Question

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

Answer 1

`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]}