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]}