# How do you find the domain and range of f(x)= sqrt(x+6)/(x-5)?

Jul 27, 2015

Domain : the fraction and the root give limitations:

#### Explanation:

$x \ge - 6$ or the argument under the square root will be negative.
$x \ne 5$ or the denominator will be $= 0$

Range : when $x = - 6 \to f \left(x\right) = 0$, but on the right side of the two-armed graph $x = 0$ is a horizontal asymtote.
${\lim}_{x \to \infty} f \left(x\right) = 0$
As $x$ nears $5$, $f \left(x\right)$ gets very large:
${\lim}_{x \to 5 -} f \left(x\right) = - \infty$ and ${\lim}_{x \to 5 +} f \left(x\right) = + \infty$

So there are no restrictions on the range.
graph{sqrt(x+6)/(x-5) [-5.04, 27.01, -8.67, 7.37]}