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#