A curve C has equation $y = k(x+1) + \dfrac{4}{x+1}$, $x \in \mathbb{R}$, $x \ne -1$, where $0 < k < 2$ is a constant. The graph of C is transformed by a reflection in the x-axis, followed by a translation of 1 unit in the positive x-direction, followed by a stretch with scale factor $\tfrac{1}{2}$ parallel to the y-axis.

1. Find the equations of the vertical and oblique asymptotes.

The vertical asymptote occurs where the denominator of the fractional term is zero:
$x + 1 = 0 \implies x = -1$
As $x \to \pm\infty$, the fractional term $\frac{4}{x+1} \to 0$. Therefore, the oblique asymptote is given by the linear part of the function:
$y = k(x+1)$

2. Determine the axial intercept(s) of the curve.

To find the $y$-intercept, set $x = 0$:
$y = k(0+1) + \frac{4}{0+1} = k + 4$
Thus, the $y$-intercept is $(0, k+4)$.

To find any $x$-intercepts, set $y = 0$:
$k(x+1) + \frac{4}{x+1} = 0 \implies k(x+1)^2 + 4 = 0$
Since we are given that $k > 0$ and $(x+1)^2 \ge 0$, it follows that $k(x+1)^2 + 4 > 0$. There are no real solutions, meaning the curve has no $x$-intercept.

3. Find the coordinates of the turning points using differentiation.

Differentiate $y$ with respect to $x$:
$\frac{dy}{dx} = k - \frac{4}{(x+1)^2}$
Set $\frac{dy}{dx} = 0$ to find the stationary points:
$k - \frac{4}{(x+1)^2} = 0 \implies (x+1)^2 = \frac{4}{k} \implies x+1 = \pm\frac{2}{\sqrt{k}}$
$x = -1 + \frac{2}{\sqrt{k}} \quad \text{or} \quad x = -1 - \frac{2}{\sqrt{k}}$
Substitute $x+1 = \pm\frac{2}{\sqrt{k}}$ back into the original equation for $y$:
When $x+1 = \frac{2}{\sqrt{k}}$:
$y = k\left(\frac{2}{\sqrt{k}}\right) + \frac{4}{\left(\frac{2}{\sqrt{k}}\right)} = 2\sqrt{k} + 2\sqrt{k} = 4\sqrt{k}$
When $x+1 = -\frac{2}{\sqrt{k}}$:
$y = k\left(-\frac{2}{\sqrt{k}}\right) + \frac{4}{\left(-\frac{2}{\sqrt{k}}\right)} = -2\sqrt{k} - 2\sqrt{k} = -4\sqrt{k}$
The turning points are $\left(-1 + \frac{2}{\sqrt{k}}, 4\sqrt{k}\right)$ and $\left(-1 - \frac{2}{\sqrt{k}}, -4\sqrt{k}\right)$.

4. Summarise the key features to be drawn on the sketch.

The curve is a rational function mapping to a hyperbola with two branches.
Since $0 < k < 2$, we know $\frac{2}{\sqrt{k}} > 1$, which places the $x$-coordinate of the first turning point at $x > 0$. Thus, the local minimum $\left(-1 + \frac{2}{\sqrt{k}}, 4\sqrt{k}\right)$ lies in the first quadrant, and the $y$-intercept $(0, k+4)$ lies slightly above and to the left of it.

• Top-right branch: Bounded by $x = -1$ on the left and approaching $y = k(x+1)$ on the right, curving downwards through $(0, k+4)$ to the local minimum before rising.
• Bottom-left branch: Bounded by the asymptotes, remaining entirely negative with a local maximum at $\left(-1 - \frac{2}{\sqrt{k}}, -4\sqrt{k}\right)$.

1. Apply the reflection in the $x$-axis to the original curve.

A reflection in the $x$-axis replaces $y$ with $-y$. The equation becomes:
$y = -\left( k(x+1) + \frac{4}{x+1} \right)$

2. Apply the translation of 1 unit in the positive $x$-direction.

A translation of $1$ unit in the positive $x$-direction replaces $x$ with $(x-1)$. The equation becomes:
$y = -\left( k(x-1+1) + \frac{4}{x-1+1} \right)$
$y = -kx - \frac{4}{x}$

3. Apply the stretch parallel to the $y$-axis to find the final equation.

A stretch with scale factor $\frac{1}{2}$ parallel to the $y$-axis multiplies the entire function by $\frac{1}{2}$. The equation becomes:
$y = \frac{1}{2} \left( -kx - \frac{4}{x} \right)$
$y = -\frac{1}{2}kx - \frac{2}{x}$