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: [