1. **State the problem:** We are given the quadratic function $y = -4x^2 + 32x - 57$ and need to analyze it. 2. **Formula and rules:** A quadratic function is generally written as $y = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. 3. **Identify coefficients:** Here, $a = -4$, $b = 32$, and $c = -57$. 4. **Find the vertex:** The vertex of a parabola $y = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Calculate: $$x = -\frac{32}{2 \times -4} = -\frac{32}{-8} = 4$$ 5. **Find the y-coordinate of the vertex:** Substitute $x=4$ into the function: $$y = -4(4)^2 + 32(4) - 57 = -4(16) + 128 - 57 = -64 + 128 - 57 = 7$$ So, the vertex is at $(4, 7)$. 6. **Determine the direction of the parabola:** Since $a = -4 < 0$, the parabola opens downward. 7. **Find the y-intercept:** Set $x=0$: $$y = -4(0)^2 + 32(0) - 57 = -57$$ So, the y-intercept is $(0, -57)$. 8. **Find the x-intercepts:** Solve $-4x^2 + 32x - 57 = 0$. Divide both sides by $-1$ for simplicity: $$4x^2 - 32x + 57 = 0$$ Use the quadratic formula: $$x = \frac{32 \pm \sqrt{(-32)^2 - 4 \times 4 \times 57}}{2 \times 4} = \frac{32 \pm \sqrt{1024 - 912}}{8} = \frac{32 \pm \sqrt{112}}{8}$$ Simplify $\sqrt{112} = \sqrt{16 \times 7} = 4\sqrt{7}$: $$x = \frac{32 \pm 4\sqrt{7}}{8} = 4 \pm \frac{\sqrt{7}}{2}$$ So, the x-intercepts are: $$\left(4 + \frac{\sqrt{7}}{2}, 0\right) \text{ and } \left(4 - \frac{\sqrt{7}}{2}, 0\right)$$ **Final answer:** - Vertex: $(4, 7)$ - Parabola opens downward - Y-intercept: $(0, -57)$ - X-intercepts: $\left(4 + \frac{\sqrt{7}}{2}, 0\right)$ and $\left(4 - \frac{\sqrt{7}}{2}, 0\right)$