What are the asymptotes for #(x^2 - 2x - 3 )/(-4x) #?

1 Answer
Jun 3, 2016

Answer:

Vertical asymptote at #x = 0#
Slant asymptote given by #y = -x/4+1/2 #

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#
#2)# horizontal asymptotes when #n le m#
#3)# slant asymptotes when #n = m + 1#
In the present case we have #x_v = 0# and #n = m+1# with #n = 2# and #m = 1#

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_n(x))/(p_{n-1}(x)) = (x^2-2x-3)/(-4x)#
#p_n(x) = p_{n-1}(x)(a x+b)+r_{n-2}(x)#
#x^2-2x-3 =(-4x)(a x + b) + c#

equating coefficients

#{ (-3 - c=0), (-2 + 4 b=0), (1 + 4 a=0) :}#

solving for #a,b,c# we get #{a = -(1/4), b = 1/2, c = -3}#
and substituting

#y = -x/4+1/2 #

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