The diagram below shows the graph of the curve $y = f(x)$, with turning point at $\left(\tfrac{k}{2},2\right)$ and asymptotes with equations $x = k$ and $y = 0$. The curve undergoes the following sequence of transformations in succession A: Translate $k$ units in the negative $x$-direction B: Reflection about the $y$-axis C: Translate $k$ units in the positive $x$-direction

1. Determine the combined effect of the three transformations.

Let $(x, y)$ be a general point on the original curve $y = f(x)$.

1. Translate $k$ units in the negative $x$-direction: $(x, y) \mapsto (x-k, y)$
2. Reflection about the $y$-axis: $(x-k, y) \mapsto (-(x-k), y) = (-x+k, y)$
3. Translate $k$ units in the positive $x$-direction: $(-x+k, y) \mapsto (-x+k+k, y) = (2k-x, y)$

The overall mapping is $(x, y) \mapsto (2k-x, y)$, which represents a reflection of the entire graph about the vertical line $x = k$.

2. Find the coordinates of the new turning point.

The original turning point is at $(\frac{k}{2}, 2)$.
Applying the transformation mapping $(x, y) \mapsto (2k-x, y)$, the new $x$-coordinate is:
$2k - \frac{k}{2} = \frac{3k}{2}$
The $y$-coordinate remains unchanged. Thus, the new turning point is $(\frac{3k}{2}, 2)$.

3. Determine the new equations of the asymptotes.

Vertical asymptote: The original line $x = k$ maps to $x = 2k - k \implies x = k$. It remains unchanged.
Horizontal asymptote: The $y$-coordinates are unchanged by purely horizontal transformations, so the horizontal asymptote remains $y = 0$.

4. Describe the shape of the transformed curve for the sketch.

Since the transformation is a horizontal reflection of the graph about $x = k$:

• The original right branch (for $x > k$, below the $x$-axis, approaching $y=0$) becomes the new left branch ($x < k$). It lies entirely below the $x$-axis, approaching $y=0$ as $x \to -\infty$ and approaching the vertical asymptote $x=k$ downwards.
• The original left branch (for $x < k$, above the $x$-axis, with a minimum trough at $x = \frac{k}{2}$) becomes the new right branch ($x > k$). It lies entirely above the $x$-axis, has its minimum at $(\frac{3k}{2}, 2)$, and approaches the vertical asymptote $x=k$ upwards.

1. Find the equation after transformation A (translation in the negative $x$-direction).

Translating $k$ units in the negative $x$-direction requires replacing $x$ with $x + k$:
$y = f(x + k)$

2. Find the equation after transformation B (reflection about the $y$-axis).

Reflecting about the $y$-axis requires replacing $x$ with $-x$:
$y = f(-x + k)$

3. Find the equation after transformation C (translation in the positive $x$-direction).

Translating $k$ units in the positive $x$-direction requires replacing $x$ with $x - k$:
$y = f(-(x - k) + k)$
$y = f(-x + k + k)$
$y = f(-x + 2k)$

1. Determine the line of symmetry from the given functional equation

A curve with equation $y = \mathrm{g}(x)$ has a line of symmetry $x = a$ if $\mathrm{g}(a + x) = \mathrm{g}(a - x)$, which can also be written as $\mathrm{g}(x) = \mathrm{g}(2a - x)$.

Comparing $\mathrm{g}(x) = \mathrm{g}(4 - x)$ with $\mathrm{g}(x) = \mathrm{g}(2a - x)$:
$2a = 4$
$a = 2$

Alternatively, the line of symmetry is at the midpoint of $x$ and $4 - x$:
$\text{Midpoint} = \frac{x + (4 - x)}{2} = 2$

Thus, the line of symmetry is $x = 2$.