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.

*These suggestions apply to both parts of this question.*

How can we determine the vertical asymptotes of a curve? The horizontal ones?

Can the curve cross an asymptote? Does the function change sign here or not?

Are there any roots to the equation \(f(x)=0\) or \(g(x)=0\)? In other words, what are the \(x\) intercepts? What are the \(y\)-intercepts?

How do the curves behave as \(x\rightarrow\infty\)?

These curves are each symmetrical about a line \(x=c\). What is this \(c\) in each case?

How can we find the stationary points of the curve?

In the second part of the question, how do the two different conditions \(b > a+2\) and \(b = a+2\) affect the asymptotes?