# What are the asymptote(s) and hole(s), if any, of # f(x) =(x^2-2x+1)/(x*(x-2))#?

##### 2 Answers

See brief explanation

#### Explanation:

To find the vertical asymptotes, set the denominator -

To find the horizontal asymptote divide the leading term of the numerator -

#### Explanation:

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(x-2)=0#

#x=0" and "x=2" 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)=(x^2/x^2-(2x)/x^2+1/x^2)/(x^2/x^2-(2x)/x^2)=(1-2/x+1/x^2)/(1-2/x)#

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

#y=1" is the asymptote"#

#"Holes occur when a common factor is cancelled on the"#

#"numerator/denominator. This is not the case here hence"#

#"there are no holes"#

graph{(x^2-2x+1)/(x(x-2)) [-10, 10, -5, 5]}