曲线 $C_{1}$ 和 $C_{2}$ 的方程分别为 $\dfrac{x^{2}}{9} - \dfrac{y^{2}}{4} = 1$ 和 $y^{2} + x^{2} = k^{2}$，其中 $k$ 为正常数。 在同一坐标系中画出 $C_{1}$ 和 $C_{2}$ 的草图，标出与坐标轴的所有交点坐标以及所有渐近线方程。 写出使 $C_{1}$ 和 $C_{2}$ 相交的 $k$ 的取值范围。 写出 $C_{1}$ 和 $C_{2}$ 的公共对称轴方程。 函数 f 的定义域为全体实数集，由下式给出： $f(x) = \begin{cases} -2x + 6 & \text{for } 0 < x \le 3, \\ 3x - 9 & \text{for } 3 < x \le 5, \end{cases}$ 且 $f(x) = f(x+5)$。 求 $f(46)$。 画出 $y = f(x)$ 在 $-5 \le x \le 5$ 范围内的图象。 由此，写出 $f(-x) = 0$ 在 $-5 \le x \le 5$ 范围内的所有根。

1. Find the intercepts and asymptotes of the hyperbola $C_1$.

For $C_1: \frac{x^2}{3^2} - \frac{y^2}{2^2} = 1$, the curve is a hyperbola with its transverse axis along the $x$-axis.
To find the $x$-intercepts, substitute $y = 0$:
$\frac{x^2}{9} = 1 \implies x^2 = 9 \implies x = \pm 3$.
The $x$-intercepts are $(3, 0)$ and $(-3, 0)$. There are no $y$-intercepts.

The equations of the asymptotes are given by $y = \pm \frac{b}{a}x$, which yields $y = \pm \frac{2}{3}x$.

2. Find the intercepts of the circle $C_2$.

For $C_2: x^2 + y^2 = k^2$, the curve is a circle centered at the origin $(0,0)$ with radius $k$.
To find the $x$-intercepts, substitute $y = 0$:
$x^2 = k^2 \implies x = \pm k$.
The $x$-intercepts are $(k, 0)$ and $(-k, 0)$.

To find the $y$-intercepts, substitute $x = 0$:
$y^2 = k^2 \implies y = \pm k$.
The $y$-intercepts are $(0, k)$ and $(0, -k)$.

3. Sketch both curves on the same diagram.

Draw the $x$ and $y$ axes.
Sketch the asymptotes $y = \frac{2}{3}x$ and $y = -\frac{2}{3}x$ as dashed lines passing through the origin.
Sketch the hyperbola $C_1$ with vertices at $(3,0)$ and $(-3,0)$, ensuring the branches approach the asymptotes as $x \to \pm \infty$.
Sketch the circle $C_2$ centered at $(0,0)$ and mark the axial intercepts $(k,0), (-k,0), (0,k), (0,-k)$ appropriately.

1. Determine the condition for intersection using the distance of the hyperbola's vertices from the origin

From the equation of the hyperbola $C_1$: $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$, the vertices are at $(\pm 3, 0)$. These are the points on the hyperbola that are closest to the origin, at a distance of $3$ units.
For the circle $C_2$: $x^2 + y^2 = k^2$ with centre $(0,0)$ and radius $k$ to intersect the hyperbola, its radius must be at least the distance from the origin to these vertices.
Therefore, $k \ge 3$.

1. Determine the axes of symmetry for the hyperbola and the circle to find their common lines of symmetry.

The hyperbola $C_1$ given by $\dfrac{x^2}{9} - \dfrac{y^2}{4} = 1$ has its transverse and conjugate axes along the coordinate axes, so its lines of symmetry are the $x$-axis and the $y$-axis.

The circle $C_2$ given by $y^2 + x^2 = k^2$ is centred at the origin, meaning any straight line passing through the origin is a line of symmetry for $C_2$.

The common lines of symmetry are therefore the $x$-axis and the $y$-axis.
Their equations are $y = 0$ and $x = 0$.

1. Use the periodic property $f(x) = f(x+5)$ to find an equivalent argument within the interval $(0, 5]$.

Since the function repeats every $5$ units, we can subtract multiples of $5$ from $46$.
$46 = 9(5) + 1$
$f(46) = f(46 - 45) = f(1)$

2. Evaluate the function using the correct expression for $x=1$.

Since $1$ is in the interval $0 < x \le 3$, we use $f(x) = -2x + 6$.
$f(1) = -2(1) + 6 = 4$

1. Determine the shape and key points of the graph for the basic interval $0 < x \le 5$.

For $0 < x \le 3$, $f(x) = -2x + 6$. This is a straight line segment. As $x \to 0^+$, $f(x) \to 6$. At $x = 3$, $f(3) = 0$.
For $3 < x \le 5$, $f(x) = 3x - 9$. This is also a straight line segment starting from $f(3) = 0$ and ending at $f(5) = 3(5) - 9 = 6$.
Thus, on $0 < x \le 5$, the graph forms a 'V' shape connecting $(0, 6)$ to $(3, 0)$ and $(3, 0)$ to $(5, 6)$.

2. Use the periodic property to determine the graph for $-5 \le x \le 0$ and sketch the full domain.

The equation $f(x) = f(x+5)$ indicates the function is periodic with a period of 5. The graph repeats horizontally every 5 units.
To find the graph for $-5 < x \le 0$, we translate the basic segment left by 5 units:

• $(5, 6) \to (0, 6)$
• $(3, 0) \to (-2, 0)$
• $(0, 6) \to (-5, 6)$
  Additionally, at the boundaries, $f(-5) = f(0) = f(5) = 6$, so the endpoints are solid and continuous.
  The full sketch consists of two repeated 'V' shapes over the domain $-5 \le x \le 5$.

1. Identify the roots of $f(x) = 0$ in the interval $[-5, 5]$ from the earlier parts

From the definition for $0 < x \le 5$, setting $f(x) = 0$ gives:
$-2x + 6 = 0 \implies x = 3$
Using the given periodicity $f(x) = f(x+5)$, the corresponding root in the negative part of the domain is:
$x = 3 - 5 = -2$
Thus, the roots of $f(x) = 0$ for $-5 \le x \le 5$ are $x = -2$ and $x = 3$.

2. Determine the roots of $f(-x) = 0$ by finding the corresponding inputs

The equation $f(-x) = 0$ represents a horizontal reflection of the graph of $f(x) = 0$ in the $y$-axis. The inputs $-x$ must therefore equal the original roots:
$-x = -2 \implies x = 2$
$-x = 3 \implies x = -3$
Both $x = 2$ and $x = -3$ lie within the required interval $-5 \le x \le 5$.