Answer:

The graph of the quadratic function \(y = -2x^2 + 8x + 5\) is a downward-opening parabola.

- Vertex: The vertex of this parabola can be found using the formula \(x = \frac{-b}{2a}\), where \(a = -2\) and \(b = 8\) in the equation \(y = ax^2 + bx + c\). Thus, the x-coordinate of the vertex is \(x = \frac{-8}{2(-2)} = 2\). To find the corresponding y-coordinate, substitute \(x = 2\) into the equation to get \(y = -2(2)^2 + 8(2) + 5 = 9\). Therefore, the vertex is at (2, 9).

- Domain: The domain of a quadratic function is all real numbers.

- Range: Since the parabola opens downwards, the range is \(\text{y} \leq 9\) (or \(\text{y} \geq 9\) if written in interval notation).

- Y-intercept: To find the y-intercept, set \(x = 0\) in the equation. Thus, \(y = -2(0)^2 + 8(0) + 5 = 5\). Therefore, the y-intercept is at (0, 5).

- Opening: The parabola opens downward because the coefficient of \(x^2\) is negative.

- Vertex Type: Since the parabola opens downward and the vertex is the highest point, it represents a maximum point.

- Axis of Symmetry: The equation of the axis of symmetry for a parabola is \(x = \frac{-b}{2a}\), which in this case is \(x = \frac{-8}{2(-2)} = 2\). Therefore, the equation of the axis of symmetry is \(x = 2\).

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