Certainly! To write the quadratic function \(f(x) = -4x^2 + 8x - 2\) in vertex form, you'll want to complete the square. Here's the process:
1. Start with the given function:
\(f(x) = -4x^2 + 8x - 2\)
2. Factor out the leading coefficient (-4) from the first two terms:
\(f(x) = -4(x^2 - 2x) - 2\)
3. To complete the square inside the parentheses, take half of the coefficient of the \(x\) term (which is -2), square it, and add it inside the parentheses. Half of -2 is -1, and (-1)^2 is 1. So, add 1 inside the parentheses and subtract 4 (since you added 4 inside the parentheses, you need to subtract 4 outside to keep the expression equivalent):
\(f(x) = -4(x^2 - 2x + 1 - 1) - 2\)
4. Now, factor the perfect square trinomial inside the parentheses:
\(f(x) = -4((x - 1)^2 - 1) - 2\)
5. Distribute the -4 through the expression inside the parentheses:
\(f(x) = -4(x - 1)^2 + 4 - 2\)
6. Simplify the constants:
\(f(x) = -4(x - 1)^2 + 2\)
So, the vertex form of the quadratic function \(f(x) = -4x^2 + 8x - 2\) is \(f(x) = -4(x - 1)^2 + 2\), and the vertex of the parabola is at the point (1, 2).