The question asks to determine the minimum value of the function $f(x) = 2x^2 + 4x - 1$.

This is a quadratic function, and the general form of a quadratic function is $f(x) = ax^2 + bx + c$, where:

• $a = 2$
• $b = 4$
• $c = -1$

The minimum or maximum value of a quadratic function occurs at its vertex. For a parabola opening upwards (since $a > 0$, it opens upwards), the vertex represents the minimum point.

The x-coordinate of the vertex is given by the formula: $x = \frac{-b}{2a}$ Substituting the values of $a$ and $b$: $x = \frac{-4}{2(2)} = \frac{-4}{4} = -1$

Now, substitute $x = -1$ into the function $f(x)$ to find the minimum value of the function: $f(-1) = 2(-1)^2 + 4(-1) - 1 = 2(1) - 4 - 1 = 2 - 4 - 1 = -3$

Therefore, the minimum value of the function is $-3$.

It seems like the correct answer isn't listed in the given options. Let me know if you need further clarification or have any questions!

Follow-up Questions:

1. What does the vertex formula represent in a quadratic function?
2. How do we determine if a quadratic function has a minimum or maximum value?
3. Why does the function $f(x) = 2x^2 + 4x - 1$ have a minimum rather than a maximum?
4. How can we rewrite the function $f(x)$ to complete the square and find its minimum value?
5. What are the applications of finding the minimum or maximum value in real-life situations?

Tip:

For any quadratic function, if $a > 0$, the function opens upwards, and the vertex gives the minimum value. If $a < 0$, the vertex gives the maximum value.