Question

How to sketch the graph f(x)=`|x^2-1|/(x^2-1)` showing any points of discontinuity, horizontal or vertical asymptotes and x and y intercepts?

`|x^2-1|/(x^2-1)`

Answer 1

First note that `absu/u = {(1,"if",u > 0),(-1,"if",u < 0):}`

Explanation:

Second observe that `x^2-1 > 0` if and only if `x < -1` OR `x > 1`.

So we have

`f(x) = abs(x^2-1)/(x^2-1) = {(1,"if",x < -1),(-1,"if",-1 < x < 1),(1,"if",1 < x):}`

So the graph is:

graph{abs(x^2-1)/(x^2-1) [-4.995, 4.87, -1.794, 3.14]}

It is now clear that there are
two discontinuities (jump),
no vertical asymptotes,
no `x`-intercepts,
`y`-intercept is `-1` and,
depending on your definition of horizontal asymptote, either none or `y=1` on both the left and right.