To find the vertex of the quadratic function $f(x) = 2x^2 - 4x - 6$, we can use the vertex formula for a parabola given by $f(x) = ax^2 + bx + c$. The x-coordinate of the vertex is calculated using:

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

For the given function $f(x) = 2x^2 - 4x - 6$:

• $a = 2$
• $b = -4$

Let's find the x-coordinate:

$x = -\frac{-4}{2 \cdot 2} = \frac{4}{4} = 1$

Now, substitute $x = 1$ back into the function to find the y-coordinate:

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

So, the vertex of the function is at $(1, -8)$.

The correct answer is:

• $(1, -8)$

Would you like more details or have any questions? Here are some related questions:

1. How do you derive the vertex formula $x = -\frac{b}{2a}$?
2. What is the significance of the vertex in the graph of a parabola?
3. How can you determine whether a parabola opens upward or downward?
4. What role does the coefficient $a$ play in the shape of a quadratic function?
5. How do you convert a quadratic function into vertex form?

Tip: The vertex form of a quadratic function, $f(x) = a(x - h)^2 + k$, makes it easier to identify the vertex, where $(h, k)$ is the vertex.