Can we sketch the curve $y=1/((x-a)^2-1)$?

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Sketch the curve \(y=f(x)\) where \[f(x)=\frac{1}{(x-a)^2-1} \qquad \qquad (x\neq a\pm 1),\] and \(a\) is a constant.

The function \(g(x)\) is defined by \[g(x)=\frac{1}{((x-a)^2-1)((x-b)^2-1)} \qquad \qquad (x\neq a\pm 1,\,x\neq b \pm 1),\] where \(a\) and \(b\) are constants, and \(b>a\). Sketch the curves \(y=g(x)\) in the two cases \(b>a+2\) and \(b=a+2\), finding the values of \(x\) at the stationary points.