# How do you find the vertical, horizontal or slant asymptotes for f(x) = (x^2 + x + 3) /( x-1)?

Jan 8, 2017

The vertical asymptote is $x = 1$
The slant asymptote is $y = x + 2$
No horizontal asymptote

#### Explanation:

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{1\right\}$

As we cannot divide by $0$, $x \ne 1$

The vertical asymptote is $x = 1$

As the degree of the numerator is $>$ the degree of the denominator, we have a slant asymptote.

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} + x + 3$$\textcolor{w h i t e}{a a a a}$∣$x - 1$

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

$\textcolor{w h i t e}{a a a a}$$0 + 2 x + 3$

$\textcolor{w h i t e}{a a a a a a}$$+ 2 x - 2$

$\textcolor{w h i t e}{a a a a a a a}$$+ 0 + 5$

Therefore,

$f \left(x\right) = \frac{{x}^{2} + x + 3}{x - 1} = \left(x + 2\right) + \frac{5}{x - 1}$

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(x + 2\right)\right) = {\lim}_{x \to - \infty} \frac{5}{x} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(x + 2\right)\right) = {\lim}_{x \to + \infty} \frac{5}{x} = {0}^{+}$

The slant asymptote is $y = x + 2$

graph{(y-(x^2+x+3)/(x-1))(y-x-2)=0 [-32.48, 32.44, -16.2, 16.32]}