一条二次曲线的对称轴为 $x = -2$，且对于所有的 $x$，该曲线都位于 $x$ 轴上方。已知点 $(-1, 4)$ 在该曲线上，求该曲线的一个可能方程，写成 $y = a(x - h)^2 + k$ 的形式，其中 $a$、$h$ 和 $k$ 是整数。

1. Identify the value of $h$ from the line of symmetry.

The line of symmetry is $x = -2$, so $h = -2$.
The equation is of the form $y = a(x + 2)^2 + k$.

2. Substitute the given point to find a relationship between $a$ and $k$.

Substitute $x = -1$ and $y = 4$ into the equation:
$4 = a(-1 + 2)^2 + k$
$4 = a(1)^2 + k$
$a + k = 4$

3. Use the condition that the curve is always above the $x$-axis to restrict $a$ and $k$.

Since the curve lies above the $x$-axis for all $x$, it must open upwards ($a > 0$) and its minimum value must be positive ($k > 0$).
Given $a$ and $k$ are integers, and $a + k = 4$, the possible integer pairs for $(a, k)$ are $(1, 3)$, $(2, 2)$, and $(3, 1)$.

4. State a possible equation.

Choosing $a = 1$ and $k = 3$:
$y = (x + 2)^2 + 3$