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

1 Answer
Jun 9, 2015

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]}