画出 $(y+2)^2 = 4(x-1)^2 - 9$ 的图像，并展示其主要特征。 由此求 $k$ 的取值范围，使得方程 $9 + (y+2)^2 = 4(k^2 - 1)^2.$ 没有实数根。 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： 画出 $y = h(x+3)$ 的图像。 画出 $y = \tfrac{1}{2}h(x)$ 的图像。

1. Rearrange the equation into the standard form of a hyperbola.

$(y+2)^2 = 4(x-1)^2 - 9$
$4(x-1)^2 - (y+2)^2 = 9$
$\frac{(x-1)^2}{\frac{9}{4}} - \frac{(y+2)^2}{9} = 1$
This represents a hyperbola with a horizontal transverse axis and center $(1, -2)$.

2. Determine the coordinates of the vertices.

The vertices lie on the horizontal line $y = -2$ passing through the center:
$\frac{(x-1)^2}{\frac{9}{4}} = 1 \implies x - 1 = \pm \frac{3}{2}$
$x = 2.5 \text{ or } x = -0.5$
The vertices are at $(2.5, -2)$ and $(-0.5, -2)$.

3. Find the equations of the asymptotes.

Set the right-hand side of the standard equation to 0:
$\frac{(x-1)^2}{\frac{9}{4}} - \frac{(y+2)^2}{9} = 0$
$(y+2)^2 = 4(x-1)^2 \implies y+2 = \pm 2(x-1)$
For $y+2 = 2(x-1)$, we get the asymptote $y = 2x - 4$.
For $y+2 = -2(x-1)$, we get the asymptote $y = -2x$ (which passes through the origin).

4. Calculate the axial intercepts.

For $x$-intercepts, set $y = 0$:
$(0+2)^2 = 4(x-1)^2 - 9 \implies 4(x-1)^2 = 13$
$x - 1 = \pm\frac{\sqrt{13}}{2} \implies x = 1 \pm \frac{\sqrt{13}}{2}$
The $x$-intercepts are $\left(1 - \frac{\sqrt{13}}{2}, 0\right)$ and $\left(1 + \frac{\sqrt{13}}{2}, 0\right)$.

For $y$-intercepts, set $x = 0$:
$(y+2)^2 = 4(0-1)^2 - 9 = -5$
Since there are no real solutions, the graph does not cross the $y$-axis.

1. Relate the given equation to the sketched graph to establish the domain condition for no real roots.

Rearranging the given equation:
$(y+2)^2 = 4(k^2 - 1)^2 - 9$
Comparing this to the equation of the graph from (a)(i), $(y+2)^2 = 4(x-1)^2 - 9$, we can see it is equivalent to substituting $x = k^2$.
From the graph's features, the hyperbola intersects its central horizontal axis ($y=-2$) when $4(x-1)^2 - 9 = 0 \implies x = -0.5 \text{ or } x = 2.5$. The curve only exists for $x \le -0.5$ and $x \ge 2.5$.
For the equation to have no real roots for $y$, the line $x = k^2$ must fall in the gap where the graph does not exist:
$-0.5 < x < 2.5$
Substituting $x = k^2$:
$-0.5 < k^2 < 2.5$

2. Solve the inequality to find the range of values for $k$.

Since $k$ is a real number, $k^2 \ge 0$, which means $-0.5 < k^2$ is automatically satisfied.
We only need to solve the upper bound:
$k^2 < 2.5$
$k^2 < \frac{5}{2}$
$-\sqrt{\frac{5}{2}} < k < \sqrt{\frac{5}{2}}$
$-\frac{\sqrt{10}}{2} < k < \frac{\sqrt{10}}{2}$

1. Relate the target function to the given function.

Let the given function be $f(x) = h(3 - x)$.
We want to sketch $y = h(x + 3)$.
Substituting $-x$ for $x$ in $f(x)$ gives:
$f(-x) = h(3 - (-x)) = h(3 + x)$
Thus, the required graph of $y = h(x + 3)$ is simply the reflection of the original graph $y = f(x)$ in the $y$-axis.

2. Determine the new positions of the key features after reflection.

Applying the transformation $x \mapsto -x$ to all key features:

• Vertical asymptote $x = 2 \implies x = -2$.
• Vertical asymptote $x = 0 \implies x = 0$ (unchanged).
• The $x$-axis asymptote as $x \to \infty \implies$ asymptote as $x \to -\infty$.
• Given point $(-3, -2) \implies (3, -2)$.
• Turning point $(1, -4) \implies (-1, -4)$.

3. Describe the final sketched graph.

Sketch the reflected branches on the Cartesian plane:

• For $x < -2$, the curve lies above the $x$-axis, approaching $y = 0$ as $x \to -\infty$ and $x = -2$ as a vertical asymptote.
• For $-2 < x < 0$, the curve is a downward loop with a turning point at $(-1, -4)$, bound by asymptotes $x = -2$ and $x = 0$.
• For $x > 0$, the curve passes through $(3, -2)$, crosses the $x$-axis, and goes to $+\infty$ as it approaches the $y$-axis.

1. Identify the relationship mapping the coordinates of the given graph $y = h(3-x)$ to the required graph $y = \frac{1}{2}h(x)$.

Let the given graph be represented by the coordinates $(x_0, y_0)$, where $y_0 = h(3-x_0)$.
We want to plot the points $(x, y)$ such that $y = \frac{1}{2}h(x)$.
Equating the arguments of the function $h$, we get $x = 3 - x_0$.
Equating the outputs gives $2y = h(x) = h(3-x_0) = y_0 \implies y = \frac{1}{2} y_0$.
Thus, every point $(x_0, y_0)$ on the original graph maps to $(3 - x_0, \frac{1}{2}y_0)$ on the new graph.
(Geometrically, this corresponds to a reflection in the line $x = 1.5$, followed by a vertical scaling by a factor of $\frac{1}{2}$.)

2. Apply the transformation to the given key points.

Using the coordinate mapping $(x_0, y_0) \mapsto (3 - x_0, \frac{1}{2}y_0)$:

• The turning point $(1, -4)$ maps to $(3 - 1, \frac{1}{2}(-4)) = (2, -2)$.
• The given reference point $(-3, -2)$ maps to $(3 - (-3), \frac{1}{2}(-2)) = (6, -1)$.
• The $x$-intercept marked at $x_0 = -1$, which is $(-1, 0)$, maps to $(3 - (-1), \frac{1}{2}(0)) = (4, 0)$.

3. Apply the transformation to the asymptotes and describe the resulting sketch.

Using the same mapping for the vertical asymptotes ($x_0 = 0$ and $x_0 = 2$):

• $x_0 = 0 \mapsto x = 3 - 0 = 3$
• $x_0 = 2 \mapsto x = 3 - 2 = 1$

The horizontal asymptote $y_0 = 0$ maps to $y = \frac{1}{2}(0) = 0$.

Reversing the horizontal orientation (due to the $-x_0$ transformation) changes the left-to-right order of the branches:

• The original right branch ($x_0 > 2$) becomes the new left branch ($x < 1$).
• The original middle loop ($0 < x_0 < 2$) becomes the new middle loop ($1 < x < 3$) with its peak turning point mapped to $(2, -2)$.
• The original left branch ($x_0 < 0$) becomes the new right branch ($x > 3$), decreasing from the asymptote $x=3$, crossing $(4, 0)$, and passing through $(6, -1)$.