Question

How do you identify all asymptotes or holes for `f(x)=(x-4)/(x^2-3x)`?

Answer 1

Vertical Asymptotes at `x=0" and "x=3`
Horizontal Asymptotes at `y=0`

Explanation:

To determine asymptotes you have understand the behaviour of the equation. Some equations are more challenging than others.

`color(blue)("Denominator approaching 0")`

It is 'bad news' to have a denominator become zero.

None technical term for this is 'not allowed'. Correct name is: undefined.

So solve for `x^2-3x=0`

`x(x-3)=0 => x=0" and "x=3`

Consider the case `x=0`

`color(brown)("Suppose "x" tended to 0 from the right (positive)")`
Then writing the equation as:

`f(x)=(x-4)/(x(x-3)) -> (x/x-4/x)/(x-3)`

Then this is almost but not quite the same as:

`lim_(xto0) f(x)=L -> (1-oo)/-3 ->+oo`

`color(brown)("Suppose "x" tended to 0 from the left (negative value"->-x")")`

`f(x)=(x-4)/(x(x-3)) -> ((-x)/(-x)-4/(-x))/((-x)-3)`

Then this is almost but not quite the same as:

`lim_(xto0) f(x)=L -> (1+oo)/-3 ->-oo`

`color(green)("You have the same situation for "x" tending to 3")`
`color(green)("See if you can work that one out using the same method as above")`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)(x" tending to "+-oo)`

`f(x)=(x-4)/(x(x-3)) -> (x/x-4/x)/(x-3)`

As `x` becomes larger and larger the 3 in `x-3` has less and less effect. In the end it becomes so insignificant you can forget about it

In the same way `4/x` becomes smaller and smaller the larger `x` becomes. In the end it tends to 0

So `lim_(x->+oo) f(x) -> 1/oo ->0`

So `lim_(x->-oo) f(x) -> 1/(-oo) ->0`