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

1 Answer
Jun 4, 2016

Answer:

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

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

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

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by #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"#

Slant 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 slant asymptotes.
graph{(x^2-81)/(x^2-4x) [-20, 20, -10, 10]}