We are given the quadratic function $f(x) = 8x + 4x^2 - 2$, and the question asks for the vertex of the function.

To find the vertex of a quadratic function $f(x) = ax^2 + bx + c$, we can use the formula for the x-coordinate of the vertex: $x = \frac{-b}{2a}$ where $a$ is the coefficient of $x^2$ and $b$ is the coefficient of $x$.

Step 1: Identify the coefficients

For the given function $f(x) = 4x^2 + 8x - 2$, we have:

• $a = 4$
• $b = 8$
• $c = -2$ (though $c$ is not needed to find the vertex)

Step 2: Apply the formula for the x-coordinate of the vertex

$x = \frac{-8}{2 \times 4} = \frac{-8}{8} = -1$

Step 3: Find the y-coordinate by substituting $x = -1$ back into the function

$f(-1) = 4(-1)^2 + 8(-1) - 2$ $f(-1) = 4(1) - 8 - 2 = 4 - 8 - 2 = -6$

Conclusion:

The vertex of the quadratic function is $(-1, -6)$.

So, the correct answer is $(-1, -6)$.

Let me know if you want more details on this. Here are some additional related questions:

1. How do you find the axis of symmetry from a quadratic function?
2. Can you explain the significance of the vertex in a parabola?
3. How can you rewrite the given quadratic function in vertex form?
4. What is the general shape of a parabola for a quadratic function with a positive $a$?
5. How would the graph change if the quadratic function was $-4x^2 + 8x - 2$?

Tip: The vertex form of a quadratic function is useful when you want to easily identify the vertex and the direction in which the parabola opens.