To rewrite the given quadratic function $f(x) = -x^2 - 4x + 2$ in standard form, we need to express it as:

$f(x) = a(x - h)^2 + k$

where $a$, $h$, and $k$ are constants, and $(h, k)$ represents the vertex of the parabola.

Step-by-step process:

1. Factor out the coefficient of the quadratic term (if necessary): In this case, the coefficient of $x^2$ is already $-1$, so we proceed without factoring for now.

   $f(x) = - (x^2 + 4x) + 2$

2. Complete the square: To complete the square inside the parentheses, take half of the coefficient of $x$ (which is $4$), square it, and add/subtract it inside the equation.

   Half of $4$ is $2$, and $2^2 = 4$. So, we add and subtract $4$ inside the parentheses:

   $f(x) = - (x^2 + 4x + 4 - 4) + 2$

   Simplifying this gives:

   $f(x) = - \left[(x + 2)^2 - 4\right] + 2$

3. Simplify the expression: Distribute the negative sign and simplify:

   $f(x) = - (x + 2)^2 + 4 + 2$

   $f(x) = - (x + 2)^2 + 6$

Final standard form:

$f(x) = - (x + 2)^2 + 6$

This is the standard form of the quadratic function. Here:

• $a = -1$
• $h = -2$
• $k = 6$

The vertex of the parabola is at $(-2, 6)$.

Would you like further details or have any questions?

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Here are 5 questions to expand on this concept:

1. How do you find the vertex of a parabola given the standard form?
2. What does the sign of the coefficient $a$ tell us about the graph of a quadratic function?
3. How does completing the square help in graphing quadratic functions?
4. Can we convert a quadratic function in standard form back to its general form?
5. How do transformations of $f(x) = x^2$ relate to changes in the quadratic function?

Tip: Completing the square is a useful technique not only for rewriting quadratics but also for solving quadratic equations!