How do you find the slant asymptote of # (x^3-1)/(x^2-6)#?

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Answer 1 · 2016-07-04T19:40:15

#y = x#

# (x^3-1)/(x^2-6)#

Making # x^3-1 = (x^2-6)(a x+b)+c x + d#

then

# (x^3-1)/(x^2-6) = ax+b + (c x + d)/(x^2-6)#

as we can observe

# (x^3-1)/(x^2-6) approx ax+b# as #abs(x)# is large.

We find #a,b,c,d# solving

#{ (-1 + 6 b - d=0), (6 a - c=0), (-b=0),( 1 - a=0) :}#

Those conditions are obtained equating

#x^3-1 -( (x^2-6)(a x+b)+c x + d)=0 forall x#

giving #{a = 1, b = 0, c = 6, d = -1}#

so

The asymptote is

#y = x#