It seems like the function you're referring to is incomplete or has some typographical errors. However, I can reconstruct the problem and solve it step by step. Based on what you’ve written, I believe the function might be similar to this:

$f(x) = 5x^2 - 50x + 2$

This is a quadratic function, which typically represents a parabola. Since the coefficient of $x^2$ is positive (5), the parabola opens upwards, meaning that it has a minimum point.

Let's solve for the value of $x$ where $f(x)$ reaches a minimum and compute the corresponding minimum value.

Step 1: Find the vertex of the parabola

For a quadratic function in the form:

$f(x) = ax^2 + bx + c$

The $x$-coordinate of the vertex, which gives the minimum (or maximum) point for the parabola, is found using the formula:

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

For our function $f(x) = 5x^2 - 50x + 2$, we have:

• $a = 5$
• $b = -50$

Substitute these values into the formula:

$x = -\frac{-50}{2(5)} = \frac{50}{10} = 5$

Thus, the value of $x$ where the function reaches its minimum is $x = 5$.

Step 2: Find the minimum value of the function

Now, substitute $x = 5$ back into the original function to find the minimum value:

$f(5) = 5(5)^2 - 50(5) + 2$ $f(5) = 5(25) - 250 + 2$ $f(5) = 125 - 250 + 2 = -123$

Conclusion:

• The function reaches its minimum value at $x = 5$.
• The minimum value of the function is $-123$.

Would you like more details or have any questions about this solution?

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Here are 5 related questions for further exploration:

1. What does the vertex of a quadratic function represent in real-world problems?
2. How can we determine if a quadratic function opens upwards or downwards?
3. What is the axis of symmetry in a quadratic function, and how is it related to the vertex?
4. How can the discriminant be used to analyze the nature of the roots of a quadratic equation?
5. What other methods can be used to find the minimum value of a quadratic function?

Tip: For any quadratic function, the vertex is always the critical point, and its coordinates can be found easily using $-\frac{b}{2a}$. This gives you valuable insights into the function's behavior.