# How do you find the vertical, horizontal and slant asymptotes of: f(x)= ( x-3) /( x^2-3x+2)?

Oct 11, 2016

$H . A \implies y = 0$

$V . A \implies x = 1$ and $x = 2$

#### Explanation:

Remember: You cannot have three asymptotes at the same time. If the Horizontal Asymptote exists, the Oblique/Slant 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,$\textcolor{v i o \le t}{\mathmr{if}}$:

$\textcolor{red}{n} \textcolor{g r e e n}{<} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A \implies y = 0}$
$\textcolor{red}{n} \textcolor{g r e e n}{=} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A \implies y = \frac{a}{b}}$
$\textcolor{red}{n} \textcolor{g r e e n}{>} \textcolor{b l u e}{m}$, $\textcolor{red}{H . A}$ $\textcolor{red}{\mathrm{do} e s n ' t}$ $\textcolor{red}{\exists}$

For this problem, $f \left(x\right) = \frac{x - 3}{{x}^{2} - 3 x + 2}$

$\textcolor{red}{n} \textcolor{g r e e n}{<} \textcolor{b l u e}{m}$, $H . A \implies y = 0$

$V . A \implies {x}^{2} - 3 x + 2 = 0$

Find the answer by using the tools that you already know. As for me, I always use $\Delta = {b}^{2} - 4 a c$, with $a = 1$, $b = - 3$ and $c = 2$

$\Delta = {\left(- 3\right)}^{2} - 4 \left(1\right) \left(2\right) = 1 \implies \sqrt{\Delta} = \pm 1$

${x}_{1} = \frac{- b + \sqrt{\Delta}}{2 a}$ and ${x}_{2} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{3 + 1}{2} = 2$ and ${x}_{2} = \frac{3 - 1}{2} = 1$

So, the $V . A$ are $x = 1$ and $x = 2$

Hope this helps :)