How do you find vertical, horizontal and oblique asymptotes for #{-9 x^3 - 9 x^2 + 28 x + 9 }/{3 x^2 + 6 x + 2}#?

1 Answer
Jun 14, 2016

Answer:

Vertical asymptotes #x_v = {1/3 (-3 - sqrt[3]), 1/3 (-3 + sqrt[3])}#
Slant asymptote #y = 3 - 3 x#.

Explanation:

In a polynomial fraction #f(x) = (p_n(x))/(p_m(x))# we have:

#1)# vertical asymptotes for #x_v# such that #p_m(x_v)=0#. Here #x_v = {1/3 (-3 - sqrt[3]), 1/3 (-3 + sqrt[3])}#
#2)# horizontal asymptotes when #n le m#
#3)# slant asymptotes when #n = m + 1#
In the present case we dont have vertical asymptotes and #n = m+1# with #n = 3# and #m = 2#

Slant asymptotes are obtained considering #(p_n(x))/(p_{n-1}(x)) approx y = a x+b# for large values of #abs(x)#

In the present case we have

#(p_3(x))/(p_2(x)) =(-9 x^3 - 9 x^2 + 28 x + 9)/(3 x^2 + 6 x + 2)#
#p_3(x)=p_2(x)(a x+b)+r_1(x)#
#r_1(x)=c x + d#
#-9 x^3 - 9 x^2 + 28 x + 9 = (a x + b) (3 x^2 + 6 x + 2) + c x + d#

equating coefficients

#{ (9 - 2 b - d=0), (28 - 2 a - 6 b - c=0), (9 + 6 a + 3 b=0), (9 + 3 a=0) :}#

solving for #a,b,c,d# we have #{a = -3, b = 3, c = 16, d = 3}#
substituting in #y = a x + b#

#y = 3 - 3 x#

Note that

#(p_3(x))/(p_2(x))=(a x+b)+(r_1(x))/(p_2(x))#

and as #abs(x) # increases #(r_1(x))/(p_2(x))->0#

Attached a figure showing the results.

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