How do you find all the asymptotes for function #f(x) = (2x^2+3x+8)/(x+3)#?

1 Answer
Shwetank Mauria
May 14, 2018

Vertical asymptote #x=3# and slanting or oblique asymptote #y=2x+3#

Explanation:

There are three types of asymptotes.

Vertical asymptotes are indicated by denominator. Here we have #(x+3)# in denominator, so vertical asymptote is given by #x+3=0# or #x=-3#. Observe that in case #x+3# is a factor of numerator, it will cancel out and we will not have an asymptote, rather we will havea hole at #x=-3#. But here #x+3# is not a factor of numerator, we do have a vertical asymptote as

#lim_(x->-3)(2x^2+3x+8)/(x+3)=oo#

Horizontal asymptotes in such cases (algebraic expressions) are there when degree of numerator is equal to that of denominator. Here it is not so we do not a horizontal asymptote. Stll assume numerator as only #3x+8# and then

#lim_(x->oo)(3x+8)/(x+3)=lim_(x->oo)(3+8/x)/(1+3/x)=3/1=3# and we would have #y=3# as horizontal asymptote.

Here we have #(2x^2+3x+8)/(x+3)# and we have

#lim_(x->oo)(2x^2+3x+8)/(x+3)=lim_(x->oo)(2x+3+8/x)/(1+3/x)#

= #2x+3#

Hence we have a slanting or oblique asymptote #y=2x+3#

It is apparent that we have this only when degree of numerator is one more than that of denominator.

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