Chapter 14.3, Problem 81AYU

$R\left(x\right)=\frac{x^3+2x^2+x}{x^4+x^3+2x+2}$

To determine

To find: The numbers at which $R$ is undefined and asymptotes or holes for $R$.

Answer to Problem 81AYU

Undefined for $x = - 1$ and $x = -\sqrt[3]{2}$.

Vertical asymptote: $x + 1 = 0$; $x = -\sqrt[3]{2}$.

Horizontal asymptote: $y = 0$.

Explanation of Solution

Given:

$R\left(x\right) = \frac{x^3 + 2x^2 + x}{x^4 + x^3 + 2x + 2}$

Calculation:

$R\left(x\right) = \frac{x^3 + 2x^2 + x}{x^4 + x^3 + 2x + 2}$

$R\left(x\right) = \frac{x\left(x + 1\right)\left(x + 1\right)}{\left(x^3 + 2\right)\left(x + 1\right)}$

The denominator of $R\left(x\right)$ becomes zero when

$\left(x^3 + 2\right)\left(x + 1\right) = 0$

$x = -\sqrt[3]{2}$; $x = -1$

Thus $R$ is undefined for $x = -\sqrt[3]{2}$; $x = -1$

This function has vertical asymptotes $x = -\sqrt[3]{2}$; $x + 1 = 0$ and horizontal asymptote$y = 0$.