To find the vertex of the quadratic function $f(x) = -x^2 + 6x - 10$ using the vertex formula, we can apply the following steps:

Step 1: Use the vertex formula

The vertex of a quadratic function $f(x) = ax^2 + bx + c$ is given by the formula: $x_v = \frac{-b}{2a}$ Where $a$ and $b$ are the coefficients of the quadratic and linear terms, respectively.

In this case, the function is: $f(x) = -x^2 + 6x - 10$ Thus, $a = -1$ and $b = 6$.

Step 2: Calculate $x_v$

Substitute $a = -1$ and $b = 6$ into the vertex formula: $x_v = \frac{-6}{2(-1)} = \frac{-6}{-2} = 3$

Step 3: Calculate $y_v$

Now, substitute $x_v = 3$ into the original equation to find $y_v$: $y_v = f(3) = -(3)^2 + 6(3) - 10$ $y_v = -9 + 18 - 10 = -1$

Step 4: Conclusion

The vertex of the function is $(x_v, y_v) = (3, -1)$.

Would you like further details on any part of the solution, or do you have any questions?

Here are some related questions you can explore:

1. How does the sign of $a$ affect the direction of the parabola?
2. What is the significance of the vertex in a quadratic function?
3. Can you use completing the square to find the vertex for this function?
4. How would the graph of this function change if the constant term changed?
5. How is the axis of symmetry related to the vertex?

Tip: The vertex formula is derived from completing the square on a general quadratic equation!