Consider the equations below. \[ \begin{array}{ll}\text { A. } y=-2 x^{2}-8 x \\ \text { B. } y=2(x+6)(x-2) \\ \text { C. } y=-2 x^{2}-8 x+16\end{array} \] Use the equations to complete the following statements. Equation \( v \) D. \( y=2(x+4)^{2}-2 \) \( \begin{array}{l}\text { reveals its extreme value without needing to be altered. The extreme value of this equation has a } \square \text { at the }\end{array} \)

The equation in statement D, \( y=2(x+4)^{2}-2 \), is in vertex form, allowing us to easily identify its extreme value. Here, the vertex, which also represents the minimum point of the parabola since it opens upwards (the coefficient of \( (x+4)^2 \) is positive), occurs at \( (-4, -2) \). Thus, the extreme value of this equation has a **minimum** at the **point (-4, -2)**. Moving on to the first three equations, equation A has a maximum since it opens downwards, while equation B opens upwards and has no extreme value displayed directly; it's a standard form that could be converted to vertex form for analysis. Equation C, also a downward-opening parabola, similarly has a maximum but would require completing the square to reveal its vertex.