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

Sep 20, 2017

$\text{vertical asymptote at } x = \frac{1}{4}$
$\text{horizontal asymptote at } y = \frac{1}{4}$

#### Explanation:

$\text{factorise and simplify}$

$y = \frac{\cancel{\left(x - 1\right)} \left(x - 4\right)}{\left(4 x - 1\right) \cancel{\left(x - 1\right)}} = \frac{x - 4}{4 x - 1}$

$\text{the removal of the factor "(x-1)" from the}$
$\text{numerator/denominator indicates a hole at x = 1}$

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve "4x-1=0rArrx=1/4" is the asymptote}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , y \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by x}$

$y = \frac{\frac{x}{x} - \frac{4}{x}}{\frac{4 x}{x} - \frac{1}{x}} = \frac{1 - \frac{4}{x}}{4 - \frac{1}{x}}$

as $x \to \pm \infty , y \to \frac{1 - 0}{4 - 0}$

$\Rightarrow y = \frac{1}{4} \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there is no slant asymptote.
graph{(x-4)/(4x-1) [-10, 10, -5, 5]}