= Solution
The graph is the upper branch of the <hyperbola> $d^2-(t+1)^2=1/2$. It is symmetric about $t=-1$, positive everywhere, and has its lowest point at $(-1,1/\sqrt2)$. Its <derivatives> are
$$
d'(t)=\frac{t+1}{d(t)},\qquad d''(t)=\frac1{2d(t)^3}>0,
$$
so it is strictly convex. Since
$$
d(t)-|t+1|=\frac{1/2}{\sqrt{(t+1)^2+1/2}+|t+1|}\longrightarrow0,
$$
its two oblique <asymptotes> are
$$
\boxed{d=t+1\quad(t\to+\infty),\qquad d=-t-1\quad(t\to-\infty).}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2-q11-distance.png]
{title=Distance from a moving point to a skew line}
{description=Original plot of $d(t)=\sqrt{(t+1)^2+1/2}$, its minimum and its two asymptotes.}
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