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

##### 1 Answer

Nov 16, 2016

#### Answer:

The vertical asymptote is

The horizontal asymptote is

A hole when

No slant asymptote

#### Explanation:

Let's factorise the numerator and the denominator

Therefore,

So, we have a hole at

As we cannot divide by 0,

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

We take the term of highest coefficient.

So

graph{(y-(x-4)/(4x-1))(y-1/4)=0 [-7.024, 7.024, -3.507, 3.52]}