If $y=(x^2+1)/(x^2-a^2)$, then $y$ cannot take which values?

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Show that, if \(y=\dfrac{x^2+1}{x^2-a^2}\), \(y\) takes all real values twice, except those for which \(-\frac{1}{a^2}\le y\le 1\). Sketch the curve \(y=\dfrac{x^2+1}{x^2-4}\), indicating its asymptotes.

Could we rearrange the equation into the form \(x^2 = f(y)\)?

Could we try sketching the graph before answering the first part of the question?