1. **State the problem:** We are given the quadratic function $f(x) = -2x^2 + 3x - 25$ and need to analyze it. 2. **Find the domain:** The domain of any quadratic function is all real numbers, so the domain is $\mathbb{R}$. 3. **Symmetry:** Quadratic functions are symmetric about their vertex. The axis of symmetry is given by the formula: $$x = -\frac{b}{2a}$$ where $a = -2$ and $b = 3$. 4. **Calculate the axis of symmetry:** $$x = -\frac{3}{2 \times (-2)} = -\frac{3}{-4} = \frac{3}{4}$$ 5. **Find the vertex:** Substitute $x = \frac{3}{4}$ into $f(x)$: $$f\left(\frac{3}{4}\right) = -2\left(\frac{3}{4}\right)^2 + 3\left(\frac{3}{4}\right) - 25$$ $$= -2 \times \frac{9}{16} + \frac{9}{4} - 25 = -\frac{18}{16} + \frac{9}{4} - 25$$ $$= -\frac{9}{8} + \frac{9}{4} - 25 = -1.125 + 2.25 - 25 = -23.875$$ So the vertex is at $\left(\frac{3}{4}, -23.875\right)$. 6. **Concavity:** Since $a = -2 < 0$, the parabola opens downward. 7. **Find the y-intercept:** Set $x=0$: $$f(0) = -2 \times 0 + 3 \times 0 - 25 = -25$$ 8. **Find the x-intercepts:** Solve $-2x^2 + 3x - 25 = 0$ using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4(-2)(-25)}}{2(-2)}$$ $$= \frac{-3 \pm \sqrt{9 - 200}}{-4} = \frac{-3 \pm \sqrt{-191}}{-4}$$ Since the discriminant is negative, there are no real x-intercepts. **Final answers:** - Domain: $\mathbb{R}$ - Axis of symmetry: $x = \frac{3}{4}$ - Vertex: $\left(\frac{3}{4}, -23.875\right)$ - Concavity: Downward - Y-intercept: $(0, -25)$ - No real x-intercepts