6.1.2 Exam: Semester 2 Exam Question 2 of 40 What are the zeros of the function \( f(x)=\frac{x^{2}+2 x-8}{x^{2}-2 x-8} \) ? \[ \begin{array}{l}\text { A. } 2,4 \\ \text { B. }-4,2 \\ \text { C. }-4,-2 \\ \text { D. }-2,4\end{array} \]

To find the zeros of the function \( f(x) = \frac{x^{2} + 2x - 8}{x^{2} - 2x - 8} \), we need to determine the values of \( x \) that make the numerator equal to zero while ensuring the denominator is not zero. 1. **Find the zeros of the numerator:** \[ x^{2} + 2x - 8 = 0 \] Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = 2\), and \(c = -8\): \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 32}}{2} = \frac{-2 \pm 6}{2} \] This gives two solutions: \[ x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-8}{2} = -4 \] 2. **Ensure the denominator is not zero at these values:** \[ x^{2} - 2x - 8 \neq 0 \] Solving: \[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm 6}{2} \] The denominator is zero at \(x = 4\) and \(x = -2\). Since our zeros \(x = 2\) and \(x = -4\) are not among these, they are valid. **Conclusion:** The zeros of the function \( f(x) \) are \( x = 2 \) and \( x = -4 \). **Answer:** **B. \(-4, 2\)**