How do you find vertical, horizontal and oblique asymptotes for #f(x)=(x^2+5x+8) / (x+3)#?

1 Answer
Jun 25, 2016

Answer:

#(x^2+5x+8)/(x+3)# has vertical asymptote #x+3=0# and oblique asymptote #y=x#

Explanation:

Asymptotes of algebraic expressions are easy to find,

First of all vertical asymptotes, if denominator has a zero at a point but numerator does not have, we have an asymptote at that point.

In the given expression, we have denominator zero for #x=-3#, but numerator is not zero for this, hence we have vertical asymptote #x+3=0#.

As regards horizontal asymptote, we have it at #y=0#, if the degree of denominator is greater than the degree of the numerator. This is not so here.

If the degrees are equal, then we have a horizontal asymptote #y=a/b#, if #a# and #b# are the coefficients of highest degree in numerator and denominator respectively. This too is not here. So we do not have any horizontal asymptote.

If the degree of numerator is just one higher than that of denominator, say they are given by #ax^n# and #bx^(n-1)#, than we have an **oblique or slanting asymptote## is at #y=(ax)/b#.

Here the degree of numerator is #2# and that of denominator is #1#, hence we have an oblique asymptote at #y=x^2/x=x# i.e. #y=x#.

Hence #(x^2+5x+8)/(x+3)# has vertical asymptote #x+3=0# and oblique asymptote #y=x#

graph{(x^2+5x+8)/(x+3) [-40, 40, -20, 20]}