How do you find vertical, horizontal and oblique asymptotes for #(6x-7)/(11x+8)#?

Question

Elias Ackerman · Accepted Answer

vertical asymptote #x=-8/11#
horizontal asymptote #y=6/11# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero. solve : 11x + 8 = 0 #rArrx=-8/11" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by x #((6x)/x-7/x)/((11x)/x+8/x)=(6-7/x)/(11+8/x)# as #xto+-oo,f(x)to(6-0)/(11+0)# #rArry=6/11" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here ( both degree 1 ). Hence there are no oblique asymptotes. graph{(6x-7)/(11x+8) [-10, 10, -5, 5]}

Chloe Alexander · Answer

undefined To find the vertical asymptote, set the denominator equal to zero and solve for $ x $. The vertical asymptote occurs where the denominator is zero.

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 degree of the numerator is equal to the degree of the denominator, divide the leading coefficients 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 oblique asymptotes, perform polynomial long division or synthetic division to divide the numerator by the denominator. The quotient represents the equation of the oblique asymptote.