How do you find vertical, horizontal and oblique asymptotes for #f(x)={ x^2 - 81 }/ {x^2 - 4x}#?

1 Answer
Aug 9, 2016

Answer:

vertical asymptotes at x = 0 , x = 4
horizontal asymptote at y = 1

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

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

#rArrx=0" and " x=4" are the asymptotes"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#(x^2/x^2-81/x^2)/(x^2/x^2-(4x)/x^2)=(1-81/x^2)/(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 2) Hence there are no oblique asymptotes.
graph{(x^2-81)/(x^2-4x) [-50.63, 50.63, -25.3, 25.33]}