# What is the domain and range of  f(x) = (x+3) /( x^2 + 8x + 15)?

Jul 15, 2018

The domain is $x \in \left(- \infty , - 5\right) \cup \left(- 5 , + \infty\right)$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

The function is

$f \left(x\right) = \frac{x + 3}{{x}^{2} + 8 x + 15} = \frac{x + 3}{\left(x + 3\right) \left(x + 5\right)} = \frac{1}{x + 5}$

The denominator must be $\ne 0$

Therefore,

$x + 5 \ne 0$

$x \ne - 5$

The domain is $x \in \left(- \infty , - 5\right) \cup \left(- 5 , + \infty\right)$

To calculate the range, let

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

$y \left(x + 5\right) = 1$

$y x + 5 y = 1$

$y x = 1 - 5 y$

$x = \frac{1 - 5 y}{y}$

The denominator must be $\ne 0$

$y \ne 0$

The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

graph{1/(x+5) [-16.14, 9.17, -6.22, 6.44]}

Jul 15, 2018

Domain: $x \in \mathbb{R} , x \ne - 5$

Range: $y \in \mathbb{R} , y \ne 0$

#### Explanation:

We can factor the denominator as $\left(x + 3\right) \left(x + 5\right)$, since $3 + 5 = 8$, and $3 \cdot 5 = 15$. This leaves us with

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

We can cancel out common factors to get

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

The only value that will make our function undefined is if the denominator is zero. We can set it equal to zero to get

$x + 5 = 0 \implies x = - 5$

Therefore, we can say the domain is

$x \in \mathbb{R} , x \ne - 5$

To think about our range, let's go back to our original function

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

Let's think about the horizontal asymptote. Since we have a higher degree on the bottom, we know we have a HA at $y = 0$. We can show this graphically:

graph{(x+3)/((x+3)(x+8)) [-17.87, 2.13, -4.76, 5.24]}

Notice, our graph never touches the $x$-axis, which is consistent with having a horizontal asymptote at $y = 0$.

We can say our range is

$y \in \mathbb{R} , y \ne 0$

Hope this helps!