text: # How do you find the Vertical, Horizontal, and Oblique Asymptote given (x^2+1)/ (3x-2x^2)?
  
  Oct 17, 2017
  
  Vertical asymptote $x = 0 , 0.5$
  Horizontal asymptote $y = - 0.5$
  No slant asymptote.
  
  #### Explanation:
  
  Vertical Asymptote: Equate Denominator to zero.
  $3 x - 2 {x}^{2} = 0$
  $x = 0 , \left(\frac{3}{2}\right)$
  
  Horizontal asymptote:
  
  Since degrees of denominator and numerator are same, horizontal asymptote is obtained by dividing leading coefficients of highest degree numerator and denominator.
  $y = \left(\frac{1}{-} 2\right) = - \left(\frac{1}{2}\right)$
  
  Slant or oblique asymptote :
  Since numerator is not one degree above the denominator polynomial, there is slant or oblique asymptote.
  
  graph{(x^2+1)/(3x-2x^2) [-10, 10, -5, 5]}
explanation: The document contains a math question and its corresponding answer. The question asks how to find the vertical, horizontal, and oblique asymptotes of the function \(\frac{x^2+1}{3x-2x^2}\). The answer provides the vertical asymptotes at \(x = 0\) and \(x = 0.5\), the horizontal asymptote at \(y = -0.5\), and explains that there is no slant or oblique asymptote.
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, and Oblique Asymptote given (x^2+1)/ (3x-2x^2)?
  
  Oct 17, 2017
  
  Vertical asymptote $x = 0 , 0.5$
  Horizontal asymptote $y = - 0.5$
  No slant asymptote.
  
  #### Explanation:
  
  Vertical Asymptote: Equate Denominator to zero.
  $3 x - 2 {x}^{2} = 0$
  $x = 0 , \left(\frac{3}{2}\right)$
  
  Horizontal asymptote:
  
  Since degrees of denominator and numerator are same, horizontal asymptote is obtained by dividing leading coefficients of highest degree numerator and denominator.
  $y = \left(\frac{1}{-} 2\right) = - \left(\frac{1}{2}\right)$
  
  Slant or oblique asymptote :
  Since numerator is not one degree above the denominator polynomial, there is slant or oblique asymptote.
  
  graph{(x^2+1)/(3x-2x^2) [-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, and oblique asymptotes of the function \(\frac{x^2+1}{3x-2x^2}\). The answer provides the vertical asymptotes at \(x = 0\) and \(x = 0.5\), the horizontal asymptote at \(y = -0.5\), and explains that there is no slant or oblique asymptote.
  
  Label: question-and-answer