= Solution
For the requested parameter value $a=2$, the <function graph> has two branches, symmetric about the vertical axis because $f$ is an <even function>. Its <derivatives> are
$$
f'(x)=\frac x2-\frac8{x^3}=\frac{x^4-16}{2x^3},\qquad f''(x)=\frac12+\frac{24}{x^4}>0.
$$
The <stationary points> are $\boxed{(-2,2)\text{ and }(2,2)}$, both <local minima> and also the lowest points of their respective branches. The <function> tends to $+\infty$ as $x\to0^\pm$, so $x=0$ is a vertical <asymptote>; it also tends to $+\infty$ as $|x|\to\infty$. The negative branch decreases to $(-2,2)$ and then rises towards the vertical <asymptote>; the positive branch falls from the vertical <asymptote> to $(2,2)$ and then rises.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1-q4-graph.png]
{title=Two branches and minima of the function in Question 4 for $a=2$}
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