To express the quadratic function $y = -32x^2 - 32x + 3$ in vertex form, we use completing the square. The general vertex form of a quadratic function is:

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

Where $(h, k)$ is the vertex of the parabola. Let's complete the square for this equation step by step.

1. Factor out the coefficient of $x^2$:

$y = -32(x^2 + x) + 3$

2. Complete the square inside the parentheses:

   • Take half of the coefficient of $x$, which is $1/2$, and square it: $(1/2)^2 = 1/4$.
   • Add and subtract $1/4$ inside the parentheses:

$y = -32\left(x^2 + x + \frac{1}{4} - \frac{1}{4}\right) + 3$

$y = -32\left(\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right) + 3$

3. Simplify the expression:

   • Distribute the $-32$ across the terms:

$y = -32\left(x + \frac{1}{2}\right)^2 + 32 \cdot \frac{1}{4} + 3$

$y = -32\left(x + \frac{1}{2}\right)^2 + 8 + 3$

$y = -32\left(x + \frac{1}{2}\right)^2 + 11$

So, the vertex form of the quadratic equation is:

$y = -32\left(x + \frac{1}{2}\right)^2 + 11$

The vertex of the parabola is $\left(-\frac{1}{2}, 11\right)$.

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Would you like more details or have any questions? Here are some related questions for further exploration:

1. How do you find the vertex of a quadratic function directly from standard form?
2. What is the significance of the vertex in a quadratic function?
3. How does completing the square help in solving quadratic equations?
4. Can you convert a quadratic function from vertex form back to standard form?
5. How does the value of "a" in the vertex form affect the graph of a quadratic function?

Tip: When completing the square, always remember to balance the equation by adding and subtracting the same value!