# What is the domain and range of y= (4x^2 - 9) / ((2x+3)(x+1))?

May 5, 2018

See below.

#### Explanation:

Notice:

$4 {x}^{2} - 9$ is the difference of two squares. This can be expressed as:

$4 {x}^{2} - 9 = \left(2 x + 3\right) \left(2 x - 3\right)$

Substituting this in numerator:

$\frac{\left(2 x + 3\right) \left(2 x - 3\right)}{\left(2 x + 3\right) \left(x + 1\right)}$

Canceling like factors:

$\frac{\cancel{\left(2 x + 3\right)} \left(2 x - 3\right)}{\cancel{\left(2 x + 3\right)} \left(x + 1\right)} = \frac{2 x - 3}{x + 1}$

We notice that for $x = - 1$ the denominator is zero. This is undefined, so our domain will be all real numbers $\boldsymbol{x}$ $x \ne - 1$

We can express this in set notation as:

$\left\{x \in \mathbb{R} | x \ne - 1\right\}$

or in interval notation:

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

To find the range:

We know the function is undefined for $x = - 1$, therefore the line $x = - 1$ is a vertical asymptote. The function will go to $\pm \infty$ at this line.

We now see what happens as $x \to \pm \infty$

Divide $\frac{2 x - 3}{x + 1}$ by $x$

$\frac{\frac{2 x}{x} - \frac{3}{x}}{\frac{x}{x} + \frac{1}{x}} = \frac{2 - \frac{3}{x}}{1 + \frac{1}{x}}$

as: $x \to \pm \infty$ $\setminus \setminus \setminus \setminus \setminus \frac{2 - \frac{3}{x}}{1 + \frac{1}{x}} = \frac{2 - 0}{1 + 0} = 2$

This shows the line $y = 2$ is a horizontal asymptote. The function can't therefore ever equal 2.

so the range can be expressed as:

$\left\{y \in \mathbb{R} | y \ne 2\right\}$

or

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

This can be seen from the graph of the function:

graph{(2x-3)/(x+1) [-32.48, 32.44, -16.23, 16.25]}