f(x)=4x^2-50x+126 
The given equation defines the function f. For what value of x
 does f(x)
 reach its minimum?

To find the value of \( x \) at which the function \( f(x) = 4x^2 - 50x + 126 \) reaches its minimum, we need to use the concept of the vertex of a parabola. 

The function \( f(x) = ax^2 + bx + c \) is a quadratic function, and its graph is a parabola. The x-coordinate of the vertex (which gives the minimum value of the function if the parabola opens upwards) can be found using the formula:

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

Here, the coefficients are:
- \( a = 4 \)
- \( b = -50 \)

Substituting these values into the formula:

\[
x = \frac{-(-50)}{2(4)} = \frac{50}{8} = 6.25
\]

So, the function \( f(x) \) reaches its minimum value when \( x = 6.25 \).

Would you like to know the minimum value of the function \( f(x) \) at this point or have any other questions?

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**Relative Questions:**
1. What is the minimum value of the function at \( x = 6.25 \)?
2. How do you determine if the parabola opens upwards or downwards?
3. How would the minimum point change if the coefficient \( a \) was negative?
4. Can you explain how the vertex formula is derived?
5. What happens to the minimum value if we add a constant \( k \) to the function \( f(x) \)?
6. How can you solve for \( x \) using the quadratic formula instead?
7. What is the significance of the discriminant in a quadratic equation?
8. How would you graph this quadratic function?

**Tip:** Remember that the vertex of a parabola provides the minimum or maximum value of a quadratic function depending on the sign of the leading coefficient \( a \).

Discover how to find the minimum value of the quadratic function f(x) = 4x^2 - 50x + 126 using the vertex formula. This problem involves quadratic functions and finding critical points. Suitable for high school students.

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