Question

What are maximum and minimum turning point of y=x^2-4x+7

Answer 1

To find the turning points of the quadratic function \( y = x^2 - 4x + 7 \), we first identify if the function has a maximum or minimum turning point. Since the coefficient of \( x^2 \) is positive, this parabola opens upwards, indicating that it has a minimum turning point.

### Step 1: Find the vertex (minimum turning point)

The vertex of a quadratic function given in the form \( y = ax^2 + bx + c \) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

In our case:
- \( a = 1 \)
- \( b = -4 \)
- \( c = 7 \)

Substituting the values:

\[
x = -\frac{-4}{2 \cdot 1} = \frac{4}{2} = 2
\]

Now that we have the x-coordinate of the vertex, we can find the corresponding y-coordinate by substituting \( x = 2 \) back into the function:

\[
y = (2)^2 - 4(2) + 7 = 4 - 8 + 7 = 3
\]

Thus, the minimum turning point is at \( (2, 3) \).

### Maximum Turning Point

Since the parabola opens upwards, there is no maximum turning point; the function tends toward infinity as \( x \) moves away from the vertex.

### Summary

- **Minimum Turning Point:** \( (2, 3) \)
- **Maximum Turning Point:** None (the function has no maximum turning point).