y) \to \left(3 - x, \frac{1}{2}y\right)$ to the specified points:

• The point $(-3, -2)$ maps to $\left(3 - (-3), \frac{1}{2}(-2)\right) = (6, -1)$.
• The minimum point of the U-shape $(1, -4)$ maps to $\left(3 - 1, \frac{1}{2}(-4)\right) = (2, -2)$.

3. Determine the new asymptotes and overall shape for the sketch.

Apply the mapping $x' = 3 - x$ to the vertical asymptotes:

• $x = 0$ maps to $x = 3$.
• $x = 2$ maps to $x = 1$.
  Horizontal asymptote $y = 0$ scales to $y = \frac{1}{2}(0) = 0$.

Reversing the $x$-direction (reflection) swaps the left and right behaviors:

• The original right tail ($x > 2$, $y \to 0$) becomes the new left tail ($x < 1$, $y \to 0$ as $x \to -\infty$).
• The U-shape is now between $x = 1$ and $x = 3$.
• The original left spike ($x \to 0^-$) becomes the new right spike ($x \to 3^+$).