1. **State the problem:** Identify the vertex, axis of symmetry, min/max value, and y-intercept of the quadratic function $$y = -\frac{1}{4}x^2 - 3x - 10$$. 2. **Recall the standard form and formulas:** The quadratic function in standard form is $$y = ax^2 + bx + c$$. - Vertex formula: $$x = -\frac{b}{2a}$$. - Axis of symmetry: the vertical line $$x = -\frac{b}{2a}$$. - Vertex coordinates: $$\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$$. - Min/Max value: the y-coordinate of the vertex; if $$a < 0$$, the parabola opens downward and vertex is a maximum; if $$a > 0$$, it opens upward and vertex is a minimum. - Y-intercept: value of $$y$$ when $$x=0$$, which is $$c$$. 3. **Identify coefficients:** $$a = -\frac{1}{4}$$, $$b = -3$$, $$c = -10$$. 4. **Calculate axis of symmetry:** $$x = -\frac{b}{2a} = -\frac{-3}{2 \times -\frac{1}{4}} = -\frac{-3}{-\frac{1}{2}} = -\frac{-3}{-0.5} = -6$$ 5. **Calculate vertex y-coordinate:** $$y = -\frac{1}{4}(-6)^2 - 3(-6) - 10 = -\frac{1}{4} \times 36 + 18 - 10 = -9 + 18 - 10 = -1$$ 6. **Vertex:** $$(-6, -1)$$. 7. **Determine min/max:** Since $$a = -\frac{1}{4} < 0$$, parabola opens downward, so vertex is a maximum point with value $$-1$$. 8. **Y-intercept:** When $$x=0$$, $$y = -10$$, so y-intercept is $$(0, -10)$$. **Final answers:** - Vertex: $$(-6, -1)$$ - Axis of symmetry: $$x = -6$$ - Max value: $$-1$$ - Y-intercept: $$(0, -10)$$