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

Answer

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Answer

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

Steps

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 $\boxed{(2, -6)}$