Question

How do you find the asympototes for `y = (-8x^3 - 2x + 12)/( 7x^5 + 9x + 12)`?

Answer 1

The horizontal asymptote is `y = 0`.

There is a vertical asymptote `x=a`, where `a ~= -0.89261052` is the only real root of `7x^5+9x+12 = 0`.

Explanation:

Since the degree of the denominator is greater than the numerator, we find `y->0` as `x->+-oo`. So `y=0` is a horizontal asymptote.

`-8x^3-2x+12 = 0` has a single real root at `x ~= 1.0720202`

`7x^5+9x+12 = 0` has a single real root at `x ~= -0.89261052`

In general, the roots of quintics cannot be expressed in terms of ordinary radicals. In this particular case, the quintic is almost in Bring-Jerrard normal form, so it's relatively easy (believe me) to express the root in terms of the Bring Radical:

Let `x_1 = (7/9)^(1/4)x`

Then:

`0 = 7x^5+9x+12`

`=7((9/7)^(1/4)x_1)^5+9(9/7)^(1/4)x^1+12`

`=9(9/7)^(1/4)x_1^5+9(9/7)^(1/4)x_1+12`

Divide through by `9(9/7)^(1/4)` to get:

`x_1^5+x_1+4/3(7/9)^(1/4) = 0`

So `x_1 = BR(4/3(7/9)^(1/4))`

and `x = (9/7)^(1/4)*x_1 = (9/7)^(1/4)BR(4/3(7/9)^(1/4))`

See: https://en.wikipedia.org/wiki/Bring_radical