# How do you find the asymptotes for y= (3(x+5))/((x+1)(x+2))?

Feb 2, 2016

Horizontal asymptotes: $x = - 1 \textcolor{w h i t e}{\text{XX")andcolor(white)("XX}} x = - 2$
Vertical asymptote: $y = 0$

#### Explanation:

The denominator of $\frac{3 \left(x + 5\right)}{\left(x + 1\right) \left(x + 2\right)}$ will become $0$ if $x = - 1$ or $x = - 2$
This results in horizontal asymptotes at these locations (provided the numerator does not also go to $0$, which it doesn't in this case).

Since the degree of the denominator is greater than the degree of the numerator, as $x \rightarrow \infty , y \rightarrow 0$,
providing the vertical asymptote $y = 0$.
This could be made explicitly clear as
$\textcolor{w h i t e}{\text{XXX}} y = \frac{3 \left(x + 5\right)}{\left(x + 1\right) \left(x + 2\right)}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{3 x + 15}{{x}^{2} + 3 x + 2}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{3 + \frac{15}{x}}{x + 3 + \frac{2}{x}}$

As $x \rightarrow \infty$ $\frac{15}{x} \rightarrow 0$ and $\frac{2}{x} \rightarrow 0$

leaving ${\lim}_{x \rightarrow \infty} \frac{3}{x + 3}$ with the constant numerator and the constant added in the denominator becoming less significant as $x \rightarrow \infty$
So ${\lim}_{x \rightarrow \infty} \frac{3}{x + 3} \rightarrow {\lim}_{x \rightarrow \infty} \frac{1}{x} \rightarrow 0$