1. **State the problem:** We are given two quadratic functions: $$y = -2x^2 - 8x - 8$$ $$y = 4x^2 - 8x - 6$$ We need to analyze their shapes, find their vertices, and understand their graph behavior. 2. **Recall the vertex formula:** For a quadratic function in the form $$y = ax^2 + bx + c$$, the vertex \((h, k)\) is found by: $$h = -\frac{b}{2a}$$ $$k = f(h) = a h^2 + b h + c$$ 3. **Analyze the first function:** $$y = -2x^2 - 8x - 8$$ - Here, $$a = -2$$, $$b = -8$$, $$c = -8$$. - Calculate $$h$$: $$h = -\frac{-8}{2 \times -2} = -\frac{-8}{-4} = -2$$ - Calculate $$k$$: $$k = -2(-2)^2 - 8(-2) - 8 = -2(4) + 16 - 8 = -8 + 16 - 8 = 0$$ - So, the vertex is at $$(-2, 0)$$. - Since $$a = -2 < 0$$, the parabola opens downward. 4. **Analyze the second function:** $$y = 4x^2 - 8x - 6$$ - Here, $$a = 4$$, $$b = -8$$, $$c = -6$$. - Calculate $$h$$: $$h = -\frac{-8}{2 \times 4} = \frac{8}{8} = 1$$ - Calculate $$k$$: $$k = 4(1)^2 - 8(1) - 6 = 4 - 8 - 6 = -10$$ - So, the vertex is at $$(1, -10)$$. - Since $$a = 4 > 0$$, the parabola opens upward. 5. **Summary:** - The first parabola opens downward with vertex at $$(-2, 0)$$. - The second parabola opens upward with vertex at $$(1, -10)$$. - Both are quadratic functions with standard parabolic shapes. 6. **Graph description:** - The first parabola peaks at $$(-2, 0)$$ and opens downward. - The second parabola has its minimum at $$(1, -10)$$ and opens upward. - These vertices help position the graphs on the coordinate plane, with the first parabola higher and to the left, and the second lower and to the right.