What are the asymptote(s) and hole(s) of # f(x) = (2x-4)/(x^2-3x+2)#?

1 Answer
May 1, 2017

Answer:

#"hole at " x=2#
#"vertical asymptote at " x=1#
#"horizontal asymptote at " y=0#

Explanation:

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

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

The removal of the factor (x - 2) indicates a hole at x = 2

The graph of #2/(x-1)# is the same as # (2x-4)/(x^2-3x+2)#

#"but without the hole"#

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 the highest power of x, that is #x^2#

#f(x)=(2/x^2)/(x^2/x^2-(3x)/x^2+2/x^2)=(2/x^2)/(1-3/x+2/x^2)#

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

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