How do you find vertical, horizontal and oblique asymptotes for #y = (4x^3 + x^2 + x + 5 )/( x^2 + 3x)#?

1 Answer
Jun 15, 2016

Answer:

Vertical asymptotes are #x=0# and #x=-3# and oblique asymptote is #y=4x#.

Explanation:

To find the asymptotes for function #(4x^3+x^2+x+5)/(x^2+3x)#, let us first start with vertical asymptotes, which are given by putting denominator equal to zero or #x^2+3x=0# i.e. #x(x+3)=0# and hence #x=-3# and #x=0# are two vertical asymptotes.

As the highest degree of numerator is #3# and is just one higher than that of denominator i.e. #2#, we would have obique asymptote given by #y=4x^3/x^2=4x# and there is no horizontal asymptote.

Hence, Vertical asymptotes are #x=0# and #x=-3# and oblique asymptote is #y=4x#.

The graph is as shown below.
graph{(4x^3+x^2+x+5)/(x^2+3x) [-40, 40, -20, 20]}