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

1 Answer
Jun 12, 2017

Answer:

#"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]}