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

Question

Lily Anderson · Accepted Answer

horizontal asymptote at y = 0

The denominator of the function cannot be zero as this would make the function 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+2=0rArrx^2=-2# 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)=((3x)/x^2)/(x^2/x^2+2/x^2)=(3/x)/(1+2/x^2)# as #xto+-oo,f(x)to0/(1+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 (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes. graph{(3x)/(x^2+2) [-10, 10, -5, 5]}

Theodore Amidon · Answer

undefined To find the vertical asymptotes, determine the values of $x$ for which the denominator of the function equals zero. These values are the vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator of the function. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both to find the horizontal asymptote. If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote.

To find the slant asymptote, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient obtained represents the equation of the slant asymptote. If the degrees of the numerator and denominator are equal, there is no slant asymptote.