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

Oct 23, 2016

A vertical asymptote is $x = - 1$
And the oblique asymptote is $y = x - 1$

Explanation:

As $f \left(x\right) = \frac{{x}^{2} + 1}{x + 1}$
Since we cannot divide by $0$
The degree of the binomial of the numerator $>$the degree of the binomial of the denominator, so we would expect an oblique asymptote
Let 's do a long division

${x}^{2}$$\textcolor{w h i t e}{a a a a a a a a}$$+ 1$$\textcolor{w h i t e}{a a a a a a}$∣$x + 1$
${x}^{2} + x$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$∣$x - 1$
$\textcolor{w h i t e}{a a a a}$$0 - x + 1$
$\textcolor{w h i t e}{a a a a a a}$$- x - 1$
$\textcolor{w h i t e}{a a a a a a a a}$$0 + 2$

And finally we get
$f \left(x\right) = x - 1 + \frac{2}{x + 1}$
So the oblique asymptote is $y = x - 1$

$\lim f \left(x\right) = - \infty$
$x \to - \infty$

$\lim f \left(x\right) = + \infty$
$x \to + \infty$