Question

How do you identify all asymptotes or holes for `f(x)=(x^2-5x+6)/(x^2-4x+3)`?

Answer 1

There is a hole at `x=3`
The vertical asymptote is `x=1`
The horizontal asymptote is `y=1`

Explanation:

Let's factorise the numerator and denominator

`x^2-5x+6=(x-3)(x-2)`

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

So,

`f(x)=(x^2-5x+6)/(x^2-4x+3)=(cancel(x-3)(x-2))/((x-1)cancel(x-3))`

`=(x-2)/(x-1)`

So, there is a hole at `x=3`

The domain of `f(x)` is `D_f(x)=RR-{1}`

As we cannot divide by `0`, `x!=1`

So, the vertical asymptote is `x=1`

To calculate the limits as `x->+-oo`, we take the terms of highest degree in the numerator and the denominator.

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

So, the horizontal asymptote is `y=1`

graph{(y-(x-2)/(x-1))(y-1)=0 [-11.25, 11.25, -5.625, 5.625]}