Question

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

Answer 1

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]}