How do you find the Vertical, Horizontal, and Oblique Asymptote given #((x^2)+3)/((9x^2)-80x-9)#?

1 Answer
Sep 13, 2016

Answer:

vertical asymptotes at #x=-1/9,x=9#
horizontal asymptote at #y=1/9#

Explanation:

The denominator of the function cannot be zero as this would make the function 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: #9x^2-80x-9=0rArr(9x+1)(x-9)=0#

#rArrx=-1/9" and " x=9" 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#

#f(x)=(x^2/x^2+3/x^2)/((9x^2)/x^2-(80x)/x^2-9/x^2)=(1+3/x^2)/(9-80/x-9/x^2)#

as #xto+-oo,f(x)to(1+0)/(9-0-0)#

#rArry=1/9" 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+3)/(9x^2-80x-9) [-20, 20, -10, 10]}