Question

write the equation of the vertical asymptote(s) of the rational function. $y = \frac{x + 4}{x^{2}-4}$
$x = 2$ and $x = - 2$
$x = 2$ and $x = 4$
$x = 4$
$x = 2$

Explanation:

Step1: Find denominator's zeros

Set $x^{2}-4 = 0$.

Step2: Solve the equation

Factor $x^{2}-4$ to $(x + 2)(x - 2)=0$. Then $x+2 = 0$ gives $x=-2$ and $x - 2=0$ gives $x = 2$. For a rational function $\frac{f(x)}{g(x)}$, vertical asymptotes occur at the values of $x$ that make $g(x)=0$ and $f(x)
eq0$. Here, for $y=\frac{x + 4}{x^{2}-4}$, when $x=\pm2$, the denominator is zero and the numerator is non - zero.

Answer:

$x = 2$ and $x=-2$