How do you identify all asymptotes or holes for #f(x)=(x^2-16)/(x^2-5x+4)#?

1 Answer
Jan 16, 2017

Answer:

hole at x = 4
vertical asymptote at x = 1
horizontal asymptote at y = 1

Explanation:

Factorise and simplify f(x)

#f(x)=(cancel((x-4))(x+4))/(cancel((x-4))(x-1))=(x+4)/(x-1)#

We have removed a duplicate factor from the numerator/denominator of f(x). This indicates there is a hole at x = 4
This hole is not in the simplified version of f(x).

The denominator of f(x) cannot be zero as this would make f(x) 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.

solve : #x-1=0rArrx=1" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"#

divide terms on numerator/denominator by x

#f(x)=(x/x+4/x)/(x/x-1/x)=(1+4/x)/(1-1/x)#

as #xto+-oo,f(x)to(1+0)/(1-0)#

#rArry=1" is the asymptote"#
graph{(x+4)/(x-1) [-10, 10, -5, 5]}