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

Question

Lincoln Anderson · Accepted Answer

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

James Allen · Answer

undefined To find the asymptotes of $ y = \frac{3(x+5)}{(x+1)(x+2)} $, you need to check for vertical and horizontal asymptotes.

Vertical asymptotes occur where the denominator equals zero, but the numerator does not. So, set the denominator equal to zero and solve for $ x $. These values of $ x $ will give you the vertical asymptotes.

Horizontal asymptotes can be found by analyzing the behavior of the function as $ x $ approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $ y = 0 $. If the degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. If the degree of the numerator is greater, there is no horizontal asymptote.

Compute these values to find the asymptotes of the given function.