How do you find the asymptotes for #f(x)= 1/(x^2-2x+1)#?
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.
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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:
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
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To find the asymptotes of ( f(x) = \frac{1}{x^2 - 2x + 1} ):
- Factor the denominator ( x^2 - 2x + 1 ) if possible.
- 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.
- Determine if there are any horizontal asymptotes by analyzing the behavior of the function as ( x ) approaches positive or negative infinity.
In this case:
- The denominator ( x^2 - 2x + 1 ) factors to ( (x - 1)^2 ).
- The denominator equals zero when ( x - 1 = 0 ), which gives ( x = 1 ). So, there is a vertical asymptote at ( x = 1 ).
- 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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To find the asymptotes for the function (f(x) = \frac{1}{x^2 - 2x + 1}):
- Factor the denominator (x^2 - 2x + 1).
- Identify any vertical asymptotes by determining where the denominator equals zero.
- Simplify or cancel any common factors in the numerator and denominator to determine if there are any holes in the graph.
- 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}):
- Factoring the denominator yields ((x - 1)^2).
- Since the denominator equals zero when (x = 1), there is a vertical asymptote at (x = 1).
- There are no common factors to simplify, so there are no holes in the graph.
- 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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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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