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

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Answer 1 · 2016-02-14T10:25:34

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

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

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

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

$x(x-1)-4(x-1)=0$ or $(x-1)(x-4)=0$

i.e. $x= 1 or 4$

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 $y=1$

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