# How do you identify all asymptotes or holes and intercepts for f(x)=(x^2+1)/((x-1)(2x-4))?

Jul 26, 2017

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

#### 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.

$\text{solve } \left(x - 1\right) \left(2 x - 4\right) = 0$

$\Rightarrow x = 1 \text{ and "x=2" are the asymptotes}$

$\text{horizontal asymptotes occur as}$

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

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

$f \left(x\right) = \frac{{x}^{2} / {x}^{2} + \frac{1}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 - \frac{6 x}{x} ^ 2 + \frac{4}{x} ^ 2} = \frac{1 + \frac{1}{x} ^ 2}{2 - \frac{6}{x} + \frac{4}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 + 0}{2 - 0 + 0}$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$

Holes occur when there are common factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(x^2+1)/(2x^2-6x+4) [-20, 20, -10, 10]}