告诉我这道题怎么做---In the Cartesian coordinate system, the parabola $y = ax^2 + bx$ (where $a, b$ are constants and $a \neq 0$) has its vertex at $(-2, -4)$. (1) Find the values of $a$ and $b$. (2) Find the coordinates of the intersection points of the parabola and the $x$-axis. (3) Two distinct points $M, B$ are on this parabola, the x-coordinate of point $M$ is $m$. $MB \perp y$-axis, point $D$ is the reflection of point $B$ about point $M$. Connect $AB, AD$. Point $A$ has coordinates $(2m + 4, -2m - 4)$. Construct parallelogram $\square ABCD$ with adjacent sides $AB, AD$. ① When point $M$ is to the right of the parabola's axis of symmetry, and point $D$ lies on the coordinate axis, find the area of $\square ABCD$. ② When the part of the parabola inside $\square ABCD$ (excluding the boundary of $\square ABCD$) has its $y$ value increasing as $x$ increases or decreasing as $x$ increases, directly write the range of $m$. Handwritten notes: $y = a(x+2)^2 - 4$ $(0,0)$ substitute $\downarrow$ (1) $a=1, b=4$ $4a - 4 = 0$ (2) $(-4,0), (0,0)$ $x = -2$