How do you find vertical, horizontal and oblique asymptotes for #(6x + 6) / (3x^2 + 1)#?

1 Answer
Jul 20, 2018

Answer:

#"horizontal asymptote at "y=0#

Explanation:

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

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

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

#"This has no real solutions 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)#

#y=0" is the asymptote"#

#"Oblique asymptotes occur when the degree of the"#
#"numerator is greater than the degree of the denominator. "#
#"This is not the case here hence no oblique asymptotes"#

graph{(6x+6)/(3x^2+1) [-10, 10, -5, 5]}