# How do you find the domain and range of f(x)=(x^2-x)/(x+1)?

Jul 11, 2017

See Below.

#### Explanation:

Before we do anything, let's see if we can simplify the function by factoring the numerator and denominator.

$\frac{{x}^{2} - x}{x + 1}$

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

The domain of a function is all of the $x$values (horizontal axis) that will give you a valid y-value (vertical axis) output.

Since the function given is a fraction, dividing by $0$ will not yield a valid $y$ value. To find the domain, let's set the denominator equal to zero and solve for $x$. The value(s) found will be excluded from the range of the function.

$x + 1 = 0$

$x = - 1$

So, the domain is all real numbers EXCEPT $- 1$. In set notation, the domain would be written as follows:

$\left(- \infty , - 1\right) \cup \left(- 1 , \infty\right)$

The range of a function is all of the $y$-values that it can take on. Let's graph the function and see what the range is.

graph{(x^2-x)/(x+1) [-10, 10, -5, 5]}