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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