To determine the vertical asymptotes of the function
[tex]\[
f(x)=\frac{x^2+4}{4 x^2-4 x-8},
\][/tex]
we need to identify the values of [tex]\( x \)[/tex] that make the denominator equal to zero since these will cause the function to be undefined.
First, we start with the denominator:
[tex]\[
4 x^2 - 4 x - 8 = 0.
\][/tex]
This is a quadratic equation, and we can solve for [tex]\( x \)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\][/tex]
where [tex]\( a = 4 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -8 \)[/tex].
First, calculate the discriminant:
[tex]\[
\text{Discriminant} = b^2 - 4ac = (-4)^2 - 4 \cdot 4 \cdot (-8) = 16 + 128 = 144.
\][/tex]
Next, take the square root of the discriminant:
[tex]\[
\sqrt{\text{Discriminant}} = \sqrt{144} = 12.
\][/tex]
Now, apply the quadratic formula:
[tex]\[
x = \frac{-(-4) \pm 12}{2 \cdot 4} = \frac{4 \pm 12}{8}.
\][/tex]
This gives us two solutions:
[tex]\[
x_1 = \frac{4 + 12}{8} = \frac{16}{8} = 2,
\][/tex]
[tex]\[
x_2 = \frac{4 - 12}{8} = \frac{-8}{8} = -1.
\][/tex]
Hence, the vertical asymptotes of the function [tex]\( f(x) \)[/tex] are [tex]\( x = 2 \)[/tex] and [tex]\( x = -1 \)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{x=-1 \text{ and } x=2}
\][/tex]
