What are the zeros of the given rational function below? \( f(x)=\frac{-x^{2}+16 x-63}{x^{2}-2 x-35} \) \( x=9 \) and \( x=5 \) \( x=9 \) only \( x=5 \) only No zeros possible for this function

To determine the zeros of the rational function \( f(x) = \frac{-x^{2} + 16x - 63}{x^{2} - 2x - 35} \), follow these steps: 1. **Find the Zeros of the Numerator:** Zeros occur where the numerator is equal to zero. \[ -x^{2} + 16x - 63 = 0 \] Multiply both sides by \(-1\) to simplify: \[ x^{2} - 16x + 63 = 0 \] Factor the quadratic: \[ (x - 9)(x - 7) = 0 \] Therefore, the potential zeros are \( x = 9 \) and \( x = 7 \). 2. **Check the Denominator for Restrictions:** The denominator cannot be zero, as this would make the function undefined. \[ x^{2} - 2x - 35 = 0 \] Factor the quadratic: \[ (x - 7)(x + 5) = 0 \] So, \( x = 7 \) and \( x = -5 \) are not in the domain of \( f(x) \). 3. **Determine the Valid Zeros:** - \( x = 9 \) is a valid zero since it does not make the denominator zero. - \( x = 7 \) is **not** a valid zero because it makes the denominator zero, resulting in an undefined value. **Conclusion:** The only zero of the function is \( x = 9 \). **Answer:** \( x=9 \) only