# How do you find vertical, horizontal and oblique asymptotes for (x^2-x+1)/(x-3)?

Dec 15, 2016

The vertical asymptote is $x = 3$
No horizontal asymptote.
The oblique asymptote is $y = x + 2$

#### Explanation:

As you cannot divide by $0$, $x \ne 3$

Therefore,

The vertical asymptote is $x = 3$

As the degree of the numerator is $>$ than the degree of the denominator, we expect an oblique asymptote.

Let's do a long division

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

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

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

$\textcolor{w h i t e}{a a a a a a a}$$+ 2 x - 6$

$\textcolor{w h i t e}{a a a a a a a}$$+ 0 + 7$

Therefore,

$\frac{{x}^{2} - x + 1}{x - 3} = x + 2 + \frac{7}{x - 3}$

The oblique asymptote is $y = x + 2$

There is no horizontal asymptote. When $x \to \pm \infty$,

$f \left(x\right) \to \pm \infty$

graph{(y-(x^2-x+1)/(x-3))(y-x-2)=0 [-41.1, 41.1, -20.56, 20.56]}