# How do you find vertical, horizontal and oblique asymptotes for (x^2+1)/(x+1)?

Aug 20, 2016

$\frac{{x}^{2} + 1}{x + 1}$ has a vertical asymptote $x = - 1$ and an oblique asymptote $y = x - 1$

#### Explanation:

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

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

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

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

As $x \to \pm \infty$, $\frac{2}{x + 1} \to 0$

So $f \left(x\right)$ has an oblique asymptote $y = \left(x - 1\right)$

When $x = - 1$ the denominator $\left(x + 1\right)$ is zero, but the numerator $\left({x}^{2} + 1\right)$ is non-zero.

Hence $f \left(x\right)$ has a vertical asymptote at $x = - 1$

graph{(y-(x^2+1)/(x+1))(y-x+1) = 0 [-20, 20, -10, 10]}