# How to find domain, asymptotes, holes, intercepts for f(x) = (x+3 )/( x^2 + 8x + 15)?

##### 1 Answer
Jun 8, 2018

Domain $\left\{x | x \in \mathbb{R} : x \ne - 3 \mathmr{and} x \ne - 5\right\}$

Range $\left\{f \left(x\right) | f \left(x\right) \in \mathbb{R} : f \left(x\right) \ne 0 \mathmr{and} f \left(x\right) \ne \frac{1}{2}\right\}$

#### Explanation:

$f \left(x\right) = \frac{x + 3}{{x}^{2} + 8 x + 15}$

$f \left(x\right) = \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)}$

since the factor $x + 3$ is cancelable that is a hole or removable discontinuity at x=-3.

the factor $x + 5$ does not cancel so that is a asymptote at $x = - 5$.

set $x = 0$ to solve y-intercept:

$f \left(x\right) = \frac{0 + 3}{\left(0 + 3\right) \left(0 + 5\right)}$

$f \left(x\right) = \frac{3}{\left(3\right) \left(5\right)} = \frac{1}{5}$

set $f \left(x\right) = 0$ and solve for roots:

$0 = \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)}$

$0 = \frac{1}{x + 5}$

$0 = 1$ so there are no $\mathbb{R}$ roots.

Finally, the domain is all real numbers except the hole and asymptote:

Domain $\left\{x | x \in \mathbb{R} : x \ne - 3 \mathmr{and} x \ne - 5\right\}$

Range is a little tricker, first:

$f \left(x\right) = \frac{\cancel{x + 3}}{\cancel{x + 3} \left(x + 5\right)} = \frac{1}{x + 5}$

so as $x \to \pm \infty , f \left(x\right) \to 0$, therefore there is a horizontal asymptote at 0.

We also need to remember the hole at -3:

as $x \to - 3 , f \left(x\right) \to \frac{1}{2}$

Range $\left\{f \left(x\right) | f \left(x\right) \in \mathbb{R} : f \left(x\right) \ne 0 \mathmr{and} f \left(x\right) \ne \frac{1}{2}\right\}$

graph{(-3+3 )/( x^2 + 8x + 15) [-10, 10, -5, 5]}