Question

How do you find the vertical, horizontal or slant asymptotes for `(6x^2+2x-1) /( x^2-1)`?

Answer 1

vertical asymptotes at `x=+-1`
horizontal asymptote at y = 6

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: `x^2-1=0rArrx=+-1`

`rArrx=-1" and " x=1" 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)=((6x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-1/x^2)=(6+2/x-1/x^2)/(1-1/x^2)`

as `xto+-oo,f(x)to(6+0-0)/(1-0)`

`rArry=6" 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{(6x^2+2x-1)/(x^2-1) [-20, 20, -10, 10]}