The diagram shows the curve $y = \text{f}(x)$. The curve passes through the points $A(-1, 0)$, $B(0, 2)$, $C(1, 0)$ and $D(3, -3)$. On separate clearly labelled diagrams, sketch the graphs of

1. Determine the new coordinates of the given points by multiplying their $y$-coordinates by $-\frac{1}{2}$.

The transformation $y = -\frac{1}{2}\text{f}(x)$ affects only the $y$-coordinates of the graph (a vertical scaling by factor $\frac{1}{2}$ and a reflection in the $x$-axis).
$A(-1, 0) \rightarrow (-1, 0)$
$B(0, 2) \rightarrow (0, -1)$
$C(1, 0) \rightarrow (1, 0)$
$D(3, -3) \rightarrow (3, 1.5)$

2. Determine the transformed horizontal asymptote and the correct shape of the curve.

The original horizontal asymptote is $y = -1$. Multiplying by $-\frac{1}{2}$ gives the new asymptote:
$y = -1 \times \left(-\frac{1}{2}\right) = 0.5$
Since the original curve approaches $y = -1$ from below (i.e. values are more negative than $-1$), the new curve will approach $y = 0.5$ from above.
The local maximum $B$ becomes a local minimum, and the local minimum $D$ becomes a local maximum. Draw a smooth curve passing through these points with the specified features.

1. Determine the transformation of coordinates.

Let a point on $y = \text{f}(x)$ be $(x, y)$. For the new graph $Y = \text{f}(1-X)$, setting $y = \text{f}(x)$ means $x = 1 - X \implies X = 1 - x$. The transformation maps every original point $(x, y)$ to $(1-x, y)$.

2. Calculate the new coordinates of the key points.

Apply $X = 1 - x$ to the given points:

• $A(-1, 0) \mapsto (1 - (-1), 0) = (2, 0)$
• Maximum $B(0, 2) \mapsto (1 - 0, 2) = (1, 2)$
• $C(1, 0) \mapsto (1 - 1, 0) = (0, 0)$
• Minimum $D(3, -3) \mapsto (1 - 3, -3) = (-2, -3)$

3. Determine the new behavior of the asymptote.

The original graph has an asymptote $y = -1$ as $x \to \infty$. For the transformed graph, as $x \to \infty$, the new $X$-coordinate becomes $1 - x \to -\infty$. Thus, the horizontal asymptote is $y = -1$ as $x \to -\infty$.