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

Question

James Allen · Accepted Answer

#"horizontal asymptote at "y=0#

#"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]}

Sadie Alexander · Answer

undefined To find vertical asymptotes, set the denominator equal to zero and solve for x. For horizontal asymptotes, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degrees are equal, divide the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote. For oblique asymptotes, use polynomial long division to divide the numerator by the denominator. The quotient will be the equation of the oblique asymptote.