How do you find vertical, horizontal and oblique asymptotes for #(x+3 )/ (x^2 + 8x + 15)#?

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Answer 1 · 2016-04-04T23:05:16

Vertical Asymptotes:#x=-5#
Horizontal Asymptote:#y=0#
No Oblique Asymptote

To obtain the Horizontal Asymptote, take the limit of the function

#lim_(x rarr oo) y=lim_(x rarr oo) (x+3)/(x^2+8x+15)=zero#

therefore, #y=0# is a Horizontal Asymptote

To obtain the Vertical Asymptote, equate the factors of the denominator to zero the solve for x.

#x+5=0#

#x=-5#

Kindly see the graph of #y=(x+3)/(x^2+8x+15)# and the location of the imaginary asymptotes.

graph{y=(x+3)/(x^2+8x+15)[-20,20,-10,10]}

God bless....I hope the explanation is useful.