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You have provided two rational functions for which we are tasked with analyzing the **domain**, **range**, and **asymptotes**. Let's go through each one step by step.

### 1. \( y = \frac{2x - 5}{2x + 5} \)

#### Domain:
The domain of a rational function is all real numbers except where the denominator is zero. 
- Set the denominator equal to zero:
  \[
  2x + 5 = 0
  \]
  Solving for \(x\):
  \[
  x = -\frac{5}{2}
  \]
Thus, the domain is:
\[
x \in (-\infty, -\frac{5}{2}) \cup (-\frac{5}{2}, \infty)
\]

#### Vertical Asymptote:
A vertical asymptote occurs where the denominator equals zero (which would make the function undefined). From the domain, we know this occurs at:
\[
x = -\frac{5}{2}
\]

#### Horizontal Asymptote:
The horizontal asymptote is found by comparing the degrees of the numerator and denominator. Both the numerator and denominator have degree 1, so we divide the leading coefficients:
\[
y = \frac{2}{2} = 1
\]
Thus, the horizontal asymptote is:
\[
y = 1
\]

#### Range:
Since the function has a horizontal asymptote at \(y = 1\), the range excludes this value. The range is:
\[
y \in (-\infty, 1) \cup (1, \infty)
\]

---

### 2. \( y = \frac{x^2 - 1}{2x^2 - 3x - 2} \)

#### Domain:
To find the domain, we set the denominator equal to zero and solve for \(x\):
\[
2x^2 - 3x - 2 = 0
\]
Using the quadratic formula:
\[
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)}
\]
\[
x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4}
\]
Thus, the solutions are:
\[
x = 2 \quad \text{and} \quad x = -\frac{1}{2}
\]
So the domain is all real numbers except \(x = 2\) and \(x = -\frac{1}{2}\):
\[
x \in (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, 2) \cup (2, \infty)
\]

#### Vertical Asymptotes:
Vertical asymptotes occur where the denominator equals zero, so:
\[
x = 2 \quad \text{and} \quad x = -\frac{1}{2}
\]
These are the vertical asymptotes.

#### Horizontal Asymptote:
The degrees of both the numerator and denominator are 2, so we again compare the leading coefficients:
\[
y = \frac{1}{2}
\]
Thus, the horizontal asymptote is:
\[
y = \frac{1}{2}
\]

#### Range:
Since the horizontal asymptote is \(y = \frac{1}{2}\), the range is:
\[


Analyze the domain, range, and asymptotes of the following rational functions: y = (2x - 5)/(2x + 5) and y = (x^2 - 1)/(2x^2 - 3x - 2).

This lesson covers how to find the domain, range, and asymptotes of two rational functions: y = (2x - 5)/(2x + 5) and y = (x^2 - 1)/(2x^2 - 3x - 2). It explains step-by-step how to determine where the function is undefined, identify vertical and horizontal asymptotes, and specify the range.

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