曲线 $C$ 的方程为 $y = \frac{1}{\sqrt{bx^2 - 2bx}}$，其中 $b < 0$。草绘 $C$ 的图像，并在适用时用含有 $b$ 的式子表示出任何渐近线的方程、对称轴方程以及转折点的坐标。

1. Determine the domain of the function to identify the vertical asymptotes.

For the curve to be defined, the expression inside the square root in the denominator must be strictly positive:
$bx^2 - 2bx > 0$
Since it is given that $b < 0$, dividing the inequality by $b$ reverses the inequality sign:
$x^2 - 2x < 0$
$x(x - 2) < 0$
Hence, the domain of the curve is $0 < x < 2$.
As $x$ approaches $0$ from the right or $2$ from the left, the denominator approaches $0$, causing $y o \infty$.
Thus, the equations of the vertical asymptotes are $x = 0$ and $x = 2$.

2. Find the coordinates of the turning point.

Let the expression in the square root be $f(x) = bx^2 - 2bx$.
Completing the square gives:
$f(x) = b(x^2 - 2x) = b[(x - 1)^2 - 1] = b(x - 1)^2 - b$
Because $b < 0$, the squared term $b(x - 1)^2 \le 0$. Hence, $f(x)$ attains a maximum value of $-b$ when $x = 1$.
The maximum of $f(x)$ results in the minimum positive value for $y = \frac{1}{\sqrt{f(x)}}$.
Substituting $x = 1$:
$y = \frac{1}{\sqrt{b(1)^2 - 2b(1)}} = \frac{1}{\sqrt{-b}}$
The coordinates of the turning point are $\left(1, \frac{1}{\sqrt{-b}}\right)$.

3. Identify the line of symmetry.

The quadratic expression $bx^2 - 2bx$ is symmetric about its vertex at $x = 1$, and the domain $0 < x < 2$ is also symmetric about $x = 1$.
Consequently, the curve $C$ is symmetric about the vertical line passing through its turning point.
The equation of the line of symmetry is $x = 1$.

4. Describe the sketch of the curve $C$.

To sketch the curve $C$:

1. Draw and label vertical dashed lines at $x = 0$ (the $y$-axis) and $x = 2$ to represent the asymptotes.
2. Plot and label the minimum turning point at $\left(1, \frac{1}{\sqrt{-b}}\right)$.
3. Draw a U-shaped curve that passes through the turning point, is visually symmetric about the line $x = 1$, and curves upwards to approach the asymptotes $x = 0$ and $x = 2$ smoothly as $y \to \infty$.