To find the value of $x$ at which the function $f(x) = 4x^2 - 50x + 126$ reaches its minimum, we need to determine the vertex of the parabola represented by this quadratic function. Since the coefficient of $x^2$ is positive (4), the parabola opens upwards, and the vertex represents the minimum point.

The $x$-coordinate of the vertex can be found using the formula:

$x = -\frac{b}{2a}$

where $a$ and $b$ are the coefficients of $x^2$ and $x$ respectively.

For the given function $f(x) = 4x^2 - 50x + 126$:

• $a = 4$
• $b = -50$

Now, calculate the $x$-coordinate of the vertex:

$x = -\frac{-50}{2(4)}$

$x = \frac{50}{8}$

$x = 6.25$

Thus, the function $f(x)$ reaches its minimum at $x = 6.25$.

Would you like further details or have any other questions?

Here are some related questions:

1. How can you determine whether a quadratic function has a minimum or maximum?
2. What is the formula to find the vertex of a quadratic function?
3. How do you calculate the $y$-coordinate of the vertex for $f(x) = 4x^2 - 50x + 126$?
4. What is the significance of the vertex in a quadratic function?
5. How does the value of the coefficient $a$ affect the graph of a quadratic function?
6. What are the axis of symmetry for a quadratic function and how is it related to the vertex?
7. How would the graph change if the coefficient of $x^2$ were negative?
8. How can you determine the range of the function $f(x) = 4x^2 - 50x + 126$?

Tip: Remember that the vertex of a parabola given by $f(x) = ax^2 + bx + c$ is the point where the function reaches its maximum or minimum, depending on the sign of $a$.