How do you find the asymptotes for # f(x) = (2x-2) /( (x-1)(x^2 + x -1))#?

1 Answer
May 2, 2017

Answer:

#"vertical asymptotes at "x~~-1.62" and " x~~0.62#

#"horizontal asymptote at " y=0#

Explanation:

#"the first step is to factorise and simplify f(x)"#

#f(x)=(2cancel((x-1)))/(cancel((x-1))(x^2+x-1))=2/(x^2+x-1)#

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

#"solve " x^2+x-1=0" using the quadratic formula"#

#x=(-1+-sqrt(1+4))/2#

#rArrx=-1/2+-1/2sqrt5#

#rArrx~~-1.62" and " x~~0.62" are the asymptotes"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#f(x)=(2/x^2)/(x^2/x^2+x/x^2-1/x^2)=(2/x^2)/(1+1/x-1/x^2)#

as #xto+-oo,f(x)to0/(1+0-0)#

#rArry=0" is the asymptote"#
graph{2/(x^2+x-1) [-10, 10, -5, 5]}