How do you find the Vertical, Horizontal, and Oblique Asymptote given #x/(x^2+x-6)#?

Question

Liam Brennan · Accepted Answer

#"vertical asymptotes at "x--3" and "x=2#
#"horizontal asymptote at "y=0# #"let "f(x)=x/(x^2+x-6)# 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 "x^2+x-6=0rArr(x+3)(x-2)=0# #x=-3" and "x=2" are the 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)=(x/x^2)/(x^2/x^2+x/x^2-6/x^2)=(1/x)/(1+1/x-6/x^2)# #"as "xto+-oo,f(x)to0/(1+0-0)# #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 there are no oblique asymptotes. graph{x/(x^2+x-6) [-10, 10, -5, 5]}

Madelyn Anderson · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $x$. In this case, the denominator $x^2 + x - 6$ factors to $(x - 2)(x + 3)$, so the vertical asymptotes are $x = 2$ and $x = -3$.

To find the horizontal asymptote, compare the degrees of the numerator and the 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 than the degree of the denominator, there is no horizontal asymptote. In this case, the degree of the denominator is greater than the degree of the numerator, so the horizontal asymptote is $y = 0$.

To find the oblique asymptote, perform long division or polynomial division on the function. If the result is a polynomial plus a proper rational function (where the degree of the numerator is less than the degree of the denominator), the oblique asymptote is the polynomial. In this case, the oblique asymptote can be found through polynomial division, which yields $y = 0$.