# How do you find the Vertical, Horizontal, and Oblique Asymptote given y=(x^2-5x+4)/ (4x^2-5x+1)?

Oct 29, 2016

The vertical asymptote is $x = \frac{1}{4}$
and the horizontal asymptote is $y = \frac{1}{4}$

#### Explanation:

Let's factorise the denominator and numerator
$4 {x}^{2} - 5 x + 1 = \left(4 x - 1\right) \left(x - 1\right)$
${x}^{2} - 5 x + 4 = \left(x - 4\right) \left(x - 1\right)$

Therefore $y = \frac{{x}^{2} - 5 x + 4}{4 {x}^{2} - 5 x + 1} = \frac{\left(x - 4\right) \cancel{x - 1}}{\left(4 x - 1\right) \cancel{x - 1}}$
so $y = \frac{x - 4}{4 x - 1}$

As we cannot divide by 0, so $x \ne \frac{1}{4}$
Therefore the vertical asymptotes are $x = \frac{1}{4}$

As the degree of the numerator and denominator are the same, there is no oblique asymptote.

Let's find the limit of y as $x \to \pm \infty$

Limit $y = \frac{x}{4 x} = \frac{1}{4}$
$x \to \pm \infty$
So $y = \frac{1}{4}$is a horizontal asymptote
graph{(x^2-5x+4)/(4x^2-5x+1) [-5.546, 5.55, -2.773, 2.774]}