# How do you find the vertical, horizontal or slant asymptotes for (x + 2)/( x + 3)?

May 28, 2016

$\frac{x + 2}{x + 3}$ has a vertical asymptote $x = - 3$ and a horizontal asymptote $y = 1$

#### Explanation:

$f \left(x\right) = \frac{x + 2}{x + 3}$

When $x = - 3$, the denominator is zero and the numerator non-zero, so $f \left(- 3\right)$ is undefined and $f \left(x\right)$ has a vertical asymptote there.

Note that:

$\frac{x + 2}{x + 3} = \frac{\left(x + 3\right) - 1}{x + 3} = 1 - \frac{1}{x + 3}$

As $x \to \pm \infty$, the expression $- \frac{1}{x + 3} \to 0$

So $f \left(x\right) \to 1 + 0 = 1$ as $x \to \pm \infty$ and $f \left(x\right)$ has a horizontal asymptote $y = 1$.

graph{(x+2)/(x+3) [-13.21, 6.79, -4.24, 5.76]}