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

##### 1 Answer

Apr 21, 2016

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]}