To find the maximum or minimum for the function $y = x^2 - 4x - 1$, we can use calculus or complete the square. Let's proceed with the completing the square method, which is often simpler for quadratic functions.

Step 1: Write the quadratic expression in a way that reveals the vertex.

Start with the equation: $y = x^2 - 4x - 1$ We want to complete the square for the $x^2 - 4x$ part.

1. Take half of the coefficient of $x$ (which is -4), square it, and then add and subtract it inside the expression.
   • Half of -4 is $-2$.
   • Squaring $-2$ gives $4$.

So we rewrite the expression as: $y = (x^2 - 4x + 4) - 4 - 1$ This simplifies to: $y = (x - 2)^2 - 5$

Step 2: Identify the vertex.

The quadratic is now in the form $y = (x - h)^2 + k$, where $(h, k)$ is the vertex of the parabola. From our equation: $y = (x - 2)^2 - 5$ The vertex is at $(2, -5)$.

Step 3: Determine whether it's a maximum or minimum.

Since the coefficient of $(x - 2)^2$ is positive (1), the parabola opens upwards, indicating that this is a minimum point.

Final Answer:

The function has a minimum at $x = 2$ with a minimum value of $y = -5$.

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Would you like more details or have any questions on this? Here are some related questions to explore:

1. How do you find the vertex of a parabola from its general quadratic form?
2. What happens to the vertex if the coefficient of $x^2$ is negative?
3. How do you use derivatives to find the maximum or minimum of a function?
4. Can the function $y = x^2 - 4x - 1$ have an inflection point?
5. How can you graph this quadratic equation?

Tip: Always check the coefficient of $x^2$ to determine if the parabola opens upwards (minimum) or downwards (maximum).