The graph of $y = f(x)$ is given below. On separate diagrams, sketch the graphs of

1. Convert the asymptotes of f into features of 1/f

The vertical asymptotes of $f$ at $x=-4$ and $x=0$ become points where $\frac{1}{f(x)}\to 0$ (the curve approaches the $x$-axis there).

The horizontal asymptote $y=2$ (as $x\to-\infty$) becomes $y=\frac{1}{2}$.

The horizontal asymptote $y=0$ (as $x\to+\infty$): since $f\to0^+$, $\frac{1}{f(x)}\to+\infty$.

2. Convert the zeros of f into vertical asymptotes of 1/f and map the turning point

$f$ crosses the $x$-axis once on the left branch (some $x<-4$) and once on the right branch (some $x>0$). At each zero $\frac{1}{f(x)}\to\pm\infty$, so these become vertical asymptotes of $y=\frac{1}{f(x)}$.

The minimum point $(-2,3)$ of $f$ maps to a maximum point $\left(-2,\tfrac13\right)$ of $\frac{1}{f(x)}$ (turning points keep their $x$-coordinate; $y\mapsto\frac1y$, max$\leftrightarrow$min).

3. Use sign preservation to draw the three correctly-shaped branches

Since $\frac{1}{f(x)}$ has the same sign as $f(x)$:

• Left branch ($x<-4$): from $y=\tfrac12^{+}$ at $-\infty$, rising to $+\infty$ at the left zero, then from $-\infty$ rising up to $0^-$ as $x\to-4^-$.
• Middle branch ($-4<x<0$): $f>0$ with the new maximum $\left(-2,\tfrac13\right)$; the curve $\to0^+$ as $x\to-4^+$ and as $x\to0^-$ (a positive hump).
• Right branch ($x>0$): from $0^-$ near $x=0^+$, falling to $-\infty$ at the right zero, then from $+\infty$ down and back up to $+\infty$ as $x\to+\infty$.

Label $y=\tfrac12$, the maximum $\left(-2,\tfrac13\right)$, and the new vertical asymptotes.

1. Set up the rules linking $f'$ to $f$ and carry over the asymptotes.

The graph of $y=f'(x)$ records the gradient of $y=f(x)$:

• where $f$ is increasing, $f'(x)>0$;
• where $f$ is decreasing, $f'(x)<0$;
• at a stationary point of $f$, $f'(x)=0$ (an $x$-intercept of $f'$).

The vertical asymptotes of $f$ at $x=-4$ and $x=0$ remain vertical asymptotes of $f'$. Where $f$ levels off towards a horizontal asymptote, the gradient $\to 0$, so $f'\to 0$.

2. Determine the sign and limiting behaviour of $f'$ on each interval.

$x<-4$: $f$ falls from $y\to 2^-$ (as $x\to-\infty$) down to $-\infty$ at $x=-4^-$, so $f$ is decreasing $\Rightarrow f'<0$. As $x\to-\infty$ the curve flattens to $y=2$, so $f'\to 0^-$; near $x=-4^-$ the fall is steep, so $f'\to-\infty$.

$-4<x<-2$: $f$ decreases towards the minimum at $(-2,3)$, so $f'<0$, reaching $f'=0$ at $x=-2$.

$-2<x<0$: $f$ increases away from the minimum towards the asymptote $x=0$, so $f'>0$, and $f'\to+\infty$ as $x\to0^-$.

$x>0$: $f$ rises from $-\infty$ (near $x=0^+$) and levels off towards $y=0$, so $f'>0$ throughout, with $f'\to+\infty$ near $x=0^+$ and $f'\to0^+$ as $x\to+\infty$.

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

Draw vertical asymptotes at $x=-4$ and $x=0$ and the horizontal asymptote $y=0$. Then:

• left branch ($x<-4$): entirely below the $x$-axis, coming up to $0^-$ as $x\to-\infty$ and plunging to $-\infty$ at $x=-4^-$;
• middle branch ($-4<x<0$): below the axis on $(-4,-2)$, crossing the $x$-axis at $x=-2$, then above the axis rising to $+\infty$ at $x=0^-$;
• right branch ($x>0$): above the axis, descending from $+\infty$ near $x=0^+$ to $0^+$ as $x\to+\infty$.

Label the $x$-intercept $(-2,0)$ and all asymptotes.

1. Find the Cartesian equations for both curve $C$ and the new curve.

For curve $C$:
$x = t$, $y = 2t^2$
Substituting $t = x$ into the second equation, we obtain the Cartesian equation for $C$:
$y = 2x^2$

For the new curve:
$x = 2t + 1 \implies t = \frac{x-1}{2}$
$y = t^2$
Substituting $t = \frac{x-1}{2}$ into $y = t^2$, we obtain:
$y = \left(\frac{x-1}{2}\right)^2 = \frac{1}{4}(x-1)^2$

2. Express the Cartesian equation of the new curve in terms of the function defining curve $C$.

Let $f(x) = 2x^2$. We want to express $y = \frac{1}{4}(x-1)^2$ in terms of $f(x)$.
Rewrite the new equation to reveal the structure of $f(x)$:
$y = \frac{1}{4}(x-1)^2$
$y = \frac{1}{8} \left[ 2(x-1)^2 \right]$
Since $f(x-1) = 2(x-1)^2$, we have:
$y = \frac{1}{8} f(x-1)$

3. Describe the sequence of transformations based on the relationship $y = \frac{1}{8} f(x-1)$.

The transformation from $y = f(x)$ to $y = \frac{1}{8} f(x-1)$ consists of two standard operations:

1. The term $(x-1)$ represents a horizontal translation of $1$ unit in the positive $x$-direction.
2. The multiplier $\frac{1}{8}$ outside the function represents a vertical scaling parallel to the $y$-axis by a scale factor of $\frac{1}{8}$.