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

1 Answer
Jun 5, 2016

Answer:

vertical asymptote #x=-8/11#
horizontal asymptote #y=6/11#

Explanation:

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]}