text: # How do you find the vertical, horizontal or slant asymptotes for y=(x^2)/(2x^2-8)?
  
  Feb 26, 2016
  
  vertical asymptotes at x = ± 2
  horizontal asymptote at $y = \frac{1}{2}$
  
  #### Explanation:
  
  Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero.
  
  solve : 2x^2 - 8 = 0 → 2(x^2 - 4) = 0 → 2(x-2)(x+2) = 0
  
  equations of vertical asymptotes are x = ± 2
  
  Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0
  
  If the degree of the numerator and denominator are equal then the equation can be found by taking the ratio of leading coefficients.
  Here they are both of degree 2.
  
  $\Rightarrow y = \frac{1}{2}$
  
  Here is the graph of the function.
  graph{x^2/(2x^2-8) [-10, 10, -5, 5]}
explanation: The document contains a math question and its corresponding answer. The question asks how to find the vertical, horizontal, or slant asymptotes for the function \( y = \frac{x^2}{2x^2 - 8} \). The answer provides the vertical asymptotes at \( x = \pm 2 \) and the horizontal asymptote at \( y = \frac{1}{2} \), along with an explanation of the steps to find these asymptotes.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # How do you find the vertical, horizontal or slant asymptotes for y=(x^2)/(2x^2-8)?
  
  Feb 26, 2016
  
  vertical asymptotes at x = ± 2
  horizontal asymptote at $y = \frac{1}{2}$
  
  #### Explanation:
  
  Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let denominator equal zero.
  
  solve : 2x^2 - 8 = 0 → 2(x^2 - 4) = 0 → 2(x-2)(x+2) = 0
  
  equations of vertical asymptotes are x = ± 2
  
  Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0
  
  If the degree of the numerator and denominator are equal then the equation can be found by taking the ratio of leading coefficients.
  Here they are both of degree 2.
  
  $\Rightarrow y = \frac{1}{2}$
  
  Here is the graph of the function.
  graph{x^2/(2x^2-8) [-10, 10, -5, 5]}
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its corresponding answer. The question asks how to find the vertical, horizontal, or slant asymptotes for the function \( y = \frac{x^2}{2x^2 - 8} \). The answer provides the vertical asymptotes at \( x = \pm 2 \) and the horizontal asymptote at \( y = \frac{1}{2} \), along with an explanation of the steps to find these asymptotes.
  
  Label: question-and-answer