consider the following equation: $f(x)=\frac{x^{2}+4}{4x^{2}-4x - 8}$. name the vertical asymptote(s).

Explanation:

Step1: Factor the denominator

Factor $4x^{2}-4x - 8$. First, factor out the greatest - common factor 4: $4(x^{2}-x - 2)$. Then factor the quadratic inside the parentheses: $4(x - 2)(x+1)$. So, $f(x)=\frac{x^{2}+4}{4(x - 2)(x + 1)}$.

Step2: Find the values that make the denominator zero

Set the denominator equal to zero: $4(x - 2)(x + 1)=0$. Using the zero - product property, if $ab = 0$, then $a = 0$ or $b = 0$. So, $x-2=0$ gives $x = 2$ and $x + 1=0$ gives $x=-1$.

Answer:

$x = 2,x=-1$