The diagram shows the curve $y = \mathrm{f}(x)$. The curve passes through the point $A(a,0)$ and the point $B(b,0)$, has a turning point at $C(c,-1)$ and asymptotes $y = \frac{1}{2}$ and $x = 0$. Sketch, on separate diagrams, the graphs of

1. Find the coordinates of the points corresponding to $A$, $B$, and $C$.

For a point $(x, y)$ on $y = \mathrm{f}(x)$, the corresponding point on the transformed graph $y = 3 - |\mathrm{f}(x)|$ has coordinates $(x, 3 - |y|)$.

• Point $A(a,0) \rightarrow (a, 3 - |0|) \implies (a, 3)$
• Point $B(b,0) \rightarrow (b, 3 - |0|) \implies (b, 3)$
• Point $C(c,-1) \rightarrow (c, 3 - |-1|) = (c, 3 - 1) \implies (c, 2)$

2. Determine the equations of the new asymptotes.

• Original horizontal asymptote: as $x \to \infty$, $\mathrm{f}(x) \to \frac{1}{2}$.
  Therefore, $y = 3 - \left|\frac{1}{2}\right| = \frac{5}{2}$. The new horizontal asymptote is $y = \frac{5}{2}$.
• Original vertical asymptote: as $x \to 0^+$, $\mathrm{f}(x) \to \infty$.
  Therefore, $y = 3 - |\infty| \to -\infty$. The vertical asymptote remains $x = 0$.

3. Determine the shape of the transformed curve.

1. Cusps: The modulus operation $|\mathrm{f}(x)|$ creates sharp corners (cusps) at the $x$-intercepts $A$ and $B$, where the graph previously crossed the axis smoothly.
2. Reflection and Translation: The negative sign in $-|\mathrm{f}(x)|$ reflects the graph so that it lies entirely below $y=0$, pointing the cusps upwards. The $+3$ shifts everything up by 3 units. Therefore, the cusps become sharp local maxima at $(a, 3)$ and $(b, 3)$.
3. Turning Point: The region between $a$ and $b$ is flipped twice (once by modulus, once by the negative sign), maintaining the smooth local minimum nature at $C(c, 2)$.
4. Asymptotic behaviour: The curve rises from $-\infty$ near $x=0$, and eventually decreases from $(b, 3)$ to approach $y = \frac{5}{2}$ from above.

1. Determine the coordinates and nature of the turning point corresponding to C.

The point $C(c, -1)$ lies on $y = \mathrm{f}(x)$ and is a local minimum.
For $y = \frac{2}{\mathrm{f}(x)}$, substitute $x = c$ and $y = \mathrm{f}(c) = -1$:
$y = \frac{2}{-1} = -2$
Since $\mathrm{f}(x)$ is negative and has a local minimum at $x=c$ (meaning its magnitude is smallest there), its reciprocal will have the largest negative magnitude. Thus, the corresponding point $(c, -2)$ is a local maximum.

2. Identify the vertical asymptotes corresponding to the points A and B.

The points $A(a, 0)$ and $B(b, 0)$ are $x$-intercepts of $y = \mathrm{f}(x)$.
When $\mathrm{f}(x) \to 0$, its reciprocal $\frac{2}{\mathrm{f}(x)} \to \pm\infty$.
Therefore, there are no coordinate points corresponding to $A$ and $B$ on the new graph; instead, they correspond to the vertical asymptotes $x = a$ and $x = b$.

3. Determine the horizontal asymptote and the boundary behavior at the y-axis.

For the horizontal asymptote, as $x \to \infty$, the given graph shows $\mathrm{f}(x) \to \frac{1}{2}$ from below (since it is increasing from the $x$-intercept $B$).
Thus, as $x \to \infty$, $y = \frac{2}{\mathrm{f}(x)} \to \frac{2}{1/2} = 4$ from above. The equation of the horizontal asymptote is $y = 4$.

For the behavior near the $y$-axis, as $x \to 0^+$, the given graph has a vertical asymptote where $\mathrm{f}(x) \to +\infty$.
Thus, $y = \frac{2}{\mathrm{f}(x)} \to 0^+$. The new graph approaches the origin $(0,0)$ as an open boundary point.

4. Synthesise the information to sketch the complete graph.

• Draw axes and indicate the asymptotes $x = a$, $x = b$, and $y = 4$ using dashed lines.
• Plot the local maximum point at $(c, -2)$ and label it.
• Sketch the branch for $0 < x < a$: starting from an open circle at $(0,0)$ and curving upwards towards $x = a$.
• Sketch the branch for $a < x < b$: an inverted U-shape passing through $(c, -2)$, approaching both $x = a$ and $x = b$ downwards.
• Sketch the branch for $x > b$: a decreasing curve starting from high up near $x = b$ and tapering off to approach $y = 4$ from above.