How do you find the asymptotes for #f(x)=(6x + 6) / (3x^2 + 1)#?

1 Answer
Aug 23, 2016

Answer:

horizontal asymptote at y = 0

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #3x^2+1=0rArrx^2=-1/3" which has no real roots"# graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}

Hence there are no vertical asymptotes.

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by highest power of x that is #x^2#

#f(x)=((6x)/x^2+6/x^2)/((3x^2)/x^2+1/x^2)=(6/x+6/x^2)/(3+1/x^2)#

as #xto+-oo,f(x)to(0+0)/(3+0)#

#rArry=0" is the asymptote"#