The diagram shows the curve $y = \text{f}(1-x)$. The curve has a maximum point at $(-1, 5)$ and two asymptotes $x = -2$ and $y = 1$. On separate diagrams, sketch the graphs of

1. Establish the algebraic mapping between the given graph and the required graph.

We are given the graph of $y = \text{f}(1-x)$ and need to sketch $y = \text{f}(x)$.
Let a point on the given graph be $(x, y)$. This means $y = \text{f}(1-x)$.
Let $X = 1 - x$. Then the equation becomes $y = \text{f}(X)$.
Therefore, a point $(x, y)$ on the given curve maps to the point $(1-x, y)$ on the target curve $y = \text{f}(x)$.

2. Apply the mapping to the key features of the graph (turning point and asymptotes).

Apply the mapping $X = 1 - x$ to the given key features:

• Maximum point: For $x = -1, y = 5 \implies X = 1 - (-1) = 2$. The new maximum point is $(2, 5)$.
• Vertical asymptote: For $x = -2 \implies X = 1 - (-2) = 3$. The new vertical asymptote is $x = 3$.
• Horizontal asymptote: The $y$-values are unaffected by horizontal transformations, so the horizontal asymptote remains $y = 1$.

3. Determine the orientation and shape of the transformed branches.

The transformation $x \mapsto 1-x$ effectively reverses the horizontal direction of the graph (a reflection in the vertical line $x = 0.5$).

• The original right branch (for $x > -2$) now becomes the new left branch (for $X < 3$). As $x \to \infty$, $y \to 1^+$, so as $X \to -\infty$, $y \to 1^+$. The branch rises to the maximum at $(2, 5)$ and plunges to $-\infty$ as $X \to 3^-$.
• The original left branch (for $x < -2$) now becomes the new right branch (for $X > 3$). As $x \to -\infty$, $y \to 1^+$, so as $X \to \infty$, $y \to 1^+$. The branch descends from $+\infty$ near $X = 3^+$ and flattens out towards $y = 1$.

1. Recognize the relationship between $\text{g}(x)$ and the given graph.

The question defines $\text{g}(x) = \text{f}(1-x)$. Since the given diagram is exactly the graph of $y = \text{f}(1-x)$, the given graph is $y = \text{g}(x)$. We must sketch the derivative of this exact graph.

2. Identify the $x$-intercepts of $y = \text{g}'(x)$.

The graph of $y = \text{g}(x)$ has a maximum turning point at $(-1, 5)$. At this point, the gradient is zero, which means $\text{g}'(-1) = 0$. Therefore, $y = \text{g}'(x)$ has an $x$-intercept at $(-1, 0)$.

3. Determine the asymptotes of $y = \text{g}'(x)$.

As $x \to -2$ from both the left and the right, the curve of $y = \text{g}(x)$ becomes increasingly steep with a positive slope, meaning the gradient approaches $+\infty$. Thus, $x = -2$ is a vertical asymptote for $y = \text{g}'(x)$, with the curve tending to $+\infty$ on both sides.

As $x \to \pm\infty$, the graph of $y = \text{g}(x)$ flattens out as it approaches the horizontal asymptote $y = 1$. This means the gradient approaches $0$. Thus, $y = 0$ (the $x$-axis) is a horizontal asymptote for $y = \text{g}'(x)$.

4. Determine the intervals of positive and negative gradient to finalize the shape.

For $x < -2$ and $-2 < x < -1$, the function $y = \text{g}(x)$ is strictly increasing, so $\text{g}'(x) > 0$ (the graph lies above the $x$-axis).
For $x > -1$, $y = \text{g}(x)$ is strictly decreasing, so $\text{g}'(x) < 0$ (the graph lies below the $x$-axis).
Because $\text{g}'(x)$ must return to $0$ as $x \to \infty$, the branch for $x > -1$ must have a local minimum point below the $x$-axis before approaching $y = 0$.