Question

lem 19
find vertical asymptotes of each curve. if there is no such asymptote, enter
\.
$y = \frac{x^3 + 1}{x^3 + x}$

Explanation:

Step1: Find denominator roots

Set denominator equal to 0:
$$x^3 + x = 0$$
Factor:
$$x(x^2 + 1) = 0$$

Step2: Solve for x-values

Solve each factor:

1. $x = 0$
2. $x^2 + 1 = 0 \implies x^2 = -1$, which has no real solutions.

Step3: Check numerator at x=0

Substitute $x=0$ into numerator:

$$0^3 + 1 = 1 eq 0$$

Since numerator is non-zero here, $x=0$ is a vertical asymptote.

Answer:

$x=0$