We are given the quadratic function:

$y = x^2 - 6x - 1$

We will complete the square to rewrite it in vertex form. The general form of a quadratic function is:

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

where $(h, k)$ is the vertex of the parabola. Here’s the step-by-step process to complete the square:

Step 1: Group the $x^2$ and $x$ terms together.

We need to focus on the quadratic and linear terms: $x^2 - 6x$. Keep the constant term ($-1$) aside for now.

$y = (x^2 - 6x) - 1$

Step 2: Complete the square for the expression $x^2 - 6x$.

To complete the square, take half of the coefficient of $x$ (which is $-6$), square it, and add it inside the parentheses.

1. Half of $-6$ is $-3$.
2. Square $-3$ to get $9$.

So, we add and subtract $9$ inside the parentheses:

$y = (x^2 - 6x + 9 - 9) - 1$

Step 3: Simplify the equation.

Now, we can rewrite the expression:

$y = (x - 3)^2 - 9 - 1$

$y = (x - 3)^2 - 10$

Final Answer:

The quadratic function in vertex form is:

$y = (x - 3)^2 - 10$

This shows that the vertex of the parabola is $(3, -10)$.

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Would you like further details or have any questions? Here are 5 related questions for you:

1. How would you find the vertex of a parabola from its equation in standard form?
2. What is the significance of the coefficient $a$ in the vertex form of a quadratic equation?
3. How do you determine whether the parabola opens upwards or downwards from its vertex form?
4. How can you solve quadratic equations by completing the square?
5. What is the relationship between the discriminant and the number of real solutions to a quadratic equation?

Tip: Remember that completing the square works best when the coefficient of $x^2$ is $1$. If it’s not, you may need to factor it out first before completing the square.