Question

How do you find vertical, horizontal and oblique asymptotes for `(3x^2+2x-1 )/(x^2-4)`?

Answer 1

`"vertical asymptotes at "x=+-2`
`"horizontal asymptote at "y=3`

Explanation:

`"the denominator of "f(x)" cannot be zero as this would"`
`"make "f(x)" 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^2-4=0rArrx=+-2" are the asymptotes"`

`"horizontal asymptotes occur as"`

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

`"divide the terms on the numerator/denominator by"`
`"the highest power of x that is "x^2`

`f(x)=((3x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-4/x^2)=(3+2/x-1/x^2)/(1-4/x^2)`

`"as "xto+-oo,f(x)to(3+0-0)/(1-0)`

`rArry=3" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence no oblique asymptote.
graph{(3x^2+2x-1)/(x^2-4) [-10, 10, -5, 5]}