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

Answer 1

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

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]}
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Answer 2

Assymptopes:
Vertical X=1,
Horizontal Y=0

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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Answer 3

To find the asymptotes of ( f(x) = \frac{1}{x^2 - 2x + 1} ):

  1. Factor the denominator ( x^2 - 2x + 1 ) if possible.
  2. Determine if there are any vertical asymptotes by finding the values of ( x ) that make the denominator equal to zero. If there are any, they represent vertical asymptotes.
  3. Determine if there are any horizontal asymptotes by analyzing the behavior of the function as ( x ) approaches positive or negative infinity.

In this case:

  1. The denominator ( x^2 - 2x + 1 ) factors to ( (x - 1)^2 ).
  2. The denominator equals zero when ( x - 1 = 0 ), which gives ( x = 1 ). So, there is a vertical asymptote at ( x = 1 ).
  3. As ( x ) approaches positive or negative infinity, ( f(x) ) approaches zero. Therefore, there are no horizontal asymptotes.

So, the vertical asymptote is ( x = 1 ), and there are no horizontal asymptotes for the function ( f(x) = \frac{1}{x^2 - 2x + 1} ).

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Answer 4

To find the asymptotes for the function (f(x) = \frac{1}{x^2 - 2x + 1}):

  1. Factor the denominator (x^2 - 2x + 1).
  2. Identify any vertical asymptotes by determining where the denominator equals zero.
  3. Simplify or cancel any common factors in the numerator and denominator to determine if there are any holes in the graph.
  4. Identify any horizontal or slant asymptotes by analyzing the behavior of the function as (x) approaches positive or negative infinity.

For the given function (f(x) = \frac{1}{x^2 - 2x + 1}):

  1. Factoring the denominator yields ((x - 1)^2).
  2. Since the denominator equals zero when (x = 1), there is a vertical asymptote at (x = 1).
  3. There are no common factors to simplify, so there are no holes in the graph.
  4. As (x) approaches positive or negative infinity, the function approaches zero, so there are no horizontal or slant asymptotes.

Therefore, the function (f(x) = \frac{1}{x^2 - 2x + 1}) has a vertical asymptote at (x = 1), and there are no horizontal or slant asymptotes.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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