= Solution
Differentiate the two powers to locate the <stationary point>:
$$
y'=\frac{x-2}{x^3}.
$$
For $x>0$, this is negative before $2$ and positive afterwards, so the <global minimum> is
$$
\boxed{x=2,\qquad y=-\frac14.}
$$
The curve crosses the horizontal axis at $x=1$. Its <limits> are $+\infty$ as $x\to0^+$ and $0$ from below as $x\to\infty$, giving the vertical <asymptote> $x=0$ and horizontal <asymptote> $y=0$. These features determine the required sketch.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2020/paper-1-2020-q8-reciprocal-curve.png]
{title=Positive branch of the reciprocal curve, with its minimum and axis crossing}
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