# How do you find the domain and range of y=(2x)/(x+9)?

Apr 9, 2018

$D : \left(- \infty , - 9\right) \cup \left(- 9 , \infty\right)$
$R : \left(- \infty , 2\right) \cup \left(2 , \infty\right)$

#### Explanation:

I know this is an extremely long answer, but hear me out.

First, to find the domain of a function, we must take note of any discontinuities that occur. In other words, we have to find impossibilities in the function. Most of the time, this will take the form of $x \div 0$ (it is impossible in mathematics to divide by 0 if you don't know). Discontinuities can either be removable or non removable.

Removable discontinuities are "holes" in the graph that are just a sudden break in the line, interrupting only one point. They are identified by a factor being present in both the numerator and denominator. For example, in the function
$y = \frac{{x}^{2} - 1}{x - 1}$

we can use the difference of squares to determine that

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

Here we now can observe that there is a factor of $\left(x - 1\right)$ in both the numerator and denominator. This creates a hole at the $x$ value of 1. In order to find the $y$ value of the point, we must cancel out the similar factors and substitute in the $x$ value of the point in for all occurrences of $x$ in the "revised" equation. Lastly, we solve for $y$, which will give us our $y$ coordinate of the "hole"

$y = x + 1 \to y = 1 + 1 \to y = 2$

Non removable discontinuities create vertical asymptotes in the graph that interrupt the points before and after the point that doesn't exist. This what the equation you stated concerns. In order to determine the location of such asymptotes. We will have to find any values of $x$ where the denominator can equal 0. In your equation, your denominator was:

$x + 9$

Using basic algebra, we can determine that the in order for the denominator to equal 0, $x$ must equal -9. -9, in this case, is the $x$ value of your vertical asymptote.

After finding all types of discontinuities in the graph, we can write our domain around them using our friend, the union sign: $\cup$.

$\left(- \infty , - 9\right) \cup \left(- 9 , \infty\right)$

For determining the range of the function, there are three rules which describe the end behavior of functions. However, there is one that applies to yours, it is, in a more casual way:

If the largest powers of the variables in the numerator and denominator are equal, then there is an asymptote at $y =$the division of the coefficients for those variables.

In terms of your equation, the powers of your largest power variables are equal, so I divide the coefficients of 2 and 1 to get $y = 2$. That is your horizontal asymptote. For most functions, it will not be crossed. Therefore, we can write the range around it:

$\left(- \infty , 2\right) \cup \left(2 , \infty\right)$