Question

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

Answer 1

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]}