Math Problem Statement

y=(x^2+x)/(x^2-1) find the asymptote

Solution

To find the asymptotes of the function y=x2+xx21y = \frac{x^2 + x}{x^2 - 1}, we'll consider both vertical and horizontal asymptotes.

1. Vertical Asymptotes

Vertical asymptotes occur where the denominator of the function is zero, provided that the numerator is not zero at those points.

The denominator is: x21=0x^2 - 1 = 0

Solve for xx: x2=1x^2 = 1 x=±1x = \pm 1

These values correspond to potential vertical asymptotes. We check the numerator at these points: Numerator: x2+x evaluated at x=1 and x=1\text{Numerator: } x^2 + x \text{ evaluated at } x = 1 \text{ and } x = -1 At x=1:12+1=2\text{At } x = 1: 1^2 + 1 = 2 At x=1:(1)2+(1)=0\text{At } x = -1: (-1)^2 + (-1) = 0

The numerator is zero at x=1x = -1, so x=1x = -1 is a removable discontinuity, not a vertical asymptote. However, at x=1x = 1, the numerator is not zero, so x=1x = 1 is a vertical asymptote.

Vertical asymptote: x=1x = 1

2. Horizontal Asymptote

To find the horizontal asymptote, we examine the behavior of the function as xx approaches infinity.

The degrees of the numerator and denominator are both 2 (since the highest power of xx in both the numerator and denominator is x2x^2). For rational functions, if the degrees are the same, the horizontal asymptote is given by the ratio of the leading coefficients.

The leading coefficient of the numerator x2+xx^2 + x is 1, and the leading coefficient of the denominator x21x^2 - 1 is also 1. Therefore, the horizontal asymptote is:

y=11=1y = \frac{1}{1} = 1

Horizontal asymptote: y=1y = 1

Summary

  • Vertical asymptote: x=1x = 1
  • Horizontal asymptote: y=1y = 1

Would you like more details on this process or have any questions?

Here are 5 related questions:

  1. What is the behavior of the function near the vertical asymptote x=1x = 1?
  2. How do you determine the presence of slant asymptotes?
  3. What happens to the function as xx approaches the removable discontinuity x=1x = -1?
  4. Can the function y=x2+xx21y = \frac{x^2 + x}{x^2 - 1} have oblique asymptotes?
  5. How would the asymptotes change if the numerator had a higher degree than the denominator?

Tip: When analyzing asymptotes, always start by factoring the numerator and denominator to check for common factors (removable discontinuities).

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Math Problem Analysis

Mathematical Concepts

Algebra
Rational Functions
Asymptotes

Formulas

Vertical Asymptote: Set denominator to 0
Horizontal Asymptote: Compare degrees of numerator and denominator

Theorems

Theorem on Rational Functions Asymptotes
Factorization and Removable Discontinuities

Suitable Grade Level

Grades 10-12