How do you find vertical, horizontal and oblique asymptotes for #(x+6) /( 2x+1 )#?

1 Answer
Apr 25, 2016

Answer:

#x = -1/2#
#y = 1/2#

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 : 2x + 1 = 0 → 2x = -1 → # x = -1/2" is the asymptote " #

Horizontal asymptotes occur as # lim_(x to +- oo) , f(x) to 0#

divide terms on numerator/denominator by x

#rArr (x/x + 6/x)/((2x)/x + 1/x) = (1 + 6/x)/(2 + 1/x)#

as # x to +- oo , y to (1+0)/(2 + 0) #

#rArr y = 1/2" is the asymptote " #

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no oblique asymptotes.
graph{(x+6)/(2x+1) [-10, 10, -5, 5]}