How do you find the asymptotes for #y= (3(x+5))/((x+1)(x+2))#?

1 Answer
Feb 2, 2016

Answer:

Horizontal asymptotes: #x=-1 color(white)("XX")andcolor(white)("XX")x=-2#
Vertical asymptote: #y=0#

Explanation:

The denominator of #(3(x+5))/((x+1)(x+2))# will become #0# if #x=-1# or #x=-2#
This results in horizontal asymptotes at these locations (provided the numerator does not also go to #0#, which it doesn't in this case).

Since the degree of the denominator is greater than the degree of the numerator, as #xrarroo, yrarr0#,
providing the vertical asymptote #y=0#.
This could be made explicitly clear as
#color(white)("XXX")y=(3(x+5))/((x+1)(x+2))#

#color(white)("XXX")=(3x+15)/(x^2+3x+2)#

#color(white)("XXX")=(3+15/x)/(x+3+2/x)#

As #xrarroo# #15/xrarr0# and #2/xrarr0#

leaving #lim_(xrarroo) 3/(x+3)# with the constant numerator and the constant added in the denominator becoming less significant as #xrarroo#
So #lim_(xrarroo) 3/(x+3) rarr lim_(xrarroo)1/x rarr 0#