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

1 Answer
Apr 1, 2017

Answer:

#"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]}