How do you find vertical, horizontal and oblique asymptotes for #y = (x-4)^2/(x^2-4)#?

1 Answer
Apr 1, 2016

Answer:

vertical asymptotes x = ± 2
horizontal asymptote y = 1

Explanation:

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

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

#rArr x = ± 2 " are the asymptotes " #

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

now numerator = #(x - 4)^2 = x^2 - 8x + 16 #

and y = #(x^2-8x+16)/(x^2-4) #

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

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

As # xtooo , 8/x , 16/x^2 " and " 4/x^2 to 0#

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

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here , hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(x-4)^2/(x^2-4) [-10, 10, -5, 5]}