How do you find the vertical, horizontal and slant asymptotes of: #f(x) = (6x + 6) / (3x^2 + 1)#?

The domain of #f(x)# is #D_f(x)=RR#

#AA x in RR, (3x^2+1)>0#

So, there is no vertical asymptote.

As the degree of the numerator is #<# than the degree of the denominator, there is no slant asymptote.

#lim_(x->+oo)f(x)=lim_(x->+oo)(6x)/(3x^2)=lim_(x->+oo)2/x=0^+#

#lim_(x->-oo)f(x)=lim_(x->-oo)(6x)/(3x^2)=lim_(x->-oo)2/x=0^-#

The horizontal asymptote is #y=0#

graph{(6x+6)/(3x^2+1) [-18.02, 18.03, -9.01, 9.01]}

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 " 3x^2+1=0rArrx^2=-1/3#

This has no real roots hence there are 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)=((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"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}