Question

How do you find the asymptotes for `y=(x^2-5x+4)/ (4x^2-5x+1)`?

Answer 1

The vertical asymptote is `x=1/4`
The horizontal asymptote is `y=1/4`
A hole when `x=1`
No slant asymptote

Explanation:

Let's factorise the numerator and the denominator

`x^2-5x+4=(x-4)(x-1)`

`4x^2-5x+1=(4x-1)(x-1)`

Therefore,

`y=(x^2-5x+4)/(4x^2-5x+1)=((x-4)cancel(x-1))/((4x-1)cancel(x-1)`

So, we have a hole at `x=1`

As we cannot divide by 0, `x!=1/4`

`x=1/4` is a vertical asymptote

The degree of the numerator = the degree of the denominator, we don't have a slant asymptote.

We take the term of highest coefficient.

`lim_(x->+-oo)y=lim_(x->+-oo)x/(4x)=1/4`

So `y=1/4` is a horizontal asymptote
graph{(y-(x-4)/(4x-1))(y-1/4)=0 [-7.024, 7.024, -3.507, 3.52]}