How do you find the Vertical, Horizontal, and Oblique Asymptote given #((x^2)+3)/((9x^2)-80x-9)#?

Question

James Allen · Accepted Answer

vertical asymptotes at #x=-1/9,x=9#
horizontal asymptote at #y=1/9# 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: #9x^2-80x-9=0rArr(9x+1)(x-9)=0# #rArrx=-1/9" and " x=9" 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^2/x^2+3/x^2)/((9x^2)/x^2-(80x)/x^2-9/x^2)=(1+3/x^2)/(9-80/x-9/x^2)# as #xto+-oo,f(x)to(1+0)/(9-0-0)# #rArry=1/9" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no oblique asymptotes. graph{(x^2+3)/(9x^2-80x-9) [-20, 20, -10, 10]}

Elijah Allen · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for x. To find the horizontal asymptote, compare the degrees of the numerator and denominator polynomials. 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 of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote. To find the oblique asymptote, perform polynomial long division to divide the numerator by the denominator. The quotient represents the equation of the oblique asymptote.