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To determine the critical points of the given function \( y = x^2 - 6x + 7 \), we need to follow these steps:

### Step 1: Find the first derivative
The first derivative of a function helps locate the critical points, where the slope is zero or undefined. Let's differentiate the given function \( y \) with respect to \( x \):

\[
y' = \frac{d}{dx} (x^2 - 6x + 7)
\]

Using basic differentiation rules, we get:

\[
y' = 2x - 6
\]

### Step 2: Set the derivative equal to zero
To find critical points, set the first derivative equal to zero and solve for \( x \):

\[
2x - 6 = 0
\]

Solving for \( x \):

\[
2x = 6 \quad \Rightarrow \quad x = 3
\]

Thus, the critical point occurs at \( x = 3 \).

### Step 3: Verify the nature of the critical point
To confirm if the critical point is a minimum or maximum, we can either use the second derivative test or analyze the behavior of the function around \( x = 3 \).

The second derivative of \( y = x^2 - 6x + 7 \) is:

\[
y'' = \frac{d}{dx}(2x - 6) = 2
\]

Since \( y'' = 2 > 0 \), the function is concave up, indicating that the critical point at \( x = 3 \) is a **local minimum**.

### Step 4: Domain Endpoints
Since the given function is a quadratic polynomial, its domain is all real numbers, \( (-\infty, \infty) \). Therefore, there are no specific domain endpoints to consider.

### Summary:
- **Critical Point**: \( x = 3 \)
  - At this point, \( y(3) = 3^2 - 6(3) + 7 = 9 - 18 + 7 = -2 \).
  - So, the critical point is \( (3, -2) \), and it's a local minimum.
- **Domain**: The domain is all real numbers.

Would you like more details on any step? Feel free to ask questions!

Here are 5 follow-up questions related to this:

1. What is the significance of the second derivative test in determining the concavity of the function?
2. How would you approach finding critical points for a more complex function, such as a cubic or quartic polynomial?
3. Can a quadratic function have more than one critical point? Why or why not?
4. What are other methods for classifying critical points besides the second derivative test?
5. How would the graph of \( y = x^2 - 6x + 7 \) look, and what does it tell us about the critical point?

**Tip**: When working with quadratic functions, always remember that they represent parabolas, and the second derivative tells you whether the parabola opens upward or downward!

Determine all critical points and all domain endpoints for the function y = x^2 - 6x + 7.

Learn how to determine the critical points and domain endpoints for the quadratic function y = x^2 - 6x + 7. This problem involves calculus concepts like first and second derivatives and is suitable for Grades 10-12.

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