Question

1. (5 points) find vertical asymptotes of $h(x) = \frac{x^2 - 2x - 3}{2x^2 + 3x + 1}$.

Explanation:

Step1: Factor denominator

$2x^2 + 3x + 1 = (2x + 1)(x + 1)$

Step2: Solve denominator=0

$2x + 1 = 0 \Rightarrow x = -\frac{1}{2}$; $x + 1 = 0 \Rightarrow x = -1$

Step3: Check numerator at these x

At $x = -\frac{1}{2}$: $(-\frac{1}{2})^2 - 2(-\frac{1}{2}) - 3 = \frac{1}{4} + 1 - 3 = -\frac{7}{4}
eq 0$
At $x = -1$: $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$ (canceled, not asymptote)

Answer:

$x = -\frac{1}{2}$