Question

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

Answer 1

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

Answer 2

`"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]}