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

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
${\left(x - 1\right)}^{2} \ne 0$
$\left(x - 1\right) \ne 0$
$\therefore x \ne 1$
Therefore, Vertical Assymptopes is: $x = 1$
Note finding the, Horizontal Assymptopes requires more logical thinking:
Since when $x \to 1$ (x values approach 1) gives a large value and when $x \to \pm \infty$ gives a very small value, this means that the curve $f \left(x\right) = \frac{1}{{x}^{2} - 2 x + 1}$ has all
f(x) > 0 thefore this means that
Horizontal Assymptope: y=0

Jun 1, 2018

$\text{vertical asymptote at } x = 1$
$\text{horizontal asymptote at } y = 0$

#### 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.

$\text{solve } {x}^{2} - 2 x + 1 = 0 \Rightarrow {\left(x - 1\right)}^{2} = 0$

$x = 1 \text{ is the asymptote}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide all terms on numerator/denominator by the }$
$\text{highest power of x that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{\frac{1}{x} ^ 2}{1 - \frac{2}{x} + \frac{1}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0}{1 - 0 + 0}$

$y = 0 \text{ is the asymptote}$
graph{1/(x^2-2x+1) [-10, 10, -5, 5]}