Question

How do I find the horizontal, vertical, and oblique asymptotes (If any exist) for f(x)=5x^3+49x^2+127x+75/(x^2+9x+18)?

Answer 1

We have a vertical asymptote at `x=-6` and oblique asymptote is `y=5x+49`

Explanation:

We can write `f(x)=(5x^3+49x^2+127x+75)/(x^2+9x+18)`

as `(5x^3+49x^2+127x+75)/(x^2+9x+18)`

= `((x+3)(5x^2+34x+25))/((x+6)(x+3))`

= `((x+3)(5x^2+34x+25))/((x+6)(x+3))`

= `(5x^2+34x+25)/(x+6)`

Although `x+3` cancels out and hence limit of `f(x)` exists at `x=-3`, as `f(x)=(5x^3+49x^2+127x+75)/(x^2+9x+18)` is not defined at `x=-3`. Hence we have a hole at `x=-3`.

We have a vertical asymptote at `x=-6`, as

`lim_(x->-6)f(x)=+-oo` depending on whether `x` approaches `-6` from left or right.

As the degree of numerator is one more than that of denominator, we have an oblique or slant asymptote (for horizontal asymptote, they should be equal) and

As `x->+-oo`, `f(x)->(5x^3+49x^2+127x+75)/(x^2+9x+18)`

or `f(x)->(5x+49+127/x+75/x^2)/(1+9/x+18/x^2)`

or `f(x)->5x+49`

Hence oblique asymptote is `y=5x+49`.

Below is the graph of `f(x)` not drawn to scale (shrunk vertically).

graph{(5x^3+49x^2+127x+75)/(x^2+9x+18) [-20, 20, -80, 80]}