Question

How do you identify all asymptotes or holes for `f(x)=(x^3-4x^2+3x)/(3x^2-15x+18)`?

Answer 1

See below.

Explanation:

First try to factor numerator and denominator:

`(x^3-4x^2+3x)/(3x^2-15x+18)=(x(x-1)(x-3))/(3(x-2)(x-3))=(x(x-1))/(3(x-2)`

Notice we can cancel the factor `(x-3)`. This shows that `x=3` is a removable discontinuity. This is sometimes called a point discontinuity or a singularity.

So there is a hole at `x=3`

Vertical asymptotes occur where the function is undefined. This occurs when `x=2` ( zero denominator )

So the line `x=2` is a vertical asymptote.

`(x(x-1))/(3(x-2))=(x^2-x)/(3x-6)`

The degree of the numerator is greater than the degree of the denominator, so there will be an oblique asymptote. Will can find the equation of this by polynomial division.

`(x^2-x)/(3x-6)=1/3x+1/3`

We only need to divide until we have the equation of a line, so the remainder of 2 in this case is irrelevant to us.

So the line `y= 1/3x +1/3` is an oblique asymptote.

For end behaviour, we only need concentrate on:

`x^2/(3x)=x/3`

as `x->oo` , `x/3->oo`

as `x->-oo` , `x/3->-oo`

GRAPH: