1. **State the problem:** We are given the quadratic function $$y = -x^2 - 6x - 3$$ and want to analyze its properties such as vertex, intercepts, and graph shape. 2. **Formula and rules:** The general form of a quadratic function is $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. - The vertex of the parabola is at $$x = -\frac{b}{2a}$$. - The parabola opens upward if $a > 0$ and downward if $a < 0$. - The y-intercept is at $y = c$. - The x-intercepts (roots) are found by solving $$ax^2 + bx + c = 0$$. 3. **Identify coefficients:** Here, $a = -1$, $b = -6$, and $c = -3$. 4. **Find the vertex:** $$x = -\frac{b}{2a} = -\frac{-6}{2 \times -1} = -\frac{-6}{-2} = -3$$ Calculate $y$ at $x = -3$: $$y = -(-3)^2 - 6(-3) - 3 = -9 + 18 - 3 = 6$$ So the vertex is at $$(-3, 6)$$. 5. **Find the y-intercept:** At $x=0$, $$y = -0 - 0 - 3 = -3$$ So the y-intercept is at $$(0, -3)$$. 6. **Find the x-intercepts:** Solve $$-x^2 - 6x - 3 = 0$$ or equivalently $$x^2 + 6x + 3 = 0$$ (multiply both sides by -1). Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-6 \pm \sqrt{36 - 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} = \frac{-6 \pm 2\sqrt{6}}{2} = -3 \pm \sqrt{6}$$ So the x-intercepts are approximately: $$x = -3 + 2.45 = -0.55$$ and $$x = -3 - 2.45 = -5.45$$. 7. **Graph shape:** Since $a = -1 < 0$, the parabola opens downward. **Final answer:** The parabola $$y = -x^2 - 6x - 3$$ has vertex at $$(-3, 6)$$, y-intercept at $$(0, -3)$$, and x-intercepts at approximately $$(-0.55, 0)$$ and $$(-5.45, 0)$$. It opens downward.