Math Problem Statement

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).

Solution

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=2x52x+5y = \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=02x + 5 = 0 Solving for xx: x=52x = -\frac{5}{2} Thus, the domain is: x(,52)(52,)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=52x = -\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=22=1y = \frac{2}{2} = 1 Thus, the horizontal asymptote is: y=1y = 1

Range:

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


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

Domain:

To find the domain, we set the denominator equal to zero and solve for xx: 2x23x2=02x^2 - 3x - 2 = 0 Using the quadratic formula: x=(3)±(3)24(2)(2)2(2)x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-2)}}{2(2)} x=3±9+164=3±254=3±54x = \frac{3 \pm \sqrt{9 + 16}}{4} = \frac{3 \pm \sqrt{25}}{4} = \frac{3 \pm 5}{4} Thus, the solutions are: x=2andx=12x = 2 \quad \text{and} \quad x = -\frac{1}{2} So the domain is all real numbers except x=2x = 2 and x=12x = -\frac{1}{2}: x(,12)(12,2)(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=2andx=12x = 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=12y = \frac{1}{2} Thus, the horizontal asymptote is: y=12y = \frac{1}{2}

Range:

Since the horizontal asymptote is y=12y = \frac{1}{2}, the range is: [

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Math Problem Analysis

Mathematical Concepts

Rational Functions
Domain and Range
Asymptotes

Formulas

Domain: Set the denominator ≠ 0
Vertical Asymptote: Set the denominator = 0
Horizontal Asymptote: Compare degrees of numerator and denominator

Theorems

Quadratic formula for solving denominators
Degree-based comparison for horizontal asymptotes

Suitable Grade Level

Grades 10-12