Let's analyze the given quadratic function $f(x) = -2x^2 - 6x + 5$ and solve the following:

1. Zeros: To find the zeros of the function, solve $f(x) = 0$: $-2x^2 - 6x + 5 = 0$ Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = -2$, $b = -6$, and $c = 5$.

2. Maximums/Minimums: The function is a downward-facing parabola because the coefficient of $x^2$ is negative. Therefore, it has a maximum at the vertex, which can be found using the formula for the x-coordinate of the vertex: $x_{\text{vertex}} = \frac{-b}{2a}$ Once we have $x_{\text{vertex}}$, substitute it into $f(x)$ to find the corresponding maximum value.

Let me calculate the zeros, maximum, and check if there are any minimums.Here are the solutions for the quadratic function $f(x) = -2x^2 - 6x + 5$:

1. Zeros (where $f(x) = 0$): $(-3.68, 0) \quad \text{and} \quad (0.68, 0)$

2. Maximum (since the parabola opens downwards, the vertex is a maximum point): $(-1.5, 9.5)$

3. Minimum: There is no minimum because the parabola opens downward, meaning there is no lower bound.

Let me know if you'd like more details or have further questions! Here are some additional related questions to explore:

1. How do you derive the quadratic formula?
2. How do you determine whether a parabola opens upwards or downwards?
3. How would you find the zeros of a function graphically?
4. What happens to the vertex if we change the coefficients $a$, $b$, and $c$?
5. How do you find the range of a quadratic function?

Tip: Always check the sign of the coefficient of $x^2$ to determine if a parabola has a maximum (downward-facing) or minimum (upward-facing).