Question

How do you graph and find the discontinuities of `(x^2-25)/(x^2+5x)`?

Answer 1

We first factorise.

Explanation:

`=((x+5)(x-5))/(x(x+5))`

The disconituities happen when the denominator `=0`
So at `x=0andx=-5`
The discontinuity at `x=5` is removable as both limits go to the same value:

`lim_(x->1^-) f(x)=lim_(x->1^+) f(x)=2`

In other cases we can cancel out the `(x+5)`'s:

`=(x-5)/x=x/x-5/x=1-5/x`

This means that if `x` gets larger `+or-` the function goes to `1` (from up or down)

graph{1-5/x [-16.02, 16, -8, 8.02]}