# How do you find vertical, horizontal and oblique asymptotes for (x^2-5x+6)/(x-4)?

Mar 7, 2017

The vertical asymptote is $x = 4$
The oblique asymptote is $y = x - 1$
No horizontal asymptote

#### Explanation:

As you cannot divide by $0$, $\implies$, $x \ne 4$

The vertical asymptote is $x = 4$

The degree of the numerator is $>$ than the degree of the denominator, there is an oblique asymptote.

Let $f \left(x\right) = \frac{{x}^{2} - 5 x + 6}{x - 4}$

Let's do a long division

$\textcolor{w h i t e}{a a a a}$${x}^{2} - 5 x + 6$$\textcolor{w h i t e}{a a a a}$$|$$x - 4$

$\textcolor{w h i t e}{a a a a}$${x}^{2} - 4 x$$\textcolor{w h i t e}{a a a a a a a a}$$|$$x - 1$

$\textcolor{w h i t e}{a a a a a}$$0 - x + 6$

$\textcolor{w h i t e}{a a a a a a a}$$- x + 4$

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

Therefore,

$f \left(x\right) = \left(x - 1\right) + \frac{2}{x - 4}$

${\lim}_{x \to - \infty} \left(f \left(x\right) - \left(x - 1\right)\right) = {\lim}_{x \to - \infty} \frac{2}{x - 4} = {0}^{-}$

${\lim}_{x \to + \infty} \left(f \left(x\right) - \left(x - 1\right)\right) = {\lim}_{x \to + \infty} \frac{2}{x - 4} = {0}^{+}$

The oblique asymptote is $y = x - 1$

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)(y-100(x-4))=0 [-18.3, 17.74, -6.74, 11.28]}