To find the vertical asymptotes of the function
[tex]\[
f(x) = \frac{x^2 + 4}{4x^2 - 4x - 8},
\][/tex]
we need to identify the values of [tex]\(x\)[/tex] that make the denominator equal to zero. Vertical asymptotes occur where the denominator of a rational function is zero and the numerator is not zero at those points.
First, let's focus on the denominator:
[tex]\[
4x^2 - 4x - 8
\][/tex]
We set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
4x^2 - 4x - 8 = 0.
\][/tex]
Next, factor the quadratic equation if possible. For this equation, it is convenient to use 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].
Substitute the values into the quadratic formula:
[tex]\[
x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot (-8)}}{2 \cdot 4}
\][/tex]
Simplify under the square root:
[tex]\[
x = \frac{4 \pm \sqrt{16 + 128}}{8}
\][/tex]
[tex]\[
x = \frac{4 \pm \sqrt{144}}{8}
\][/tex]
[tex]\[
x = \frac{4 \pm 12}{8}
\][/tex]
This gives us two solutions:
[tex]\[
x = \frac{4 + 12}{8} = \frac{16}{8} = 2
\][/tex]
[tex]\[
x = \frac{4 - 12}{8} = \frac{-8}{8} = -1
\][/tex]
Thus, the values of [tex]\(x\)[/tex] that make the denominator zero are [tex]\(x = 2\)[/tex] and [tex]\(x = -1\)[/tex].
We need to ensure these values are not zeros of the numerator [tex]\(x^2 + 4\)[/tex]. In this case, [tex]\(x^2 + 4\)[/tex] is always positive since [tex]\(x^2\)[/tex] is non-negative and adding 4 ensures it is never zero.
Therefore, the vertical asymptotes of the function [tex]\(f(x) = \frac{x^2 + 4}{4x^2 - 4x - 8}\)[/tex] occur at:
[tex]\[
x = -1 \quad \text{and} \quad x = 2
\][/tex]
