How do you find the asymptotes for #(8x^2-x+2)/(4x^2-16)#?

1 Answer
Apr 21, 2016

Answer:

vertical asymptotes x = ± 2
horizontal asymptote y = 2

Explanation:

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

solve: # 4x^2-16 = 0 → 4(x^2-4) = 0 → 4(x-2)(x+2) = 0 #

#rArr x = -2 , x = 2 " are the asymptotes " #

Horizontal asymptotes occur as #lim_(xto+-oo) f(x) to 0 #

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

#rArr ((8x^2)/x^2 - x/x^2 + 2/x^2)/((4x^2)/x^2 - 16/x^2)=(8-1/x+2/x^2)/(4-16/x^2) #

as# xto+-oo , 1/x , 2/x^2" and " 16/x^2 to 0#

and # y to (8-0+0)/(4-0) = 8/4 = 2 #

#rArr y = 2 " is the asymptote " #
graph{(8x^2-x+2)/(4x^2-16) [-10, 10, -5, 5]}