How do you find the asymptotes for b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)?

Aug 11, 2018

$b \left(x\right)$ has two vertical asymptotes and one horizontal asymptote...

Explanation:

First let us find the zeros of the numerator:

Given:

${x}^{3} - x + 1 = 0$

Let $x = u + v$, to get:

${u}^{3} + {v}^{3} + \left(3 u v - 1\right) \left(u + v\right) + 1 = 0$

Add the constraint $v = \frac{1}{3 u}$ to eliminate the expression in $\left(u + v\right)$ and get:

${u}^{3} + \frac{1}{27 {u}^{3}} + 1 = 0$

Multiply by $27 {u}^{3}$ and rearrange slightly to get:

$27 {\left({u}^{3}\right)}^{2} + 27 \left({u}^{3}\right) + 1 = 0$

Hence using the quadratic formula, we find:

${u}^{3} = \frac{- 27 \pm \sqrt{{27}^{2} - 4 \cdot 27}}{54}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{- 27 \pm 3 \sqrt{81 - 12}}{54}$

$\textcolor{w h i t e}{{u}^{3}} = \frac{- 27 \pm 3 \sqrt{69}}{54}$

Since this is real and the derivation is symmetric in $u$ and $v$, we can deduce that the only real zero of ${x}^{3} - x + 1$ is:

$\sqrt{\frac{- 27 + 3 \sqrt{69}}{54}} + \sqrt{\frac{- 27 - 3 \sqrt{69}}{54}} \approx - 1.324717957$

The quartic is somewhat more tedious to solve algebraically, so suffice it to say that its two real zeros are approximately:

$- 1.20291275$

$0.86169984$

These can be found numerically using a Newton-Raphson method, or a Durand-Kerner method.

Since these are distinct from the real zero of the cubic, these are points of vertical asymptotes and $b \left(x\right)$ has no holes.

Since the numerator is of degree $3$ which is less than that of the denominator, the given rational expression has a horizontal asymptote $y = 0$ and no oblique asymptotes.

graph{(x^3-x+1)/(2x^4+x^3-x^2-1) [-5, 5, -2.5, 2.5]}