To find the vertex of a quadratic function $f(x) = ax^2 + bx + c$ without completing the square, we can use the vertex formula for the x-coordinate:

$x_{\text{vertex}} = \frac{-b}{2a}$

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

• $a = 1$
• $b = -6$

Substituting into the formula:

$x_{\text{vertex}} = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3$

Now, to find the y-coordinate of the vertex, substitute $x = 3$ into the original function $f(x)$:

$f(3) = (3)^2 - 6(3) + 4 = 9 - 18 + 4 = -5$

Thus, the vertex of the quadratic function is $(3, -5)$.

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How does completing the square confirm the vertex we found?
2. What is the significance of the vertex in the graph of a parabola?
3. Can you explain how the direction of the parabola is determined by the sign of $a$?
4. How would you solve for the x-intercepts of this quadratic function?
5. What is the axis of symmetry for this quadratic function?

Tip: The vertex represents the maximum or minimum point of a parabola, depending on the direction it opens.