= Solution
The <absolute value> is nonnegative, so a real point must satisfy $1-x^2\geq0$, or $-1\leq x\leq1$. On that interval the two branches are
$$
\boxed{y=1-x^2\quad\text{or}\quad y=x^2-1,\qquad -1\leq x\leq1.}
$$
They form a closed lens symmetric about both coordinate axes, meeting at $(\pm1,0)$ and reaching $(0,\pm1)$. The <parabolas> outside the stated interval do not belong to the original <locus>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-absolute-value.png]
{title=The absolute-value locus and its intersections with the horizontal line y = 1/2}
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