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]}