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