1. **State the problem:** We are given the quadratic function $$y = -x^2 - 6x - 10$$ and want to analyze its properties such as shape, vertex, and intercepts. 2. **Formula and rules:** A quadratic function is generally written as $$y = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. Here, $a = -1$, $b = -6$, and $c = -10$. 3. **Shape of the graph:** Since $a = -1 < 0$, the parabola opens downward. 4. **Find the vertex:** The vertex $x$-coordinate is given by $$x = -\frac{b}{2a} = -\frac{-6}{2 \times -1} = -\frac{-6}{-2} = -3.$$ 5. **Calculate the vertex $y$-coordinate:** Substitute $x = -3$ into the function: $$y = -(-3)^2 - 6(-3) - 10 = -9 + 18 - 10 = -1.$$ So the vertex is at $$(-3, -1).$$ 6. **Find the $y$-intercept:** Set $x=0$: $$y = -0 - 0 - 10 = -10.$$ So the $y$-intercept is at $$(0, -10).$$ 7. **Find the $x$-intercepts:** Set $y=0$ and solve for $x$: $$0 = -x^2 - 6x - 10$$ Multiply both sides by $-1$: $$0 = x^2 + 6x + 10$$ Use the quadratic formula: $$x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 10}}{2 \times 1} = \frac{-6 \pm \sqrt{36 - 40}}{2} = \frac{-6 \pm \sqrt{-4}}{2}.$$ Since the discriminant is negative, there are no real $x$-intercepts. **Final answer:** The parabola opens downward with vertex at $$(-3, -1)$$, $y$-intercept at $$(0, -10)$$, and no real $x$-intercepts.