How do you find the asymptotes for #f(x) = (x^2+4x-2)/(x^2-x-7)#?
1 Answer
Jun 9, 2015
Explanation:
- Calculate Horizontal Asymptotes
#lim_(x->+-oo)(f(x))=1#
So we have a horizontal asymptote that is#y=1# . - 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# - There aren't Oblique Asymptotes because there is a horizontal asymptote.
- 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]}