The given quadratic function is:

$f(x) = -2x^2 + 2x - 6$

Step 1: Identify whether the graph is concave up or concave down.

The coefficient of $x^2$ is $-2$, which is negative. Therefore, the parabola opens downward, meaning the graph is concave down.

Step 2: Find the axis of symmetry.

For a quadratic function in the form $f(x) = ax^2 + bx + c$, the formula for the axis of symmetry is:

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

Here, $a = -2$ and $b = 2$. Substituting these values:

$x = \frac{-2}{2(-2)} = \frac{-2}{-4} = \frac{1}{2}$

Thus, the axis of symmetry is $x = \frac{1}{2}$.

Step 3: Find the vertex.

To find the vertex, substitute $x = \frac{1}{2}$ into the quadratic function $f(x)$ to find the corresponding $y$-coordinate.

$f\left(\frac{1}{2}\right) = -2\left(\frac{1}{2}\right)^2 + 2\left(\frac{1}{2}\right) - 6$

Simplifying:

$f\left(\frac{1}{2}\right) = -2\left(\frac{1}{4}\right) + 2\left(\frac{1}{2}\right) - 6$ $= -\frac{2}{4} + 1 - 6 = -\frac{1}{2} + 1 - 6 = \frac{1}{2} - 6 = -\frac{11}{2}$

Thus, the vertex is $\left(\frac{1}{2}, -\frac{11}{2}\right)$.

Final Answer:

• The vertex is $\left(\frac{1}{2}, -\frac{11}{2}\right)$.
• The axis of symmetry is $x = \frac{1}{2}$.
• The graph is concave down.

Would you like more details or have any questions? Here are 5 follow-up questions to explore this further:

1. How do you find the y-intercept of a quadratic function?
2. What is the significance of the vertex in a quadratic graph?
3. Can you derive the formula for the vertex from completing the square?
4. How does the sign of the leading coefficient affect the shape of the parabola?
5. What are the properties of the axis of symmetry in any quadratic function?

Tip: The vertex form of a quadratic equation can make finding the vertex much easier!