# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= 1/(x^2-2x+1)?

Jul 30, 2016

vertical asymptote x = 1
horizontal asymptote y = 0

#### Explanation:

The denominator of f(x) cannot be zero as this would be 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: ${x}^{2} - 2 x + 1 = 0 \Rightarrow {\left(x - 1\right)}^{2} = 0 \Rightarrow x = 1$

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

Horizontal asymptotes occur as

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

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

$\frac{\frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2 + \frac{1}{x} ^ 2} = \frac{\frac{1}{x} ^ 2}{1 - \frac{2}{x} + \frac{1}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 0, denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{(1)/(x^2-2x+1) [-10, 10, -5, 5]}