Question

How to Identified maximum and minimum turning point of y=x^2-4x+7

Answer 1

To identify the maximum and minimum turning points of the quadratic function \( y = x^2 - 4x + 7 \), we follow these steps:

1. **Identify the coefficients**: The general form of a quadratic equation is \( y = ax^2 + bx + c \). Here, we have:
- \( a = 1 \)
- \( b = -4 \)
- \( c = 7 \)

2. **Determine the vertex**: The turning point (vertex) of a quadratic function can be found using the formula:
\[
x = -\frac{b}{2a}
\]
Substituting the values of \( a \) and \( b \):
\[
x = -\frac{-4}{2 \times 1} = \frac{4}{2} = 2
\]

3. **Calculate the \( y \)-coordinate of the vertex**: Substitute \( x = 2 \) back into the original equation to find the corresponding \( y \)-coordinate:
\[
y = (2)^2 - 4(2) + 7
\]
Simplifying this:
\[
y = 4 - 8 + 7 = 3
\]

4. **Determine the nature of the turning point**: Since the coefficient \( a = 1 \) is positive, the parabola opens upwards, indicating that the vertex at \( (2, 3) \) is a minimum turning point.

5. **Conclusion**: The quadratic function \( y = x^2 - 4x + 7 \) has a minimum turning point at \( (2, 3) \). There is no maximum turning point for this function since it opens upwards.

### Summary
- Minimum turning point: \( (2, 3) \)
- Maximum turning point: None (the function opens upwards).