Find the vertex. \[ f(x)=4 x^{2}-16 x+10 \] $([?],[])$

Step 1 ：Given the function \(f(x)=4 x^{2}-16 x+10\)

Step 2 ：The vertex of a parabola given by the equation \(f(x) = ax^2 + bx + c\) is given by the point \(-\frac{b}{2a}, f(-\frac{b}{2a})\)

Step 3 ：In this case, \(a = 4\) and \(b = -16\), so the x-coordinate of the vertex is \(-\frac{-16}{2*4} = 2\)

Step 4 ：We can substitute \(x = 2\) into the equation to find the y-coordinate of the vertex

Step 5 ：The vertex of the function \(f(x)=4 x^{2}-16 x+10\) is \(\boxed{(2, -6)}\)