Question

What are the asymptote(s) and hole(s), if any, of `f(x) =(3x^2)/(x^2-x-1)`?

Answer 1

`"vertical asymptotes at "x~~-0.62" and "x~~1.62`
`"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-x-1=0`

`"here " a=1,b-1" and " c=-1`

`"solve using the "color(blue)"quadratic formula"`

`x=(1+-sqrt(1+4))/2=(1+-sqrt5)/2`

`rArrx~~1.62,x~~-0.62" 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)=((3x^2)/x^2)/(x^2/x^2-x/x^2-1/x^2)=3/(1-1/x-1/x^2)`

as `xto+-oo,f(x)to3/(1-0-0)`

`rArry=3" is the asymptote"`

Holes occur when there is a duplicate factor on the numerator/denominator. This is not the case here hence there are no holes.
graph{(3x^2)/(x^2-x-1) [-10, 10, -5, 5]}