What are the asymptote(s) and hole(s), if any, of f(x) =(3x^2)/(x^2-x-1)f(x)=3x2x2−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=0solve x2−x−1=0
"here " a=1,b-1" and " c=-1here a=1,b−1 and c=−1
"solve using the "color(blue)"quadratic formula"solve using the quadratic formula
x=(1+-sqrt(1+4))/2=(1+-sqrt5)/2x=1±√1+42=1±√52
rArrx~~1.62,x~~-0.62" are the asymptotes"⇒x≈1.62,x≈−0.62 are the asymptotes
"Horizontal asymptotes occur as"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]}
