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

1 Answer
Feb 22, 2017

Answer:

Vertical asymptote: #x = 0#
Horizontal asymptote: #y = 2#
Oblique asymptote: none

Explanation:

Find a common denominator for the function:
#y = 3/x +(2x)/x = (3+2x)/x = (2x+3)/x#

Rational Function: #(N(x))/(D(x))#, when #N(x) = 0# gives x-intercepts,
when #D(x) = 0# you find vertical asymptotes.

Vertical asymptote at # x = 0#

When #(N(x))/(D(x)) =( a_nx^n+....)/(b_mx^m+....)# where #n, m# are the degrees of the polynomials.

If #n=m# horizontal asymptote is at #y=a_n/b_m#

#n = m = 1#, so Horizontal asymptote: #y = 2/1; y = 2#

To have an oblique or slant asymptote, #m+1 = n# which is not the case, so there is no oblique asymptote.