Sure, let's analyze the function [tex]\( f(x) = \frac{x^2 + 4}{4x^2 - 4x - 8} \)[/tex] to find its vertical and horizontal asymptotes.
### Vertical Asymptotes
To find the vertical asymptotes of the function, we need to determine where the function is undefined, i.e., where the denominator equals zero.
The denominator is [tex]\( 4x^2 - 4x - 8 \)[/tex]. We set this equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
4x^2 - 4x - 8 = 0
\][/tex]
We can simplify this quadratic equation by dividing all terms by 4:
[tex]\[
x^2 - x - 2 = 0
\][/tex]
Next, we factor the quadratic equation:
[tex]\[
x^2 - x - 2 = (x - 2)(x + 1) = 0
\][/tex]
Setting each factor equal to zero gives the solutions:
[tex]\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\][/tex]
[tex]\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\][/tex]
Therefore, the vertical asymptotes are:
[tex]\[
x = 2 \quad \text{and} \quad x = -1
\][/tex]
### Horizontal Asymptotes
To find the horizontal asymptotes, we compare the degrees of the numerator and the denominator.
The degree of the numerator [tex]\( x^2 + 4 \)[/tex] is 2.
The degree of the denominator [tex]\( 4x^2 - 4x - 8 \)[/tex] is also 2.
When the degrees of the numerator and the denominator are the same, the horizontal asymptote is given by the ratio of the leading coefficients. The leading coefficient of the numerator [tex]\( x^2 + 4 \)[/tex] is 1, and the leading coefficient of the denominator [tex]\( 4x^2 - 4x - 8 \)[/tex] is 4.
Thus, the horizontal asymptote is:
[tex]\[
y = \frac{1}{4}
\][/tex]
### Summary
- The vertical asymptotes are [tex]\( x = 2 \)[/tex] and [tex]\( x = -1 \)[/tex].
- The horizontal asymptote is [tex]\( y = \frac{1}{4} \)[/tex].
