What are the \( x \)-intercepts of the given rational function below? \( f(x)=\frac{16 x(x+3)(x+6)}{x^{2}-36} \) \( x=0 \) and \( x=-3 \) \( x=0, x=-3 \), and \( x=-6 \) \( x=-6 \) \( x=0 \) only x

To determine the \( x \)-intercepts of the rational function \( f(x) = \frac{16x(x+3)(x+6)}{x^2 - 36} \), follow these steps: 1. **Find the zeros of the numerator:** The \( x \)-intercepts occur where \( f(x) = 0 \), which happens when the numerator is zero (provided the denominator is not zero at those points). \[ 16x(x + 3)(x + 6) = 0 \] - \( x = 0 \) - \( x + 3 = 0 \Rightarrow x = -3 \) - \( x + 6 = 0 \Rightarrow x = -6 \) 2. **Identify where the denominator is zero:** The function is undefined where the denominator is zero, so these points cannot be \( x \)-intercepts. \[ x^2 - 36 = 0 \Rightarrow (x - 6)(x + 6) = 0 \] - \( x = 6 \) - \( x = -6 \) 3. **Determine valid \( x \)-intercepts:** - \( x = 0 \) and \( x = -3 \) make the numerator zero and do not make the denominator zero, so they are valid \( x \)-intercepts. - \( x = -6 \) makes both the numerator and denominator zero, which means there is a hole in the graph at \( x = -6 \), not an \( x \)-intercept. **Conclusion:** The \( x \)-intercepts of the function are at \( x = 0 \) and \( x = -3 \). **Answer:** \( x=0 \) and \( x=-3 \)