1. **State the problem:** Find the number of zeros for the quadratic function given by $$y = -2(x - 3)^2 - 4$$ 2. **Recall the vertex form and zeros:** The function is in vertex form $$y = a(x - h)^2 + k$$ where the vertex is at $$(h, k)$$. 3. **Number of zeros depends on the vertex and the value of $a$:** - If $a > 0$ and $k > 0$, no zeros. - If $a > 0$ and $k = 0$, one zero. - If $a > 0$ and $k < 0$, two zeros. - If $a < 0$ and $k < 0$, no zeros. - If $a < 0$ and $k = 0$, one zero. - If $a < 0$ and $k > 0$, two zeros. 4. **Identify $a$, $h$, and $k$:** - $a = -2$ - $h = 3$ - $k = -4$ 5. Since $a = -2 < 0$ and $k = -4 < 0$, the parabola opens downward and the vertex is below the x-axis. 6. This means the parabola is entirely below the x-axis and does not cross it, so there are **no zeros**. **Final answer:** The function has **0 zeros**.