Question

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

Answer 1

`"horizontal asymptote at "y=1/2`

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 "2x^2-x+1=0`

`"here " a=2,b=-1" and "c=1`

checking the `color(blue)"discriminant"`

`Delta=b^2-4ac=(-1)^2-(4xx2xx1)=-7`

Since `Delta<0` there are no real solutions hence no vertical 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)/((2x^2)/x^2-x/x^2+1/x^2)=1/(2-1/x+1/x^2)`

as `xto+-oo,f(x)to1/(2-0+0)`

`rArry=0" 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{(x^2)/(2x^2-x+1) [-10, 10, -5, 5]}