Math Problem Statement

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

Solution

The image contains two rational functions for which you need to determine the domain, range, and asymptotes. Here are the functions:

  1. y=2x52x+5y = \frac{2x - 5}{2x + 5}
  2. y=x212x23x2y = \frac{x^2 - 1}{2x^2 - 3x - 2}

Let's break them down one by one.

1. y=2x52x+5y = \frac{2x - 5}{2x + 5}

  • Domain: The domain is all real numbers except where the denominator is zero. Solve 2x+5=02x + 5 = 0: 2x+5=0    x=522x + 5 = 0 \implies x = -\frac{5}{2} So, the domain is xRx \in \mathbb{R}, x52x \neq -\frac{5}{2}.

  • Vertical Asymptote: A vertical asymptote occurs where the denominator is zero. In this case, there is a vertical asymptote at x=52x = -\frac{5}{2}.

  • Horizontal Asymptote: Compare the degrees of the numerator and the denominator. Since both are linear (degree 1), the horizontal asymptote is determined by the ratio of the leading coefficients: y=22=1y = \frac{2}{2} = 1 Thus, the horizontal asymptote is y=1y = 1.

  • Range: The range of the function excludes the horizontal asymptote y=1y = 1, so the range is yR,y1y \in \mathbb{R}, y \neq 1.


2. y=x212x23x2y = \frac{x^2 - 1}{2x^2 - 3x - 2}

  • Domain: The domain excludes values where the denominator is zero. Factor the denominator: 2x23x2=(2x+1)(x2)2x^2 - 3x - 2 = (2x + 1)(x - 2) So, the domain is all real numbers except x=12x = -\frac{1}{2} and x=2x = 2.

  • Vertical Asymptotes: Vertical asymptotes occur where the denominator is zero. From the factored form, we get vertical asymptotes at: x=12,x=2x = -\frac{1}{2}, \quad x = 2

  • Horizontal Asymptote: Since the degrees of the numerator and denominator are the same (degree 2), the horizontal asymptote is determined by the ratio of the leading coefficients: y=12y = \frac{1}{2} So, the horizontal asymptote is y=12y = \frac{1}{2}.

  • Range: The range excludes the horizontal asymptote y=12y = \frac{1}{2}, so the range is yR,y12y \in \mathbb{R}, y \neq \frac{1}{2}.


Would you like more details or explanations on any of these points?

Related Questions:

  1. How do you determine slant (oblique) asymptotes for rational functions?
  2. What is the general method to find the domain of any rational function?
  3. Can a rational function have both a horizontal and slant asymptote?
  4. What are the conditions for a rational function to have no horizontal asymptote?
  5. How do you determine the behavior of a function near its vertical asymptote?

Tip: Always check if a rational function can be simplified before analyzing its asymptotes and other characteristics.

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Domain and Range
Asymptotes

Formulas

Domain: Set of all possible x-values except where the denominator is 0.
Vertical Asymptotes: Occur where the denominator equals 0.
Horizontal Asymptotes: Determined by the degrees of the numerator and denominator.

Theorems

Rational Function Theorem
Asymptote Theorem

Suitable Grade Level

Grades 10-12