1. **State the problem:** We are given the quadratic function $f(x) = -2x^2 - 4x + 5$ and want to analyze it. 2. **Formula and rules:** A quadratic function is generally written as $f(x) = ax^2 + bx + c$ where $a$, $b$, and $c$ are constants. Here, $a = -2$, $b = -4$, and $c = 5$. 3. **Find the vertex:** The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is at $x = -\frac{b}{2a}$. Calculate: $$x = -\frac{-4}{2 \times -2} = -\frac{-4}{-4} = -1$$ 4. **Find the vertex's y-coordinate:** Substitute $x = -1$ into $f(x)$: $$f(-1) = -2(-1)^2 - 4(-1) + 5 = -2(1) + 4 + 5 = -2 + 4 + 5 = 7$$ So the vertex is at $(-1, 7)$. 5. **Determine the direction of the parabola:** Since $a = -2 < 0$, the parabola opens downward, so the vertex is a maximum point. 6. **Find the x-intercepts:** Solve $f(x) = 0$: $$-2x^2 - 4x + 5 = 0$$ Divide both sides by $-1$ to simplify: $$\cancel{-}2x^2 - \cancel{-}4x + \cancel{5} = \cancel{0}$$ Actually, dividing by $-1$: $$2x^2 + 4x - 5 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{4^2 - 4 \times 2 \times (-5)}}{2 \times 2} = \frac{-4 \pm \sqrt{16 + 40}}{4} = \frac{-4 \pm \sqrt{56}}{4}$$ Simplify $\sqrt{56} = \sqrt{4 \times 14} = 2\sqrt{14}$: $$x = \frac{-4 \pm 2\sqrt{14}}{4} = \frac{\cancel{-4} \pm \cancel{2}\sqrt{14}}{\cancel{4}} = \frac{-2 \pm \sqrt{14}}{2}$$ So the x-intercepts are: $$x = \frac{-2 + \sqrt{14}}{2} \quad \text{and} \quad x = \frac{-2 - \sqrt{14}}{2}$$ 7. **Find the y-intercept:** Set $x=0$: $$f(0) = -2(0)^2 - 4(0) + 5 = 5$$ **Final summary:** - Vertex at $(-1, 7)$ (maximum point) - Opens downward - X-intercepts at $\frac{-2 + \sqrt{14}}{2}$ and $\frac{-2 - \sqrt{14}}{2}$ - Y-intercept at $(0, 5)$