How do you find the vertical, horizontal or slant asymptotes for f(x)={ x^2 - 81 }/ {x^2 - 4x}f(x)=x2−81x2−4x?
1 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=4x2−4x=0⇒x(x−4)=0⇒x=0,x=4
rArrx=0,x=4" are the asymptotes"⇒x=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]}
