What are the zeros of the given rational function below? \( f(x)=\frac{x^{3}-4 x^{2}}{x^{2}-9} \) \( x=0 \) only \( x=4 \) only \( x=0 \) and \( x=4 \) No zeros possible for this function

To determine the zeros of the rational function \( f(x) = \frac{x^{3} - 4x^{2}}{x^{2} - 9} \), we need to find the values of \( x \) that make the numerator zero while ensuring the denominator is not zero. 1. **Find the zeros of the numerator:** \[ x^{3} - 4x^{2} = 0 \\ x^{2}(x - 4) = 0 \] This gives: \[ x = 0 \quad \text{and} \quad x = 4 \] 2. **Check the denominator for these values:** \[ x^{2} - 9 = (x + 3)(x - 3) \] - For \( x = 0 \): \[ 0^{2} - 9 = -9 \neq 0 \] - For \( x = 4 \): \[ 4^{2} - 9 = 16 - 9 = 7 \neq 0 \] Both \( x = 0 \) and \( x = 4 \) make the numerator zero without making the denominator zero. Therefore, these are the zeros of the function. **Answer:** \( x = 0 \) and \( x = 4 \)