Question

What are the asymptote(s) and hole(s), if any, of `f(x) =3/x-(8x)/(x^2-3x)`?

Answer 1

Asymptotes: `x=3, x=0, y=0`

Explanation:

`f(x)=3/x-(8x)/(x^2-3x)`

`f(x)=(3(x^2-3x)-8x*x)/(x(x^2-3x)`

For the asymptotes, we look at the denominator.
Since the denominator cannot equal to `0`

ie `x(x^2-3x)=0`

`x^2(x-3)=0`

therefore `x!=0,3`

For the y asymptotes, we use the limit as `x -> 0`

`lim x->0 (3(x^2-3x)-8x*x)/(x(x^2-3x)`

=`lim x->0 (3x^2-9x-8x^2)/(x(x^2-3x))`

=`lim x->0 (-5x^2-9x)/(x^3-3x^2)`

=`lim x->0 ((-5/x-9/x^2))/(1-3/x)`

=`0`

therefore `y!=0`