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

Nov 16, 2016

The vertical asymptote is $x = \frac{1}{4}$
The horizontal asymptote is $y = \frac{1}{4}$
A hole when $x = 1$
No slant asymptote

#### Explanation:

Let's factorise the numerator and the denominator

${x}^{2} - 5 x + 4 = \left(x - 4\right) \left(x - 1\right)$

$4 {x}^{2} - 5 x + 1 = \left(4 x - 1\right) \left(x - 1\right)$

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, $x \ne \frac{1}{4}$

$x = \frac{1}{4}$ 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.

${\lim}_{x \to \pm \infty} y = {\lim}_{x \to \pm \infty} \frac{x}{4 x} = \frac{1}{4}$

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