# How do you find vertical, horizontal and oblique asymptotes for (x+6)/(x^2-9x+18)?

Dec 14, 2016

The vertical asymptotes are $x = 3$ and $x = 6$
No oblique asymptote.
The horizontal asymptote is $y = 0$

#### Explanation:

Let's factorise the denominator

${x}^{2} - 9 x + 18 = \left(x - 6\right) \left(x - 3\right)$

Let

$f \left(x\right) = \frac{x + 6}{{x}^{2} - 9 x + 18} = \frac{x + 6}{\left(x - 6\right) \left(x - 3\right)}$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{3 , 6\right\}$

As we cannot divide by $0$, $x \ne 3$ and $x \ne 6$

The vertical asymptotes are $x = 3$ and $x = 6$

As the degree of the numerator is $<$ than the degree of the denominator, there is no oblique asymptote.

To calculate the limits as $x \to \pm \infty$, we take the highest coefficients in the numerator and denominator

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

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

The horizontal asymptote is $y = 0$

graph{(y-(x+6)/(x^2-9x+18))(y)=0 [-22.8, 22.8, -11.4, 11.4]}

Dec 14, 2016

Vertical asymototes x=6 and x=3

Horizontal asymptote y=0

#### Explanation:

$f \left(x\right) = \frac{x + 6}{{x}^{2} - 9 x + 18}$. Factorise the denominator:

$= \frac{x + 6}{\left(x - 6\right) \left(x - 3\right)}$

f(x) has no oblique asymptotes. It has two vertical asymptotes given by x-6=0 and x-3=0.

Since the degree of numerator is less than that of denominator, divide both by x , giving $f \left(x\right) = \frac{1 + \frac{6}{x}}{x - 9 + \frac{18}{x}}$ . Now find the limit as x$\to \infty$ giving f(x)=y=0, which is a horizontal asymptotes