To identify the values of [tex]\(x\)[/tex] where the function [tex]\( \frac{x^2 + 8x + 4}{x^2 - x - 6} \)[/tex] is discontinuous, we need to find the points where the denominator equals zero. This is because a function is discontinuous at values that make the denominator zero (if the numerator is not also zero at those points).
The denominator of the function is [tex]\(x^2 - x - 6\)[/tex].
We start by finding the roots of the denominator:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
To solve this quadratic equation, we can factor it. We look for two numbers that multiply to [tex]\(-6\)[/tex] (the constant term) and add up to [tex]\(-1\)[/tex] (the coefficient of the [tex]\(x\)[/tex] term).
The factors of [tex]\(-6\)[/tex] that add up to [tex]\(-1\)[/tex] are [tex]\(-3\)[/tex] and [tex]\(2\)[/tex]. So, we can factor the quadratic as follows:
[tex]\[ (x - 3)(x + 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \][/tex]
[tex]\[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \][/tex]
Therefore, the function [tex]\( \frac{x^2 + 8x + 4}{x^2 - x - 6} \)[/tex] is discontinuous at [tex]\( x = -2 \)[/tex] and [tex]\( x = 3 \)[/tex].
Among the given options:
- [tex]\(x = -1\)[/tex]
- [tex]\(x = -4\)[/tex]
- [tex]\(x = -3\)[/tex]
- [tex]\(x = -2\)[/tex]
The correct value that is a discontinuity of the function is [tex]\( x = -2 \)[/tex].
