# How do you find the asymptotes for f(x) =(x^2 - 16)/(x^2- 5x + 4)?

Feb 14, 2016

Vertical asymptotes are $x = 1$ and $x = 4$ and horizontal asymptote is $y = 1$

#### Explanation:

The vertical asymptotes are given by restrictions on the domain and hence come from the zeroes of the denominator.

Putting ${x}^{2} - 5 x + 4 = 0$ ad solving this will give domain

${x}^{2} - 5 x + 4 = 0$ i.e. ${x}^{2} - x - 4 x + 4 = 0$ or

$x \left(x - 1\right) - 4 \left(x - 1\right) = 0$ or $\left(x - 1\right) \left(x - 4\right) = 0$

i.e. $x = 1 \mathmr{and} 4$

Hence $x$ can take all values other than $1$ and $4$

and vertical asymptotes are $x = 1$ and $x = 4$.

Since, numerator and denominator of the function are of same degree, horizontal asymptote is found by dividing the leading terms

i.e. ${x}^{2} / {x}^{2}$ i.e. $1$.

Hence horizontal asymptote is $y = 1$