# Question #ef8d7

Aug 21, 2017

$\text{vertical asymptote at } x = 2$
$\text{slant asymptote } y = x + 2$

#### 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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

$\text{solve "x-2=0rArrx=2" is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator $\le$ the degree of the denominator. This is not the case here hence there are no horizontal asymptotes.

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is the case here hence there is a slant asymptote.

$\text{dividing out gives}$

$x \left(x - 2\right) + 2 \left(x - 2\right) + 5$

$\Rightarrow f \left(x\right) = \frac{{x}^{2} + 1}{x - 2} = x + 2 + \frac{5}{x - 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to x + 2$

$\Rightarrow y = x + 2 \text{ is the asymptote}$
graph{(x^2+1)/(x-2) [-40, 40, -20, 20]}