# How do you find vertical, horizontal and oblique asymptotes for (x^2 - 5x + 6)/ (x - 3)?

Oct 7, 2016

Remember: You cannot have three asymptotes at the same time. If the Horizontal Asymptote exists, the Oblique Asymptote doesn't exist. Also, $\textcolor{red}{H . A}$ $\textcolor{red}{f o l l o w}$ $\textcolor{red}{t h r e e}$ $\textcolor{red}{p r o c e \mathrm{du} r e s} .$ Let's say $\textcolor{red}{n}$ = highest degree of the numerator and $\textcolor{b l u e}{m}$ = highest degree of the denominator,$\textcolor{v i o \le t}{\mathmr{if}}$:
$\textcolor{red}{n} \textcolor{g r e e n}{<} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A \implies y = 0}$
$\textcolor{red}{n} \textcolor{g r e e n}{=} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A \implies y = \frac{a}{b}}$
$\textcolor{red}{n} \textcolor{g r e e n}{>} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A}$ $\textcolor{red}{\mathrm{do} e s n ' t}$ $\textcolor{red}{\exists}$

Here, $\frac{{x}^{2} - 5 x + 6}{x - 3}$

$V . A : x - 3 = 0 \implies x = 3$
$O . A : y = x - 2$

Please, take a look at the picture.

The oblique/slant asymptote is found by dividing the numerator by the denominator (long division.)

Notice that I did not do the long division in the way some people excepted me to. I always use the "French" way because I've never understood the English way, also I'm a francophone :) but it is the same answer.

Hope this helps :)