To find the vertical asymptotes of the function [tex]\( f(x) = \frac{2x^2 + 3x + 6}{x^2 - 1} \)[/tex], we must determine where the denominator of the function is equal to zero. Vertical asymptotes occur at the values of [tex]\( x \)[/tex] that make the denominator zero, provided that these values do not make the numerator zero simultaneously.
The denominator of [tex]\( f(x) \)[/tex] is [tex]\( x^2 - 1 \)[/tex]. We need to solve the equation [tex]\( x^2 - 1 = 0 \)[/tex].
Step-by-step, we solve:
1. Set the denominator equal to zero:
[tex]\[
x^2 - 1 = 0
\][/tex]
2. Factorize the quadratic equation:
[tex]\[
x^2 - 1 = (x + 1)(x - 1)
\][/tex]
3. Set each factor equal to zero:
[tex]\[
x + 1 = 0 \quad \text{or} \quad x - 1 = 0
\][/tex]
4. Solve for [tex]\( x \)[/tex]:
[tex]\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\][/tex]
[tex]\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\][/tex]
Hence, the values of [tex]\( x \)[/tex] at which the denominator becomes zero are [tex]\( x = -1 \)[/tex] and [tex]\( x = 1 \)[/tex].
Therefore, the vertical asymptotes of the function [tex]\( f(x) = \frac{2x^2 + 3x + 6}{x^2 - 1} \)[/tex] are at:
[tex]\[
x = -1 \quad \text{and} \quad x = 1
\][/tex]
So, the correct answer is:
[tex]\[ x = -1, \, 1 \][/tex]
