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

2 Answers
Jun 1, 2018

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

"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]}