How do you find the asymptotes for #Q(x) = (2x^2)/ (x^2 - 5x - 6)#?

1 Answer
Mar 26, 2016

Answer:

vertical asymptotes x = -1 , x = 6
horizontal asymptote y = 2

Explanation:

Vertical asymptotes occur when the denominator of a rational function tends to zero. To find the equation/s let the denominator equal zero.

solve : #x^2 - 5x - 6 = 0 → (x-6)(x+1) = 0#

# rArr x = -1 , x = 6" are the asymptotes " #

Horizontal asymptotes occur as # lim_(x→±∞) f(x) → 0 #

divide all terms on numerator/denominator by # x^2 #

# ((2x^2)/x^2)/(x^2/x^2 - (5x)/x^2 - 6/x^2)= 2/(1-5/x -6/x^2) #

As x → ∞ , # 5/x" and " 6/x^2 → 0 #

#rArr y = 2/1 = 2 " is the asymptote " #

Here is the graph of the function.
graph{(2x^2)/(x^2-5x-6) [-20, 20, -10, 10]}