The image contains two rational functions for which we need to determine the domain, range, and asymptotes.

Function 1: $y = \frac{2x - 5}{2x + 5}$

1. Domain: The domain of a rational function is all real numbers except where the denominator is zero. $2x + 5 = 0 \quad \Rightarrow \quad x = -\frac{5}{2}$ So, the domain is $x \in \mathbb{R} \setminus \left\{ -\frac{5}{2} \right\}$.

2. Range: To find the range, we can use the fact that horizontal asymptotes can give insight into the behavior of the function as $x \to \pm \infty$. Since the degrees of the numerator and denominator are the same, the horizontal asymptote is determined by the ratio of the leading coefficients: $\text{Horizontal Asymptote: } y = \frac{2}{2} = 1$ However, since the function never actually reaches the horizontal asymptote but can approach any value above or below 1, the range is $y \in \mathbb{R} \setminus \{ 1 \}$.

3. Asymptotes:

   • Vertical Asymptote: Occurs where the denominator is zero, which is $x = -\frac{5}{2}$.
   • Horizontal Asymptote: $y = 1$.

Function 2: $y = \frac{x^2 - 1}{2x^2 - 3x - 2}$

1. Domain: We need to find where the denominator equals zero to exclude these values from the domain. $2x^2 - 3x - 2 = 0$ Solving this quadratic equation using the quadratic formula: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4}$ $x = \frac{3 \pm 5}{4} \quad \Rightarrow \quad x = 2 \quad \text{or} \quad x = -\frac{1}{2}$ So, the domain is $x \in \mathbb{R} \setminus \{ 2, -\frac{1}{2} \}$.

2. Range: Finding the range analytically for more complex rational functions is challenging, but we can use a numerical or graphical approach. Generally, it involves investigating the behavior of the function as $x \to \pm \infty$.

3. Asymptotes:

   • Vertical Asymptotes: These occur where the denominator is zero, which gives vertical asymptotes at $x = 2$ and $x = -\frac{1}{2}$.
   • Horizontal Asymptote: Since the degrees of the numerator and denominator are equal, the horizontal asymptote is determined by the ratio of the leading coefficients: $y = \frac{1}{2}$

Summary:

• For $y = \frac{2x - 5}{2x + 5}$:

  • Domain: $x \in \mathbb{R} \setminus \left\{ -\frac{5}{2} \right\}$
  • Range: $y \in \mathbb{R} \setminus \{ 1 \}$
  • Vertical Asymptote: $x = -\frac{5}{2}$
  • Horizontal Asymptote: $y = 1$

• For $y = \frac{x^2 - 1}{2x^2 - 3x - 2}$:

  • Domain: $x \in \mathbb{R} \setminus \{ 2, -\frac{1}{2} \}$
  • Range: Needs further analysis or graphical investigation.
  • Vertical Asymptotes: $x = 2$ and $x = -\frac{1}{2}$
  • Horizontal Asymptote: $y = \frac{1}{2}$

Let me know if you'd like further details or a graphical analysis!

Here are five related questions to expand your understanding:

1. How do we determine slant (oblique) asymptotes for rational functions?
2. What methods can be used to find the range of a rational function besides graphical approaches?
3. How do horizontal asymptotes affect the long-term behavior of a rational function?
4. What are removable discontinuities in rational functions, and how can they be identified?
5. How does the degree of the numerator and denominator influence the behavior of a rational function?

Tip: Always check for common factors in the numerator and denominator before determining vertical asymptotes. Sometimes a factor can cancel, resulting in a removable discontinuity instead of an asymptote.