How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=(x^2-4x-32)/(x^2-16)#?

1 Answer
Sep 8, 2016

Answer:

vertical asymptote at x = 4
horizontal asymptote at y = 1

Explanation:

The first step is to factorise and simplify f(x)

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x-4=0rArrx=4" is the asymptote"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

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

as #xto+-oo,f(x)to(1-0)/(1-0)#

#rArry=1" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no oblique asymptotes.
graph{(x-8)/(x-4) [-20, 20, -10, 10]}