How do you find the asymptotes for #f(x) = (x^2+4x-2)/(x^2-x-7)#?

1 Answer
Jun 9, 2015

Answer:

#y=1#
#x=(-1-sqrt(29))/2#
#x=(-1+sqrt(29))/2#

Explanation:

  1. Calculate Horizontal Asymptotes
    #lim_(x->+-oo)(f(x))=1#
    So we have a horizontal asymptote that is #y=1#.
  2. Calculate Vertical Asymptotes
    #x^2-x-7=0 ? #
    #Delta_x=29#
    #x_1= (-1+sqrt(29))/2 ; x_2= (-1-sqrt(29))/2#
    #lim_(x->x_1)(f(x))=+oo#
    #lim_(x->x_2)(f(x))=-oo#
    So we have two vertical asymptotes in #x=x_1# and in #x=x_2#
  3. There aren't Oblique Asymptotes because there is a horizontal asymptote.
  4. For sure... Let's see the graph of #f(x)=(x^2-4x-2)/(x^2-x-7)#

graph{(x^2-4x-2)/(x^2-x-7) [-10, 10, -5, 5]}