Question

What are the asymptotes of `f(x)=-x/((x^2-8)(5x+2))`?

Answer 1

vertical asymptotes: `x = +- 2sqrt(2); x = -2/5`
horizontal asymptotes: `y = 0`

Explanation:

Given: `-x/((x^2-8)(5x+2))`

Factor the difference of squares in the denominator `a^2 - b^2 = (a-b)(a+b):`

`(x^2 - (sqrt(8))^2) = (x - sqrt(8))(x + sqrt(8))`

`sqrt(8) = sqrt(4*2) = 2sqrt(2)`

Therefore:
`-x/((x^2-8)(5x+2)) = -x/((x-2sqrt(2))(x+2sqrt(2))(5x+2))`

Find vertical asymptotes `D(x) = 0`:

vertical asymptotes: `x = +- 2sqrt(2); x = -2/5`

Find horizontal asymptotes using Precalculus for a rational function: `(N(x))/(D(x)) = (a_nx^n + ...)/(b_mx^m + ...)`

When `n < m` horizontal asymptote is `y = 0`

Using Calculus to find horizontal asymptotes:

`lim x->oo -x/((x^2-8)(5x+2)) =`

`lim x->oo -x/(5x^3+2x^2-40x+16)`

Divide all by the largest `x` in the denominator:

`lim x->oo (-x/x^3)/((5x^3)/x^3 + (2x^2)/x^3 - (40x)/x^3 + 16/x^3)`

`lim x->oo (-1/x^2)/(5 + 2/x - 40/x^2 + 16/x^3) = 0/5 = 0`