How do you find vertical, horizontal and oblique asymptotes for #f(x)= (2x+3)/(3x+4)#?

1 Answer
Mar 16, 2016

Answer:

Vertical asymptote #x = -4/3#
Horizontal asymptote # y = 2/3 #

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve : 3x + 4 = 0 → 3x = -4 #→ x = - 4/3" is the asymptote "#

Horizontal asymptotes occur as #lim_(x→±∞) f(x) → 0 #

divide all terms on numerator/ denominator by x

#(2x+3)/(3x+4) = ((2x)/x + 3/x )/((3x)/x + 4/x) = (2 + 3/x)/(3+4/x)#

now as x →∞ ,# 3/x" and " 4/x → 0 #

#rArr y = 2/3 " 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 is no oblique asymptote.

Here is the graph of the function.
graph{(2x+3)/(3x+4) [-10, 10, -5, 5]}