A curve $C$ has equation $y = \frac{1}{\sqrt{bx^2 - 2bx}}$, where $b < 0$. Sketch $C$ and give, in terms of $b$ where appropriate, the equations of any asymptotes and the line of symmetry, and coordinates of turning point.

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 $y \to \infty$.