Write an equation in vertex form of the parabola that has the same shape as the graph of $f(x)=4 x^{2}$, but with the point $(-4,2)$ as the vertex. \[ f(x)= \]

Step 1 ：The vertex form of a parabola is given by \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola.

Step 2 ：In this case, we know that the vertex is \((-4, 2)\), so \(h = -4\) and \(k = 2\).

Step 3 ：We also know that the shape of the parabola is the same as that of \(f(x) = 4x^2\), which means that the value of \(a\) is the same, i.e., \(a = 4\).

Step 4 ：Substitute \(h = -4\), \(k = 2\), and \(a = 4\) into the vertex form equation, we get \(f(x) = 4(x - -4)^2 + 2\).

Step 5 ：Simplify the equation to get the final answer: \(\boxed{f(x) = 4(x + 4)^2 + 2}\).