The diagram shows the graphs of $y = f(|x|)$ and $y = f'(x)$. The graph of $y = f(|x|)$ has turning points at $(-3,4)$ and $(3,4)$, crosses the x-axis at $(-2,0)$ and $(2,0)$, and the equations of asymptotes are $x = -1$, $x = 1$ and $y = 2$. The graph of $y = f'(x)$ crosses the x-axis at $(3,0)$, and the equations of asymptotes are $x = 1$ and $y = 0$. On separate diagrams, sketch the graphs of the following, labelling clearly the equation(s) of any asymptote(s), coordinates of any axial intercept(s) and turning point(s) where applicable.

1. Deduce the graph of $y = f(x)$ for $x \ge 0$ from $y = f(|x|)$.

For $x \ge 0$, the transformation $y = f(|x|)$ simplifies to $y = f(x)$. Thus, the right portion of the graph of $y = f(x)$ is identical to the right portion of the given $y = f(|x|)$ graph.
From the given information, we deduce the following features for $x \ge 0$:

• Vertical asymptote: $x = 1$
• Horizontal asymptote: $y = 2$ (as $x \to \infty$)
• Axial intercept: $(2, 0)$
• Turning point: Local maximum at $(3, 4)$

To connect the intercept at $x=2$ up to the peak at $x=3$, the graph must approach $-\infty$ as $x \to 1^+$.

2. Deduce the graph of $y = f(x)$ for $x < 1$ using $y = f'(x)$.

The vertical asymptote $x = -1$ on $f(|x|)$ is merely a reflection; $f(x)$ itself only has a vertical asymptote at $x = 1$.
From the description of $y = f'(x)$, its only $x$-intercept is at $x = 3$. This means $f(x)$ has no turning points for $x < 1$.
Since $f'(x)$ has an asymptote at $y = 0$, $f(x)$ approaches a horizontal asymptote as $x \to -\infty$, which must be $y = 2$ (consistent with $f(|x|)$).
Because $f(x)$ approaches $y=2$ and has no turning points for $x<1$, it is strictly monotonic. Given that typical rational functions with a single vertical asymptote and a horizontal asymptote have an even-multiplicity pole (to go to the same infinity on both sides and avoid slant asymptotes), $f(x) \to -\infty$ as $x \to 1^-$.
Consequently, $f(x)$ monotonically decreases from $y=2$ down to $-\infty$, crossing the negative $x$-axis and negative $y$-axis.

3. Sketch the complete graph of $y = f(x)$.

Combine the deduced features to sketch the full graph of $y = f(x)$:

1. Draw and label the asymptotes $x = 1$ and $y = 2$ with dashed lines.
2. Plot and label the known coordinates: the $x$-intercept at $(2,0)$ and the turning point at $(3,4)$.
3. Sketch the right branch ($x > 1$) coming from $-\infty$ at $x=1^+$, passing through $(2,0)$, turning smoothly at $(3,4)$, and approaching $y=2$ from above.
4. Sketch the left branch ($x < 1$) as a strictly decreasing curve starting from just below $y=2$ at the far left, crossing the negative axes, and going down to $-\infty$ at $x=1^-$.

1. Identify the axial intercepts of the reciprocal graph.

The graph of $f'(x)$ has a vertical asymptote at $x=1$, meaning $|f'(x)| \to \infty$ as $x \to 1$. Consequently, its reciprocal $\frac{1}{f'(x)} \to 0$.
This provides an x-intercept for the reciprocal graph at the point $(1, 0)$.

2. Identify the vertical and horizontal asymptotes.

The graph of $f'(x)$ crosses the x-axis at $x=3$, so $f'(3) = 0$. The reciprocal $\frac{1}{f'(x)}$ will therefore have a vertical asymptote at $x=3$.
Furthermore, since $f'(x)$ has a horizontal asymptote of $y=0$ (meaning $f'(x) \to 0$ as $x \to \pm\infty$), the reciprocal $\frac{1}{f'(x)} \to \pm\infty$ as $x \to \pm\infty$. Thus, the reciprocal graph has no horizontal asymptotes.

3. Determine the shape of the curve and any unmarked turning points.

• For $1 < x < 3$: $f'(x)$ is positive and decreases from $+\infty$ to $0$. Thus, $\frac{1}{f'(x)}$ increases from $0$ (at $x=1$) to $+\infty$ (as $x \to 3^-$).
• For $x > 3$: $f'(x)$ crosses into negative values but must approach $0$ as $x \to \infty$. By Rolle's Theorem/logic, it must reach a negative local minimum before turning back towards $y=0$. Consequently, $\frac{1}{f'(x)}$ starts from $-\infty$ at $x=3^+$, goes up to reach a negative local maximum, and then decreases back to $-\infty$ as $x \to \infty$.
• For $x < 1$: $f'(x)$ does not cross the x-axis and approaches $0$ as $x \to -\infty$. Thus $\frac{1}{f'(x)}$ connects from $(1,0)$ and monotonically tends towards $\pm\infty$ as $x \to -\infty$.

1. Determine the sequence of transformations required.

Let $g(x) = f(|x|)$. The target function is $y = -f(|x-1|) = -g(x-1)$.
This requires two transformations applied to the graph of $g(x)$:

1. Replace $x$ with $x-1$: a horizontal translation of $1$ unit in the positive $x$-direction.
2. Multiply the function by $-1$: a reflection about the $x$-axis.

2. Apply the transformations to the turning points and $x$-intercepts.

Turning points:
Start: $(-3, 4)$ and $(3, 4)$
Translate right by 1: $(-2, 4)$ and $(4, 4)$
Reflect in $x$-axis: $(-2, -4)$ and $(4, -4)$

$x$-axis intercepts:
Start: $(-2, 0)$ and $(2, 0)$
Translate right by 1: $(-1, 0)$ and $(3, 0)$
Reflect in $x$-axis: The $y$-coordinates are $0$, so they remain at $(-1, 0)$ and $(3, 0)$.

3. Apply the transformations to the asymptotes.

Vertical Asymptotes:
Start: $x = -1$ and $x = 1$
Translate right by 1: $x = 0$ and $x = 2$
Reflect in $x$-axis: Vertical lines are unchanged, remaining $x = 0$ and $x = 2$.

Horizontal Asymptote:
Start: $y = 2$
Translate right by 1: Unchanged, $y = 2$
Reflect in $x$-axis: $y = -2$