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

##### 2 Answers

#### Answer:

Assymptopes:

Vertical X=1,

Horizontal Y=0

#### Explanation:

The Assymptopes of this function is found when the

Denominator expression isn't equal to zero:

Vertical Asymptotes given when denominator isn't equal to zero:

Therefore, Vertical Assymptopes is:

Note finding the, Horizontal Assymptopes requires more logical thinking:

Since when

f(x) > 0 thefore this means that

Horizontal Assymptope: y=0

#### Answer:

#### 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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

#"solve "x^2-2x+1=0rArr(x-1)^2=0#

#x=1" is the asymptote"#

#"horizontal asymptotes occur as"#

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

#"divide all terms on numerator/denominator by the "#

#"highest power of x that is "x^2#

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

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

#y=0" is the asymptote"#

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