How do you find the vertical, horizontal or slant asymptotes for #(x^2-4x-32)/(x^2-16)#?

1 Answer
Apr 9, 2016

Answer:

vertical asymptote x = 4
horizontal asymptote y = 1

Explanation:

First step is to factorise the function.

#rArr( (x - 8)(x + 4))/((x - 4)(x + 4)) = ((x-8)cancel((x+4)))/((x-4)cancel((x+4))#

# = (x - 8)/(x - 4 ) #

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

solve : x - 4 = 0 → x = 4 is the asymptote

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

divide terms on numerator/denominator by x

#(x/x - 8/x)/(x/x - 4/x) = (1 - 8/x)/(1 - 4/x) #

As #x to+-oo , 8/x" and 4/x to 0 #

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here , hence there are no slant asymptotes.

Here is the graph.
graph{(x-8)/(x-4) [-10, 10, -5, 5]}