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

2 Answers
Jun 1, 2018

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:
#1/(x^2-2x+1)= 1/((x-1)^2#
#(x-1)^2 != 0#
#(x-1) != 0#
#:.x!= 1 #
Therefore, Vertical Assymptopes is: #x= 1 #
Note finding the, Horizontal Assymptopes requires more logical thinking:
Since when #x->1# (x values approach 1) gives a large value and when #x->+-oo # gives a very small value, this means that the curve #f(x)=1/(x^2-2x+1)# has all
f(x) > 0 thefore this means that
Horizontal Assymptope: y=0

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Jun 1, 2018

Answer:

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

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]}