# What are the asymptote(s) and hole(s), if any, of  f(x) =(x^2-2x+1)/(x*(x-2))?

Jun 2, 2018

See brief explanation

#### Explanation:

To find the vertical asymptotes, set the denominator - $x \left(x - 2\right)$ - equal to zero and solve. There are two roots, points where the function goes to infinity. If either of those two roots also have zero in the numerators, then they are a hole. But they don't, so this function has no holes.

To find the horizontal asymptote divide the leading term of the numerator - ${x}^{2}$ by the leading term of the denominator - also ${x}^{2}$. The answer is a constant. This is because when x goes to infinity (or minus infinity), the highest order terms become infinitely larger than any other terms.

Jun 2, 2018

$\text{vertical asymptotes at "x=0" and } x = 2$
$\text{horizontal asymptote at } y = 1$

#### 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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

$\text{solve } x \left(x - 2\right) = 0$

$x = 0 \text{ and "x=2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

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

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

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

$\text{Holes occur when a common factor is cancelled on the}$
$\text{numerator/denominator. This is not the case here hence}$
$\text{there are no holes}$
graph{(x^2-2x+1)/(x(x-2)) [-10, 10, -5, 5]}