Question

1. find the vertical asymptote(s) for $y = \frac{x + 5}{x^2 - 16}$

\bigcirc $x = 0, x = 16$
\bigcirc none
\bigcirc $x = -4, x = 4$
\bigcirc $x = -5$

Explanation:

Step1: Factor denominator

$x^2 - 16 = (x-4)(x+4)$

Step2: Find undefined x-values

Set denominator to 0: $(x-4)(x+4)=0 \Rightarrow x=4$ or $x=-4$

Step3: Check numerator at these x

At $x=4$: $4+5=9≠0$; at $x=-4$: $-4+5=1≠0$

Step4: Confirm asymptotes

Vertical asymptotes at $x=4$ and $x=-4$

Answer:

x = -4, x = 4